[
    {
        "id": "Physics_248",
        "problem": "The characterization of point object motion, when both radial and tangential forces are applied, is usually rather complicated, and requires advanced mathematical tools. However, some systems such as a motion of a point charge near an electric dipole, which have a very specific electrostatic field, provide interesting results, even for the case when an angular momentum is not conserved.\n\nAssume that the relativistic and the electromagnetic radiation effects can be neglected, unless otherwise stated.\n\nIn this part, analyze a situation when an angular momentum is not conserved. The system is the same as in the previous part with the only difference that the dipole is fixed and the charged small object with a mass $2 m$ is moving around the dipole. The electrostatic field of the dipole is easier to describe in the polar system of coordinates, which is defined with the distance $r$ from the center of the dipole, and angle $\\theta$ counted counterclockwise, as shown in Figure 3.\n\n[figure1]\n\nFigure 3: The system analyzed in Part 2. (Direction of the vector $\\mathbf{E}_{\\mathbf{n}}$ and $\\mathbf{E}_{\\mathbf{t}}$ could be wrong)\n\nFind the components of the electric field $E_{n}$ and $E_{\\tau}$ in terms of $r, \\theta, q$ and $d$.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nThe characterization of point object motion, when both radial and tangential forces are applied, is usually rather complicated, and requires advanced mathematical tools. However, some systems such as a motion of a point charge near an electric dipole, which have a very specific electrostatic field, provide interesting results, even for the case when an angular momentum is not conserved.\n\nAssume that the relativistic and the electromagnetic radiation effects can be neglected, unless otherwise stated.\n\nIn this part, analyze a situation when an angular momentum is not conserved. The system is the same as in the previous part with the only difference that the dipole is fixed and the charged small object with a mass $2 m$ is moving around the dipole. The electrostatic field of the dipole is easier to describe in the polar system of coordinates, which is defined with the distance $r$ from the center of the dipole, and angle $\\theta$ counted counterclockwise, as shown in Figure 3.\n\n[figure1]\n\nFigure 3: The system analyzed in Part 2. (Direction of the vector $\\mathbf{E}_{\\mathbf{n}}$ and $\\mathbf{E}_{\\mathbf{t}}$ could be wrong)\n\nFind the components of the electric field $E_{n}$ and $E_{\\tau}$ in terms of $r, \\theta, q$ and $d$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [$E_{n}$, $E_{t}$].\nTheir answer types are, in order, [expression, expression].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_ed4e92416bdbac30298dg-2.jpg?height=477&width=1442&top_left_y=1694&top_left_x=336"
        ],
        "answer": [
            "$-\\frac{q}{2 \\pi \\epsilon_{0}} \\frac{d \\cos \\theta}{r^{3}}$",
            "$-\\frac{q}{4 \\pi \\epsilon_{0}} \\frac{d \\sin \\theta}{r^{3}}$"
        ],
        "solution": "The distance between the charge and the respective charges of the dipole using first-order approximations in $d$ :\n\n$$\nr_{-}=\\sqrt{r^{2}+\\frac{d^{2}}{4}-d r \\cos \\theta}=r-\\frac{d \\cos \\theta}{2} ; \\quad r_{+}=\\sqrt{r^{2}+\\frac{d^{2}}{4}+d r \\cos \\theta}=r+\\frac{d \\cos \\theta}{2} ;\n$$\n\nusing these expressions we obtain the formula of the electric potential:\n\n$$\n\\varphi=\\frac{q}{4 \\pi \\epsilon_{0}}\\left(\\frac{1}{r+d \\cos \\theta / 2}-\\frac{1}{r-d \\cos \\theta / 2}\\right)=-\\frac{q}{4 \\pi \\epsilon_{0}} \\frac{d \\cos \\theta}{r^{2}}\n$$\n\n$2 \\mathbf{E}=-\\nabla \\varphi$; using the cylindrical expression of the gradient:\n\n$$\nE_{n}=-\\frac{\\partial \\varphi}{\\partial r}=-\\frac{q}{2 \\pi \\epsilon_{0}} \\frac{d \\cos \\theta}{r^{3}} ; \\quad \\quad E_{t}=-\\frac{1}{r} \\frac{\\partial \\varphi}{\\partial \\theta}=-\\frac{q}{4 \\pi \\epsilon_{0}} \\frac{d \\sin \\theta}{r^{3}}\n$$",
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "$E_{n}$",
            "$E_{t}$"
        ],
        "type_sequence": [
            "EX",
            "EX"
        ],
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1712",
        "problem": "如图, 半径为 $R$ 、质量为 $M$ 的半球静置于光滑水平桌面上, 在半球顶点上有一质量为 $m$ 、半径为 $r$ 的匀质小球。某时刻, 小球受到微小的扰动后由静止开始沿半球表面运动。在运动过程中, 小球相对于半球的位置由角位置 $\\theta$ 描述, $\\theta$ 为两球心的连线与坚直方向之间的夹角。已知小球绕其对称轴的转动惯量为 $\\frac{2}{5} m r^{2}$, 小球与半球之间的动摩擦因数为 $\\mu$, 假定最大静摩擦力等于滑动摩擦力。重力加速度大小为 $g$ 。\n\n[图1]小球开始运动后在一段时间内做纯滚动, 求在此过程中, 当小球的角位置为 $\\theta_{1}$ 时, 半球运动的速度大小 $V_{M}\\left(\\theta_{1}\\right)$ 和加速度大小 $a_{M}\\left(\\theta_{1}\\right)$;",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题包含多个待求解的量。\n这里是一些可能会帮助你解决问题的先验信息提示：\n如图, 半径为 $R$ 、质量为 $M$ 的半球静置于光滑水平桌面上, 在半球顶点上有一质量为 $m$ 、半径为 $r$ 的匀质小球。某时刻, 小球受到微小的扰动后由静止开始沿半球表面运动。在运动过程中, 小球相对于半球的位置由角位置 $\\theta$ 描述, $\\theta$ 为两球心的连线与坚直方向之间的夹角。已知小球绕其对称轴的转动惯量为 $\\frac{2}{5} m r^{2}$, 小球与半球之间的动摩擦因数为 $\\mu$, 假定最大静摩擦力等于滑动摩擦力。重力加速度大小为 $g$ 。\n\n[图1]\n\n问题:\n小球开始运动后在一段时间内做纯滚动, 求在此过程中, 当小球的角位置为 $\\theta_{1}$ 时, 半球运动的速度大小 $V_{M}\\left(\\theta_{1}\\right)$ 和加速度大小 $a_{M}\\left(\\theta_{1}\\right)$;\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你的最终解答的量应该按以下顺序输出：[当小球的角位置为 $\\theta_{1}$ 时, 半球运动的速度大小 $V_{M}\\left(\\theta_{1}\\right)$ , 当小球的角位置为 $\\theta_{1}$ 时, 半球运动的加速度大小 $a_{M}\\left(\\theta_{1}\\right)$]\n它们的答案类型依次是[表达式, 表达式]\n你需要在输出的最后用以下格式总结答案：“最终答案是\\boxed{ANSWER}”，其中ANSWER应为你的最终答案序列，用英文逗号分隔，例如：5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_35bc41298eef336dfdafg-01.jpg?height=337&width=486&top_left_y=351&top_left_x=1276"
        ],
        "answer": [
            "$\\sqrt{\\frac{10 m^{2}(R+r) g\\left(1-\\cos \\theta_{1}\\right) \\cos ^{2} \\theta_{1}}{\\left[7(M+m)-5 m \\cos ^{2} \\theta_{1}\\right](M+m)}}$",
            "$-\\frac{5 m g \\sin \\theta_{1}\\left[14(M+m)-21(M+m) \\cos \\theta_{1}+5 m \\cos ^{3} \\theta_{1}\\right]}{\\left[7(M+m)-5 m \\cos ^{2} \\theta_{1}\\right]^{2}}$"
        ],
        "solution": "半球和小球组成的系统在水平方向上没有受到外力作用, 系统在水平方向上动量守恒\n\n$$\n-M V_{M}+m\\left[(R+r) \\dot{\\theta} \\cos \\theta-V_{M}\\right]=0\n$$\n\n设小球转动角速度大小为 $\\omega$, 小球做纯滚动, 故有\n\n$$\nr \\omega=(R+r) \\dot{\\theta}\n$$\n\n无耗散力做功，系统的机械能守恒\n\n$$\n\\begin{aligned}\n& m g(R+r)(1-\\cos \\theta) \\\\\n& =\\frac{1}{2} M V_{M}^{2}+\\frac{1}{2} m\\left\\{\\left[(R+r) \\dot{\\theta} \\cos \\theta-V_{M}\\right]^{2}+[(R+r) \\dot{\\theta} \\sin \\theta]^{2}\\right\\}+\\frac{1}{2} I \\omega^{2}\n\\end{aligned}\n$$\n\n式中 $I=\\frac{2}{5} m r^{2}$ 。联立(1)(2)(3)式得, 小球运动到角位置 $\\theta_{1}$ 时半球速度大小\n\n$$\nV_{M}=\\sqrt{\\frac{10 m^{2}(R+r) g\\left(1-\\cos \\theta_{1}\\right) \\cos ^{2} \\theta_{1}}{\\left[7(M+m)-5 m \\cos ^{2} \\theta_{1}\\right](M+m)}}\n$$\n\n或\n\n$$\nV_{M}^{2}=\\frac{10 m^{2} g(R+r)\\left(1-\\cos \\theta_{1}\\right) \\cos ^{2} \\theta_{1}}{\\left[7 M+\\left(5 \\sin ^{2} \\theta_{1}+2\\right) m\\right](m+M)}\n$$\n将上式两边对时间 $t$ 微商得\n\n$$\n\\begin{aligned}\n2 V_{M} a_{M}= & \\frac{10 m g\\left(-2 \\cos \\theta+3 \\cos ^{2} \\theta\\right)\\left[7(M+m)-5 m \\cos ^{2} \\theta\\right]-100 m^{2} g(1-\\cos \\theta) \\cos ^{3} \\theta}{\\left[7(M+m)-5 m \\cos ^{2} \\theta\\right]^{2}} \\\\\n& \\cdot \\frac{m(R+r) \\sin \\theta \\cdot \\dot{\\theta}}{M+m}\n\\end{aligned}\n$$\n\n由(1)式可知\n\n$$\n\\frac{m(R+r) \\dot{\\theta}}{M+m}=\\frac{V_{M}}{\\cos \\theta}\n$$\n\n由以上两式得, 小球运动到角位置 $\\theta_{1}$ 时, 半球的加速度大小为\n\n$$\na_{M}\\left(\\theta_{1}\\right)=-\\frac{5 m g \\sin \\theta_{1}\\left[14(M+m)-21(M+m) \\cos \\theta_{1}+5 m \\cos ^{3} \\theta_{1}\\right]}{\\left[7(M+m)-5 m \\cos ^{2} \\theta_{1}\\right]^{2}}\n$$",
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "当小球的角位置为 $\\theta_{1}$ 时, 半球运动的速度大小 $V_{M}\\left(\\theta_{1}\\right)$ ",
            "当小球的角位置为 $\\theta_{1}$ 时, 半球运动的加速度大小 $a_{M}\\left(\\theta_{1}\\right)$"
        ],
        "type_sequence": [
            "EX",
            "EX"
        ],
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1301",
        "problem": "用薄膜制备技术在某均质硅基片上沉积一层均匀等厚氮化镓薄膜,制备出一个硅基氮化镓样品, 如图 I 所示。\n\n[图1]当用波长范围为 $450 \\sim 1200 \\mathrm{~nm}$ 的光垂直均匀照射该\n\n样品氮化镓表面, 观察到其反射光谱仅有两种波长的光获得最大相干加强, 其中之一波长为 $600 \\mathrm{~nm}$; 氮化镓的折射率 $n$ 与入射光在真空中波长 $\\lambda$ (单位 $\\mathrm{nm}$ ) 之间的关系（色散关系）为\n\n$$\nn^{2}=2.26^{2}+\\frac{330.1^{2}}{\\lambda^{2}-265.7^{2}}\n$$\n\n硅的折射率随波长在 3.49 5.49 范围内变化。只考虑氮化镓表面和氮化镓-硅基片界面的反射, 求氮化镓薄膜的厚度 $d$ 和另一种获得最大相干加强光的波长。",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个数值。\n这里是一些可能会帮助你解决问题的先验信息提示：\n用薄膜制备技术在某均质硅基片上沉积一层均匀等厚氮化镓薄膜,制备出一个硅基氮化镓样品, 如图 I 所示。\n\n[图1]\n\n问题:\n当用波长范围为 $450 \\sim 1200 \\mathrm{~nm}$ 的光垂直均匀照射该\n\n样品氮化镓表面, 观察到其反射光谱仅有两种波长的光获得最大相干加强, 其中之一波长为 $600 \\mathrm{~nm}$; 氮化镓的折射率 $n$ 与入射光在真空中波长 $\\lambda$ (单位 $\\mathrm{nm}$ ) 之间的关系（色散关系）为\n\n$$\nn^{2}=2.26^{2}+\\frac{330.1^{2}}{\\lambda^{2}-265.7^{2}}\n$$\n\n硅的折射率随波长在 3.49 5.49 范围内变化。只考虑氮化镓表面和氮化镓-硅基片界面的反射, 求氮化镓薄膜的厚度 $d$ 和另一种获得最大相干加强光的波长。\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n请记住，你的答案应以$\\mathrm{~nm}$为单位计算，但在给出最终答案时，请不要包含单位。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是不包含任何单位的数值。",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a47de6806e8da0a0f86dg-04.jpg?height=71&width=423&top_left_y=1238&top_left_x=1225"
        ],
        "answer": [
            "$1168$"
        ],
        "solution": "已知入射光的波长范围为 $450-1200 \\mathrm{~nm}$ ，由色散关系可得到折射率范围\n\n$$\n2.278 \\leq n \\leq 2.436\n$$\n\n可见, 氮化镓的折射率小于硅的折射率, 在空气-氮化镓和氮化镓-硅界面处均要考虑半波损失，但对于两界面反射光的干涉不需要考虑半波损失。\n\n设对于波长为 $600 \\mathrm{~nm}$ 的入射光, 薄膜的折射率为 $n_{1}$; 对于波长 $\\lambda$ 的入射光, 薄膜的折射率为 $n$ 。对于波长为 $600 \\mathrm{~nm}$ 的入射光, 光经厚度为 $d$ 的氮化镓薄膜的前、后面反射后,\n在表面相遇, 相干加强条件为\n\n$$\n2 n_{1} d=k_{1} \\lambda_{1} \\quad\\left(k_{1}=1,2, \\cdots\\right)\n$$\n\n同理, 对于波长 $\\lambda$ 的入射光，相干加强的条件有为\n\n$$\n2 n d=k \\lambda \\quad(k=1,2, \\cdots)\n$$\n\n由(1)(2)式有\n\n$$\n\\frac{k_{1}}{k} \\frac{n}{n_{1}}=\\frac{\\lambda}{\\lambda_{1}} \\quad \\text { 或 } \\quad \\frac{k_{1}}{k}=\\frac{n_{1}}{n} \\frac{\\lambda}{\\lambda_{1}}\n$$\n\n由(3)式, 根据题意, 设 $\\lambda^{\\prime}$ 和 $\\lambda^{\\prime \\prime}$ 分别对应 $450 \\mathrm{~nm}$ 和 $1200 \\mathrm{~nm}$ 的入射光波长, 薄膜相应的折射率分别为 $n^{\\prime}$ 和 $n^{\\prime \\prime}$, 根据题给色散关系得\n\n$$\nn_{1}(600 \\mathrm{~nm}) \\approx 2.342, \\quad n^{\\prime}(450 \\mathrm{~nm}) \\approx 2.436, \\quad n^{\\prime \\prime}(1200 \\mathrm{~nm}) \\approx 2.278\n$$\n\n由(4)式有\n\n$$\n\\frac{k_{1}}{k^{\\prime}}=\\frac{n_{1}(600)}{n^{\\prime}(450)} \\frac{450}{600} \\approx 0.721, \\quad \\frac{k_{1}}{k^{\\prime \\prime}}=\\frac{n_{1}(600)}{n^{\\prime \\prime}(1200)} \\frac{1200}{600} \\approx 2.056\n$$\n\n由(5)式和题意得\n\n$$\n\\frac{k_{1}}{k^{\\prime}}<\\frac{k_{1}}{k}<\\frac{k_{1}}{k^{\\prime \\prime}}\n$$\n\n由(6)式和题给条件知仅有\n\n$$\nk_{1}=2, k=1\n$$\n\n符合条件。由(4)(7)式代入(1)式得\n\n$$\nd \\approx 256 \\mathrm{~nm}\n$$\n\n将(7)(8)式代入(2)式得\n\n$$\n\\frac{\\lambda}{n} \\approx 512.38\n$$\n\n由题给色散关系式和(9)式得\n\n$$\n2.41 n^{4}-12.97 n^{2}+2.32=0\n$$\n\n或\n\n$$\n2.41\\left(\\frac{\\lambda}{512.38}\\right)^{4}-12.97\\left(\\frac{\\lambda}{512.38}\\right)^{2}+2.32=0\n$$\n\n由此方程解得\n\n$$\n\\lambda=1168 \\mathrm{~nm}\n$$\n\n另一解 $\\lambda=220 \\mathrm{~nm}$ 不在题给波长范围内, 舍去。",
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~nm}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1075",
        "problem": "A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q$ ' placed inside the sphere (you do not have to prove it). **Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface).**\n\n[figure1]\n\nConsider a point charge $q$ placed at a distance $d$ from the center of a grounded metallic sphere of radius $R$. We are interested in how the grounded metallic sphere affects the electric field at point $A$ on the opposite side of the sphere (see Fig. 2). Point $A$ is on the line connecting charge $q$ and the center of the sphere; its distance from the point charge $q$ is $r$.\n\nFor a very large distance $r>>d$, find the expression for the electric field by using the approximation $(1+a)^{-2} \\approx 1-2 a$, where $a<<1$.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nA point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q$ ' placed inside the sphere (you do not have to prove it). **Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface).**\n\n[figure1]\n\nConsider a point charge $q$ placed at a distance $d$ from the center of a grounded metallic sphere of radius $R$. We are interested in how the grounded metallic sphere affects the electric field at point $A$ on the opposite side of the sphere (see Fig. 2). Point $A$ is on the line connecting charge $q$ and the center of the sphere; its distance from the point charge $q$ is $r$.\n\nFor a very large distance $r>>d$, find the expression for the electric field by using the approximation $(1+a)^{-2} \\approx 1-2 a$, where $a<<1$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_843c51f18af1a9802d4bg-1.jpg?height=774&width=1627&top_left_y=1046&top_left_x=220"
        ],
        "answer": [
            "$\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{\\left(1-\\frac{R}{d}\\right) q}{r^{2}} \\hat{r}-\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{2 q \\frac{R}{d}\\left(d-\\frac{R^{2}}{d}\\right)}{r^{3}}  \\hat{r}$"
        ],
        "solution": "For very large distances $r$ we can apply approximate formula $(1+a)^{-2} \\approx 1-2 a$ to the expression (11) what leads us to\n\n$$\n\\vec{E}_{A}=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{\\left(1-\\frac{R}{d}\\right) q}{r^{2}} \\hat{r}-\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{2 q \\frac{R}{d}\\left(d-\\frac{R^{2}}{d}\\right)}{r^{3}} \\hat{r}\n$$\n\nIn general a grounded metallic sphere cannot completely screen a point charge $q$ at a distance $d$ (even in the sense that its electric field would decrease with distance faster than $1 / r^{2}$ ) and the dominant dependence of the electric field on the distance $r$ is as in standard Coulomb law.",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1453",
        "problem": "平行板电容器极板 1 和 2 的面积均为 $S$, 水平固定放置,它们之间的距离为 $d$, 接入如图所示的电路中, 电源的电动势记为 $U$ 。不带电的导体薄平板 3 的质量为 $m$ 、尺寸与电容器极板相同。平板 3 平放在极板 2 的正上方, 且与极板 2 有良好的电接触。整个系统置于真空室内, 真空的介电常量为 $\\varepsilon_{0}$ 。闭合电键 $\\mathrm{K}$ 后, 平板 3 与极板 1 和 2 相继碰撞, 上下往复运动。假设导体板之间的电场均可视为匀强电场; 导线电阻和电源内阻足够小, 充放电时间可忽略不计; 平板 3 与极板 1 或 2 碰撞后立即在极短时间内达到静电平衡; 所有碰撞都是完全非弹性的。重力加速度大小为 $g$ 。\n\n[图1]\n\n已知积分公式\n\n$$\n\\int \\frac{\\mathrm{d} x}{\\sqrt{a x^{2}+b x}}=\\frac{1}{\\sqrt{a}} \\ln \\left(2 a x+b+2 \\sqrt{a} \\sqrt{a x^{2}+b x}\\right)+C \\text {, 其中 } a>0, C \\text { 为积分常数。 }\n$$电源电动势 $U$ 至少为多大?",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n这里是一些可能会帮助你解决问题的先验信息提示：\n平行板电容器极板 1 和 2 的面积均为 $S$, 水平固定放置,它们之间的距离为 $d$, 接入如图所示的电路中, 电源的电动势记为 $U$ 。不带电的导体薄平板 3 的质量为 $m$ 、尺寸与电容器极板相同。平板 3 平放在极板 2 的正上方, 且与极板 2 有良好的电接触。整个系统置于真空室内, 真空的介电常量为 $\\varepsilon_{0}$ 。闭合电键 $\\mathrm{K}$ 后, 平板 3 与极板 1 和 2 相继碰撞, 上下往复运动。假设导体板之间的电场均可视为匀强电场; 导线电阻和电源内阻足够小, 充放电时间可忽略不计; 平板 3 与极板 1 或 2 碰撞后立即在极短时间内达到静电平衡; 所有碰撞都是完全非弹性的。重力加速度大小为 $g$ 。\n\n[图1]\n\n已知积分公式\n\n$$\n\\int \\frac{\\mathrm{d} x}{\\sqrt{a x^{2}+b x}}=\\frac{1}{\\sqrt{a}} \\ln \\left(2 a x+b+2 \\sqrt{a} \\sqrt{a x^{2}+b x}\\right)+C \\text {, 其中 } a>0, C \\text { 为积分常数。 }\n$$\n\n问题:\n电源电动势 $U$ 至少为多大?\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_35bc41298eef336dfdafg-01.jpg?height=317&width=448&top_left_y=1966&top_left_x=1318"
        ],
        "answer": [
            "$\\sqrt{\\frac{2 m d^{2} g}{\\varepsilon_{0} S}}$"
        ],
        "solution": "在平板3离开极板 2 之前, 平板3的带电量为\n\n$$\nQ=C_{0} U=\\frac{\\varepsilon_{0} S}{d} U\n$$\n\n设平板3离开极板 2 之后, 各板电荷面密度如图a所示。由电荷守恒有\n\n$$\n\\sigma_{1}-\\sigma_{2}=\\sigma \\equiv \\frac{Q}{S}=\\frac{\\varepsilon_{0}}{d} U\n$$\n\n设上、下两电容器各自两极板间电场的场强分别为 $E_{1} 、 E_{2}$ （见图a）, 有\n\n$$\nE_{1}=\\frac{\\sigma_{1}}{\\varepsilon_{0}}, \\quad E_{2}=\\frac{\\sigma_{2}}{\\varepsilon_{0}}\n$$\n\n将上式代入(1)式得\n\n$$\n\\varepsilon_{0} E_{1}-\\varepsilon_{0} E_{2}=\\frac{\\varepsilon_{0}}{d} U\n$$\n\n即\n\n$$\nE_{1}-E_{2}=\\frac{U}{d}\n$$\n\n另外, 两个串联电容器的总电势差为 $U$, 故\n\n$$\nE_{2} x+E_{1}(d-x)=U\n$$\n\n联立(2)(3)式得\n\n$$\n\\begin{aligned}\n& E_{1}=\\frac{U}{d} \\frac{d+x}{d} \\\\\n& E_{2}=\\frac{U}{d} \\frac{x}{d}\n\\end{aligned}\n$$\n\n由(4)(5)式得，极板 $1 、 2$ 上电荷面密度分别为\n\n$$\n\\sigma_{1}=\\varepsilon_{0} E_{1}, \\quad \\sigma_{2}=\\varepsilon_{0} E_{2}\n$$\n\n平板3受到的电场力 (向上为正方向, 下同) 为\n\n$$\n\\begin{aligned}\nF_{\\mathrm{e}} & =-\\sigma_{2} S \\cdot \\frac{E_{2}}{2}+\\sigma_{1} S \\cdot \\frac{E_{1}}{2}=\\frac{\\varepsilon_{0} S}{2}\\left(E_{1}^{2}-E_{2}^{2}\\right)=\\varepsilon_{0} S\\left(E_{1}-E_{2}\\right)\\left(\\frac{E_{1}+E_{2}}{2}\\right) \\\\\n& =\\varepsilon_{0} S \\frac{U}{d} \\cdot\\left(E_{1}+E_{2}\\right) / 2=\\varepsilon_{0} S \\frac{U}{d} \\cdot \\frac{U}{d}\\left(\\frac{2 x+d}{2 d}\\right)=\\frac{\\varepsilon_{0} S U^{2}}{2 d^{3}}(2 x+d)\n\\end{aligned}\n$$\n\n平板3受到的坚直方向的合力为\n\n$$\nF_{\\text {total }}=F_{\\mathrm{e}}-m g=\\frac{\\varepsilon_{0} S U^{2}}{2 d^{3}}(2 x+d)-m g=\\left(\\frac{\\varepsilon_{0} S U^{2}}{2 d^{2}}-m g\\right)+\\frac{\\varepsilon_{0} S U^{2}}{d^{3}} x\n$$\n\n由此得, 平板3在图a所示位置的加速度为\n\n$$\na=\\frac{F_{\\text {total }}}{m}=\\left(\\frac{\\varepsilon_{0} S U^{2}}{2 m d^{2}}-g\\right)+\\frac{\\varepsilon_{0} S U^{2}}{m d^{3}} x\n$$\n\n为使平板 3 向上运动，应有条件\n\n$$\n\\frac{\\varepsilon_{0} S U^{2}}{2 m d^{2}} \\geq g\n$$\n\n且开始运动之后加速度始终为正, 因此最终将撞到极板 2 上。上述条件意味着电源电动势 $U$ 至\n少应为\n\n$$\nU_{\\min }=\\sqrt{\\frac{2 m d^{2} g}{\\varepsilon_{0} S}}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_779",
        "problem": "If a string of linear mass density $\\mu$ (measured in $\\mathrm{kg} \\mathrm{m}^{-1}$ ) is placed under a tension $T$ (a force, measured in newtons, $\\mathrm{N}$ ), then the fundamental oscillation frequency $f$ (measured in hertz, $\\mathrm{Hz}$, equivalent to cycles per second) is related to the length $L$ of the fundamental oscillation mode of the string (measured in metres, $\\mathrm{m}$ ) by $f=\\frac{1}{2 L} \\sqrt{\\frac{T}{\\mu}}$. If you plot a graph with a series of applied tensions $(T)$ on the $x$-axis, and the square of the length of the fundamental oscillation mode $\\left(L^{2}\\right)$ on the $y$-axis, what would you expect to see if the fundamental oscillation frequency is kept constant?\nA: A straight line through the origin, with a positive slope.\nB: A straight line through the origin, with a negative slope.\nC: A straight line, parallel to the $\\mathrm{x}$-axis.\nD: A parabolic curve, with a minimum at $x=0$.\nE: A parabolic curve, with a maximum at $x=0$.\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf a string of linear mass density $\\mu$ (measured in $\\mathrm{kg} \\mathrm{m}^{-1}$ ) is placed under a tension $T$ (a force, measured in newtons, $\\mathrm{N}$ ), then the fundamental oscillation frequency $f$ (measured in hertz, $\\mathrm{Hz}$, equivalent to cycles per second) is related to the length $L$ of the fundamental oscillation mode of the string (measured in metres, $\\mathrm{m}$ ) by $f=\\frac{1}{2 L} \\sqrt{\\frac{T}{\\mu}}$. If you plot a graph with a series of applied tensions $(T)$ on the $x$-axis, and the square of the length of the fundamental oscillation mode $\\left(L^{2}\\right)$ on the $y$-axis, what would you expect to see if the fundamental oscillation frequency is kept constant?\n\nA: A straight line through the origin, with a positive slope.\nB: A straight line through the origin, with a negative slope.\nC: A straight line, parallel to the $\\mathrm{x}$-axis.\nD: A parabolic curve, with a minimum at $x=0$.\nE: A parabolic curve, with a maximum at $x=0$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "A"
        ],
        "solution": "The relationship given can be rearranged to be $L^{2}=\\left(\\frac{1}{2 f}\\right)^{2} \\frac{T}{\\mu}$. This shows that the relationship between $L^{2}$ and $T$ is linear with a positive slope, as the coefficient of $T,\\left(\\frac{1}{2 f}\\right)^{2} \\frac{1}{\\mu}$, is positive. As there is no constant term the line is expected to pass through the origin.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_131",
        "problem": "A light, uniform, ideal spring is fixed at one end. If a mass is attached to the other end, the system oscillates with angular frequency $\\omega$. Now suppose the spring is fixed at the other end, then cut in half. The mass is attached between the two half springs.\n\n[figure1]\n\nThe new angular frequency of oscillations is\nA: $\\omega / 2$\nB: $\\omega$\nC: $\\sqrt{2} \\omega$\nD: $2 \\omega $ \nE: $4 \\omega$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA light, uniform, ideal spring is fixed at one end. If a mass is attached to the other end, the system oscillates with angular frequency $\\omega$. Now suppose the spring is fixed at the other end, then cut in half. The mass is attached between the two half springs.\n\n[figure1]\n\nThe new angular frequency of oscillations is\n\nA: $\\omega / 2$\nB: $\\omega$\nC: $\\sqrt{2} \\omega$\nD: $2 \\omega $ \nE: $4 \\omega$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://i.postimg.cc/G2084xWG/image.png"
        ],
        "answer": [
            "D"
        ],
        "solution": "Cutting a spring in half doubles its spring constant; if a force $F$ would elongate the original spring by an amount $x$, it elongates the new spring by an amount $x^{\\prime}=x / 2$, and using $F=-k x$ gives $k^{\\prime}=2 k$. The new system thus consists of two springs, each of whose spring constant is twice that of the original spring; the effective spring constant has thus increased by a factor of four. Since the angular frequency goes as the square root of the spring constant, the angular frequency increases by a factor of two.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_21",
        "problem": "The figure to the right shows the position-time graph of an object traveling in a straight line, starting out moving to the right (the positive $\\mathrm{x}$-direction). At which points is the object moving to the left?[figure1]\nA: $\\mathrm{B}$ and $\\mathrm{F}$\nB: $\\mathrm{E}$ and $\\mathrm{F}$\nC: A, B, and C\nD: C, D, and E\nE: None of the above\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe figure to the right shows the position-time graph of an object traveling in a straight line, starting out moving to the right (the positive $\\mathrm{x}$-direction). At which points is the object moving to the left?[figure1]\n\nA: $\\mathrm{B}$ and $\\mathrm{F}$\nB: $\\mathrm{E}$ and $\\mathrm{F}$\nC: A, B, and C\nD: C, D, and E\nE: None of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_7379c76224c2ea8ab8f4g-02.jpg?height=325&width=667&top_left_y=1553&top_left_x=1203"
        ],
        "answer": [
            "D"
        ],
        "solution": "At these points, the object is traveling in the opposite direction of the starting direction.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1406",
        "problem": "考虑水平地面上一辆自行车和骑车人构成的系统, 如图所示 (骑车人未画出）。取后轮着地点为原点 $\\mathrm{O}, x$ 轴过原点水平向右, $y$轴过原点坚直向上, 前轮着地点为 $\\mathrm{A}(l, 0)$, 系统 (假设为刚体) 质量为 $m$, 质心为 $\\mathrm{C}\\left(x_{\\mathrm{C}}, y_{\\mathrm{C}}\\right)$ 。已知重力加速度大小为 $g$, 假定车轮与地面之间的静摩擦系数足够大, 且前轮的滚动摩擦力可忽略。求\n\n[图1]骑车人开始蹬踏自行车时, 在保证安全的条件下自行车可达到的最大加速度。",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n这里是一些可能会帮助你解决问题的先验信息提示：\n考虑水平地面上一辆自行车和骑车人构成的系统, 如图所示 (骑车人未画出）。取后轮着地点为原点 $\\mathrm{O}, x$ 轴过原点水平向右, $y$轴过原点坚直向上, 前轮着地点为 $\\mathrm{A}(l, 0)$, 系统 (假设为刚体) 质量为 $m$, 质心为 $\\mathrm{C}\\left(x_{\\mathrm{C}}, y_{\\mathrm{C}}\\right)$ 。已知重力加速度大小为 $g$, 假定车轮与地面之间的静摩擦系数足够大, 且前轮的滚动摩擦力可忽略。求\n\n[图1]\n\n问题:\n骑车人开始蹬踏自行车时, 在保证安全的条件下自行车可达到的最大加速度。\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_65f5c8a7e52f16edf8b7g-03.jpg?height=477&width=676&top_left_y=2417&top_left_x=1384"
        ],
        "answer": [
            "$\\frac{x_{\\mathrm{C}}}{y_{\\mathrm{C}}} g$"
        ],
        "solution": "设骑车人开始蹬踏自行车时, 在保证安全的条件下自行车的加速度为 $a$ （朝前为正）。设地面给车后轮施加的静摩擦力大小为 $f$ 。按照牛顿第二定律, 由题意有\n\n$$\nf=m a\n$$\n\n系统相对于其质心没有转动, 由力矩平衡有\n\n$$\ny_{\\mathrm{C}} f-x_{\\mathrm{C}} F_{1}+\\left(l-x_{\\mathrm{C}}\\right) F_{2}=0\n$$\n\n由(1)(5)(6)式得\n\n$$\n\\begin{aligned}\n& F_{1}=\\left(\\frac{l-x_{C}+\\frac{a}{g} y_{C}}{l}\\right) m g \\\\\n& F_{2}=\\left(\\frac{x_{\\mathrm{C}}}{l}-\\frac{y_{\\mathrm{C}}}{l} \\frac{a}{g}\\right) m g\n\\end{aligned}\n$$\n\n由(5)式和关于加速度正方向的约定，\n\n$$\na>0\n$$\n\n(7)(8)式表明, 随着加速度 $a$ 增大, 后轮与地面间的正压力 $F_{1}$ 增大, 同时前轮与地面间的正压力 $F_{2}$ 减小。一旦 $F_{2}=0$, 系统就失去平衡。为了保证安全, 应有\n\n$$\nF_{2} \\geq 0\n$$\n\n由(8)(9)式知, 在保证安全的条件下自行车可达到的最大加速度大小为\n\n$$\na_{\\text {max }}=\\frac{x_{\\mathrm{C}}}{y_{\\mathrm{C}}} g\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1436",
        "problem": "如图, $1 \\mathrm{~mol}$ 单原子理想气体构成的系统分别经历循环过程 $a b c d a$ 和 $a b c^{\\prime} a$ 。已知理想气体在任一缓慢变化过程中, 压强 $p$ 和体积 $V$ 满足函数关系 $p=f(V)$ 。\n\n[图1]分别计算系统经 $b c^{\\prime}$ 直线过程中升降温的转折点在的坐标 $A$ 和吸放热的转折点在 $p-V$ 图中的坐标 $B$",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题包含多个待求解的量。\n这里是一些可能会帮助你解决问题的先验信息提示：\n如图, $1 \\mathrm{~mol}$ 单原子理想气体构成的系统分别经历循环过程 $a b c d a$ 和 $a b c^{\\prime} a$ 。已知理想气体在任一缓慢变化过程中, 压强 $p$ 和体积 $V$ 满足函数关系 $p=f(V)$ 。\n\n[图1]\n\n问题:\n分别计算系统经 $b c^{\\prime}$ 直线过程中升降温的转折点在的坐标 $A$ 和吸放热的转折点在 $p-V$ 图中的坐标 $B$\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你的最终解答的量应该按以下顺序输出：[坐标 $A$, 坐标 $B$]\n它们的答案类型依次是[数值, 数值]\n你需要在输出的最后用以下格式总结答案：“最终答案是\\boxed{ANSWER}”，其中ANSWER应为你的最终答案序列，用英文逗号分隔，例如：5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_e3a9fdbbef225ad3aefbg-03.jpg?height=982&width=785&top_left_y=1682&top_left_x=1092"
        ],
        "answer": [
            "$$ A\\left(\\frac{7}{2} V_{1}, \\frac{7}{4} p_{1}\\right) $$",
            "$$ B\\left(\\frac{35 V_{1}}{8}, \\frac{21 p_{1}}{16}\\right) $$"
        ],
        "solution": "根据过程热容的定义有\n\n$$\nC_{\\pi}=\\frac{\\Delta Q}{\\Delta T}\n$$\n\n式中, $\\Delta Q$ 是气体在此直线过程中, 温度升高 $\\Delta T$ 时从外界吸收的热量。由(10)11式得\n\n$$\n\\begin{aligned}\n\\Delta T & =\\frac{4 V-14 V_{1}}{8 V-35 V_{1}} R \\Delta Q \\\\\n\\Delta Q & =\\frac{8 V-35 V_{1}}{4 V-14 V_{1}} \\frac{\\Delta T}{R}\n\\end{aligned}\n$$\n\n由(12)式可知, $b c^{\\prime}$ 过程中的升降温的转折点 $A$ 在 $p-V$ 图上的坐标为\n\n$$\nA\\left(\\frac{7}{2} V_{1}, \\frac{7}{4} p_{1}\\right)\n$$\n\n由(10)式可知, $b c^{\\prime}$ 过程中的吸放热的转折点 $B$ 在 $p-V$ 图上的坐标为\n\n$$\nB\\left(\\frac{35 V_{1}}{8}, \\frac{21 p_{1}}{16}\\right)\n$$",
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "坐标 $A$",
            "坐标 $B$"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1727",
        "problem": "假定月球绕地球作圆周运动, 地球绕太阳也作圆周运动, 且轨道都在同一平面内. 已知地球表面处的重力加速度 $g=9.80 \\mathrm{~m} / \\mathrm{s}^{2}$, 地球半径 $R_{\\mathrm{e}}=6.37 \\times 10^{6} \\mathrm{~m}$, 月球质量 $m_{\\mathrm{m}}=7.3 \\times 10^{22} \\mathrm{~kg}$, 月球半径 $R_{\\mathrm{m}}=1.7 \\times 10^{6} \\mathrm{~m}$, 引力恒量 $G=6.67 \\times 10^{-11} \\mathrm{~N} \\cdot \\mathrm{m}^{2} \\cdot \\mathrm{kg}^{-2}$, 月心地心间的距离约为 $r_{\\mathrm{em}}=3.84 \\times 10^{8} \\mathrm{~m}$.月球的月心绕地球的地心运动一周需多少天?",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个数值。\n这里是一些可能会帮助你解决问题的先验信息提示：\n假定月球绕地球作圆周运动, 地球绕太阳也作圆周运动, 且轨道都在同一平面内. 已知地球表面处的重力加速度 $g=9.80 \\mathrm{~m} / \\mathrm{s}^{2}$, 地球半径 $R_{\\mathrm{e}}=6.37 \\times 10^{6} \\mathrm{~m}$, 月球质量 $m_{\\mathrm{m}}=7.3 \\times 10^{22} \\mathrm{~kg}$, 月球半径 $R_{\\mathrm{m}}=1.7 \\times 10^{6} \\mathrm{~m}$, 引力恒量 $G=6.67 \\times 10^{-11} \\mathrm{~N} \\cdot \\mathrm{m}^{2} \\cdot \\mathrm{kg}^{-2}$, 月心地心间的距离约为 $r_{\\mathrm{em}}=3.84 \\times 10^{8} \\mathrm{~m}$.\n\n问题:\n月球的月心绕地球的地心运动一周需多少天?\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n请记住，你的答案应以天为单位计算，但在给出最终答案时，请不要包含单位。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是不包含任何单位的数值。",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ff26c2613277b882b9edg-13.jpg?height=411&width=734&top_left_y=271&top_left_x=550"
        ],
        "answer": [
            "27.4"
        ],
        "solution": "月球在地球引力作用下绕地心作圆周运动, 设地球的质量为 $m_{\\mathrm{e}}$, 月球绕地心作圆周运动的角速度为 $\\omega_{\\mathrm{m}}$, 由万有引力定律和牛顿定律有\n\n$$\nG \\frac{m_{e} m_{m}}{r_{e m}^{2}}=m_{m} r_{\\mathrm{m}} \\omega_{m}\n$$\n\n另有\n\n$$\nG \\frac{m_{e}}{R_{e}^{2}}=g\n$$\n\n月球绕地球一周的时间\n\n$$\nT_{m}=\\frac{2 \\pi}{\\omega_{m}}\n$$\n\n解(1)、(2)、(3)三式得\n\n$$\nT_{m}=2 \\pi \\sqrt{\\frac{r_{e m}^{3}}{g R_{e}^{2}}}\n$$\n\n代入有关数据得\n\n$$\nT_{m}=2.37 \\times 10^{6} \\mathrm{~s}=27.4 \\text { 天 }\n$$\n\n[图1]",
        "answer_type": "NV",
        "unit": [
            "天"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Physics_1710",
        "problem": "质子是由更小的所谓“部分子”构成的。欧洲大型强子对撞机 (LHC) 是高能质子-质子对撞机，质子束内单个质子能量为 $E=7.0 \\mathrm{TeV}\\left(1 \\mathrm{TeV}=10^{3} \\mathrm{GeV}=10^{12} \\mathrm{eV}\\right)$, 两束能量相同的质子相向而行对撞碎裂, 其中相撞的两个部分子 $\\mathrm{a} 、 \\mathrm{~b}$ 相互作用湮灭产生一个新粒子。设部分子 $\\mathrm{a} 、 \\mathrm{~b}$ 的动能在质子能量中所占的比值分别为 $x_{\\mathrm{a}} 、 x_{\\mathrm{b}}$, 且远大于其静能。假设两个部分子 $\\mathrm{a} 、 \\mathrm{~b}$ 对撞湮灭产生了一个静质量为 $m_{\\mathrm{s}}=1.0 \\mathrm{TeV} / \\mathrm{c}^{2}$ 的新粒子 $\\mathrm{S}$, 求 $x_{\\mathrm{a}}$ 和 $x_{\\mathrm{b}}$ 的乘积 $X_{\\mathrm{a}} X_{\\mathrm{b}}$;",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个数值。\n这里是一些可能会帮助你解决问题的先验信息提示：\n质子是由更小的所谓“部分子”构成的。欧洲大型强子对撞机 (LHC) 是高能质子-质子对撞机，质子束内单个质子能量为 $E=7.0 \\mathrm{TeV}\\left(1 \\mathrm{TeV}=10^{3} \\mathrm{GeV}=10^{12} \\mathrm{eV}\\right)$, 两束能量相同的质子相向而行对撞碎裂, 其中相撞的两个部分子 $\\mathrm{a} 、 \\mathrm{~b}$ 相互作用湮灭产生一个新粒子。设部分子 $\\mathrm{a} 、 \\mathrm{~b}$ 的动能在质子能量中所占的比值分别为 $x_{\\mathrm{a}} 、 x_{\\mathrm{b}}$, 且远大于其静能。\n\n问题:\n假设两个部分子 $\\mathrm{a} 、 \\mathrm{~b}$ 对撞湮灭产生了一个静质量为 $m_{\\mathrm{s}}=1.0 \\mathrm{TeV} / \\mathrm{c}^{2}$ 的新粒子 $\\mathrm{S}$, 求 $x_{\\mathrm{a}}$ 和 $x_{\\mathrm{b}}$ 的乘积 $X_{\\mathrm{a}} X_{\\mathrm{b}}$;\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是数值。",
        "figure_urls": null,
        "answer": [
            "0.0051"
        ],
        "solution": "部分子 $\\mathrm{a} 、 \\mathrm{~b}$ 的动能分别为\n\n$$\nE_{\\mathrm{a}}=x_{\\mathrm{a}} E, \\quad E_{\\mathrm{b}}=x_{\\mathrm{b}} E\n$$\n\n它们远大于其静能, 所以部分子 $a 、 b$ 的质量可忽略, 其动量大小分别为\n\n$$\np_{\\mathrm{a}}=\\frac{E_{\\mathrm{a}}}{c}, \\quad p_{\\mathrm{b}}=-\\frac{E_{\\mathrm{b}}}{c}\n$$\n负号表明部分子 $a 、 b$ 的动量方向相反, 在同一条直线上。部分子 $a 、 b$ 对撞湮灭产生新粒子 $s$,根据能量守恒定律, $\\mathrm{S}$ 的能量为\n\n$$\nE_{\\mathrm{S}}=E_{\\mathrm{a}}+E_{\\mathrm{b}}\n$$\n\n根据动量守恒定律, $\\mathrm{S}$ 的动量大小为\n\n$$\np_{\\mathrm{s}}=p_{\\mathrm{a}}+p_{\\mathrm{b}}\n$$\n\n由自由粒子的动量和能量关系有\n\n$$\nE_{\\mathrm{S}}^{2}-p_{\\mathrm{S}}^{2} c^{2}=m_{\\mathrm{S}}^{2} c^{4}\n$$\n\n由(1)(2)(3)(4)(5)式得\n\n$$\nx_{\\mathrm{a}} x_{\\mathrm{b}}=\\frac{m_{\\mathrm{s}}^{2} c^{4}}{4 E^{2}}\n$$\n\n由(6)式和题给数据得\n\n$$\nx_{\\mathrm{a}} x_{\\mathrm{b}}=0.0051\n$$",
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Physics_1721",
        "problem": "某电视节目中演示了一个用三根火柴棍和细棉线悬挂起一瓶或多瓶矿泉水的实验,如图所示。图中, $\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{C}$ 为三根相同的火柴棍, 火柴棍长为 $l$, 细实线为棉线, 棉线的直径为 $d(d<<l)$ 。火柴棍 $\\mathrm{A}$ 的一半在水平桌面内, 另一半在桌面外, 火柴棍 $\\mathrm{A}$与桌面上表面的边沿垂直; 桌面厚度为 $h$; $\\mathrm{O}$ 是火柴棍 $\\mathrm{A}$ 的中点与桌面边沿的接触点;棉线紧贴桌沿绕过 A, 其两端与水平火柴棍 C 的两端连接; 火柴棍 B 的一端顶在火柴棍 A 的圆球状头部（可近似忽略圆球状头部的尺度）, 另一端顶在火柴棍 C 的中点。\n\n[图1]\n这样的结构可以稳定地悬挂起一瓶或多瓶矿泉水。加上火柴棍 B 和 C、小心挂上重物时, 火柴棍 $\\mathrm{A}$ 会绕 $\\mathrm{O}$ 点在过 $\\mathrm{A}$ 的坚直平面内转动,通过火柴棍 B 的带动, 绕过 A 并与火柴棍 C 两端相连的棉线将绕桌面下表面的边沿转动一个很小的角度; 只要角度大小合适, 可使整个系统达到稳定平衡。求平衡时该角度的大小。\n\n已知火柴棍与桌沿、火柴棍与棉线以及火柴棍之间都足够粗糙（即可以没有滑动），而且它们的质量与重物相比均可忽略。",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n这里是一些可能会帮助你解决问题的先验信息提示：\n某电视节目中演示了一个用三根火柴棍和细棉线悬挂起一瓶或多瓶矿泉水的实验,如图所示。图中, $\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{C}$ 为三根相同的火柴棍, 火柴棍长为 $l$, 细实线为棉线, 棉线的直径为 $d(d<<l)$ 。火柴棍 $\\mathrm{A}$ 的一半在水平桌面内, 另一半在桌面外, 火柴棍 $\\mathrm{A}$与桌面上表面的边沿垂直; 桌面厚度为 $h$; $\\mathrm{O}$ 是火柴棍 $\\mathrm{A}$ 的中点与桌面边沿的接触点;棉线紧贴桌沿绕过 A, 其两端与水平火柴棍 C 的两端连接; 火柴棍 B 的一端顶在火柴棍 A 的圆球状头部（可近似忽略圆球状头部的尺度）, 另一端顶在火柴棍 C 的中点。\n\n[图1]\n这样的结构可以稳定地悬挂起一瓶或多瓶矿泉水。\n\n问题:\n加上火柴棍 B 和 C、小心挂上重物时, 火柴棍 $\\mathrm{A}$ 会绕 $\\mathrm{O}$ 点在过 $\\mathrm{A}$ 的坚直平面内转动,通过火柴棍 B 的带动, 绕过 A 并与火柴棍 C 两端相连的棉线将绕桌面下表面的边沿转动一个很小的角度; 只要角度大小合适, 可使整个系统达到稳定平衡。求平衡时该角度的大小。\n\n已知火柴棍与桌沿、火柴棍与棉线以及火柴棍之间都足够粗糙（即可以没有滑动），而且它们的质量与重物相比均可忽略。\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d716ce15f03757bb482eg-03.jpg?height=477&width=300&top_left_y=447&top_left_x=1706",
            "https://cdn.mathpix.com/cropped/2024_03_31_d716ce15f03757bb482eg-06.jpg?height=398&width=169&top_left_y=2048&top_left_x=998",
            "https://cdn.mathpix.com/cropped/2024_03_31_d716ce15f03757bb482eg-06.jpg?height=271&width=228&top_left_y=2035&top_left_x=1254",
            "https://cdn.mathpix.com/cropped/2024_03_31_d716ce15f03757bb482eg-06.jpg?height=496&width=217&top_left_y=1945&top_left_x=1528",
            "https://cdn.mathpix.com/cropped/2024_03_31_d716ce15f03757bb482eg-07.jpg?height=294&width=228&top_left_y=1372&top_left_x=1508"
        ],
        "answer": [
            "$$ \\arcsin \\frac{d}{\\sqrt{3} l-2 h} $$"
        ],
        "solution": "如题图所示的结构可以稳定地悬挂起重物，当重物和火柴棍的质量趋于零时（未挂重物时）时, 在过火柴棍 A 的坚直平面内的情景应如图 2.a 所示。由题设, 火柴棍与桌沿、火柴棍与棉线以及火柴棍之间都足够粗䊁（可以没有滑动），可形成一个稳定的三脚架结构。由于火柴棍 $\\mathrm{A}$ 水平, 火柴棍 B 的下端正好在 A 的中点的正下方, 由几何关系知,火柴棍 $\\mathrm{A}$ 和 $\\mathrm{B}$ 之间的夹角为\n\n$$\n\\alpha=60^{\\circ}\n$$\n\n[图2]\n\n图1\n\n[图3]\n\n图2.a\n\n[图4]\n\n图2.b\n\n（由于 $d<<l$, 此时忽略了棉线的粗细）,桌面上表面边沿 $\\mathrm{O}$ 点到火柴棍 $\\mathrm{B}$ 的下端（即火柴棍 $\\mathrm{C}$ 的中点）的距离为\n\n$$\nL=\\frac{\\sqrt{3}}{2} l\n$$\n\n又由于火柴棍 $\\mathrm{C}$ 水平, 由几何关系知, 从 $\\mathrm{O}$ 点到火柴棍 C 两端的距离均为 $l$ 。\n\n值得指出的是, 据题意, 前面提及的三脚架结构在重物质量逐渐增大时是稳定的: 因为根据题意, 火柴棍与棉线之间是足够粗䊁的, 以至于跨过火柴棍 A 的棉线与它所连接的火柴棍 C 形成的正三角形能得以保持, 且火柴棍 C 也继续保持水平, 因而火柴棍 $\\mathrm{B}$ 与 $\\mathrm{A}$ 之间的夹角也能得以保持不变, 从而在图 2.b 中, 火柴棍 $\\mathrm{A}$ 和 $\\mathrm{B}$ 之间的夹角仍然为 $60^{\\circ}$ 。\n\n小心地挂上重物, 在过火柴棍 A 的坚直平面内的情景如图 2.b 所示。这时, 火柴棍 A 受到四个力的作用: 火柴棍 B 对它的头部的斜向上的推力、跨过火柴棍 A 的中部的细棉线对它的坚直向下的压力、桌面上边沿 $\\mathrm{O}$ 点对它的正压力和静摩擦力。火柴棍 $\\mathrm{A}$ 与桌面之间静摩擦力在一定范围内可以保证火柴棍 A 所受合力为零。同时, 由于力矩 $L$ 的作用, 火柴棍 $\\mathrm{A}$ 会绕 $\\mathrm{O}$ 点转动, 通过火柴棍 $\\mathrm{B}$ 的带动, 绕过 $\\mathrm{A}$ 连结火柴棍 $\\mathrm{C}$ 两端的棉线将绕桌面下表面的边沿转过一小角度, 使重物向左平移; 当重物向左平移的距离达到 $\\frac{d}{2}$ 时, 重物处于桌沿 $\\mathrm{O}$ 的正下方, 重力作用线恰好过支点 $\\mathrm{D}$ (见图 3), 因而对 $\\mathrm{D}$ 点的力矩为零, 整个系统不会转动, 从而达到稳定平衡。设此时过 $\\mathrm{A}$ 连结火柴棍 $\\mathrm{C}$ 两端的棉线绕桌面下表面的边沿转过的角度 $\\angle \\mathrm{DW}^{\\prime} \\mathrm{D}^{\\prime}=\\theta$, 如示意图 3 所示; 图中 D 是桌面下表面边沿, 且\n\n$$\n\\begin{aligned}\n& \\mathrm{OO}^{\\prime}=\\mathrm{DD}^{\\prime}=\\frac{d}{2}, \\\\\n& \\mathrm{OD}=\\mathrm{O}^{\\prime} \\mathrm{D}^{\\prime}=h\n\\end{aligned}\n$$\n\n$\\mathrm{W}^{\\prime}$ 是火柴棍 $\\mathrm{C}$ 的中点, 显然\n\n$$\n\\mathrm{D}^{\\prime} \\mathrm{W}^{\\prime}=L-h=\\frac{\\sqrt{3}}{2} l-h\n$$\n\n[图5]\n\n图3\n\n在直角 $\\Delta \\mathrm{DW}^{\\prime} \\mathrm{D}^{\\prime}$ 中, 由几何关系得\n\n$$\n\\sin \\theta=\\frac{\\frac{d}{2}}{\\frac{\\sqrt{3}}{2} l-h}\n$$\n\n或\n\n$$\n\\theta=\\arcsin \\frac{d}{\\sqrt{3} l-2 h}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_41",
        "problem": "A ticker tape timer is operating at $60 \\mathrm{~Hz}$. It was used to analyze the motion of a batterypowered car. The following displacements were measured for five intervals on the ticker tape.\n\n|  | Interval 1 | Interval 2 | Interval 3 | Interval 4 | Interval 5 |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| Displacement | $0.055 \\mathrm{~m}$ | $0.065 \\mathrm{~m}$ | $0.055 \\mathrm{~m}$ | $0.070 \\mathrm{~m}$ | $0.060 \\mathrm{~m}$ |\n\nWhat is the average speed of the car?\nA: $0.612 \\mathrm{~m} / \\mathrm{s}$\nB: $3.66 \\mathrm{~m} / \\mathrm{s}$\nC: $0.366 \\mathrm{~m} / \\mathrm{s}$\nD: $0.947 \\mathrm{~m} / \\mathrm{s}$\nE: $6.32 \\mathrm{~m} / \\mathrm{s}$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA ticker tape timer is operating at $60 \\mathrm{~Hz}$. It was used to analyze the motion of a batterypowered car. The following displacements were measured for five intervals on the ticker tape.\n\n|  | Interval 1 | Interval 2 | Interval 3 | Interval 4 | Interval 5 |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| Displacement | $0.055 \\mathrm{~m}$ | $0.065 \\mathrm{~m}$ | $0.055 \\mathrm{~m}$ | $0.070 \\mathrm{~m}$ | $0.060 \\mathrm{~m}$ |\n\nWhat is the average speed of the car?\n\nA: $0.612 \\mathrm{~m} / \\mathrm{s}$\nB: $3.66 \\mathrm{~m} / \\mathrm{s}$\nC: $0.366 \\mathrm{~m} / \\mathrm{s}$\nD: $0.947 \\mathrm{~m} / \\mathrm{s}$\nE: $6.32 \\mathrm{~m} / \\mathrm{s}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "B"
        ],
        "solution": "$t=\\frac{1}{60} s$ for each of the five displacement intervals.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_545",
        "problem": "An ideal rocket when empty of fuel has a mass $m_{r}$ and will carry a mass of fuel $m_{f}$. The fuel burns and is ejected with an exhaust speed of $v_{e}$ relative to the rocket. The fuel burns at a constant mass rate for a total time $T_{b}$. Ignore gravity; assume the rocket is far from any other body.\n\nDetermine an equation for the acceleration of the rocket as a function of time $t$ in terms of any or all of $t, m_{f}, m_{r}, v_{e}, T_{b}$, and any relevant fundamental constants.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nAn ideal rocket when empty of fuel has a mass $m_{r}$ and will carry a mass of fuel $m_{f}$. The fuel burns and is ejected with an exhaust speed of $v_{e}$ relative to the rocket. The fuel burns at a constant mass rate for a total time $T_{b}$. Ignore gravity; assume the rocket is far from any other body.\n\nDetermine an equation for the acceleration of the rocket as a function of time $t$ in terms of any or all of $t, m_{f}, m_{r}, v_{e}, T_{b}$, and any relevant fundamental constants.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": [
            "$\\frac{v_{e}}{m_{r}+m_{f}(1-t / T)} \\frac{m_{f}}{T}$"
        ],
        "solution": "Since there are no external forces on the system,\n\n$$\n0=\\frac{d p}{d t}=\\frac{d m}{d t} v+m \\frac{d v}{d t}\n$$\n\nwhich means\n\n$$\na=-\\frac{1}{m(t)} v_{e} \\frac{d m}{d t}=\\frac{v_{e}}{m_{r}+m_{f}(1-t / T)} \\frac{m_{f}}{T}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_122",
        "problem": "A launcher is designed to shoot objects horizontally across an ice rink. It consists of two boards of negligible mass connected via a spring-loaded hinge, which exerts a constant torque $\\tau$ on each board to keep them together. For both problems, neglect friction with either the ice or the launch boards.\n\n17. A hard disc is pushed into the launcher between the boards until the boards make contact with it a distance $\\ell$ from the hinge and are open to an angle $\\theta$, as shown in the figure. What is the minimum force necessary to hold the disc in this position?\n\n[figure1]\nA: $2 \\tau \\sin (\\theta / 2) / \\ell $ \nB: $2 \\tau \\cos (\\theta / 2) / \\ell$\nC: $\\tau \\cos (\\theta) / \\ell$\nD: $\\tau \\tan (\\theta) / \\ell$\nE: $2 \\tau \\tan (\\theta) / \\ell$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA launcher is designed to shoot objects horizontally across an ice rink. It consists of two boards of negligible mass connected via a spring-loaded hinge, which exerts a constant torque $\\tau$ on each board to keep them together. For both problems, neglect friction with either the ice or the launch boards.\n\n17. A hard disc is pushed into the launcher between the boards until the boards make contact with it a distance $\\ell$ from the hinge and are open to an angle $\\theta$, as shown in the figure. What is the minimum force necessary to hold the disc in this position?\n\n[figure1]\n\nA: $2 \\tau \\sin (\\theta / 2) / \\ell $ \nB: $2 \\tau \\cos (\\theta / 2) / \\ell$\nC: $\\tau \\cos (\\theta) / \\ell$\nD: $\\tau \\tan (\\theta) / \\ell$\nE: $2 \\tau \\tan (\\theta) / \\ell$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_17b50ad575c0c7f654b7g-14.jpg?height=328&width=444&top_left_y=606&top_left_x=843"
        ],
        "answer": [
            "A"
        ],
        "solution": "Each board exerts a force with magnitude $F=\\tau / \\ell$, directed toward the center of the disc. We can decompose these forces into a component $F_{\\|}=(\\tau / \\ell) \\sin (\\theta / 2)$ that is parallel to the symmetry axis of the launcher, and a perpendicular component $F_{\\perp}$. The perpendicular components from the two forces cancel, while the parallel components add together. The minimal force $F_{\\text {tot }}$ required to keep the disc in place must equal the sum of the two parallel components, which gives\n\n$$\nF_{\\text {tot }}=2 F_{\\|}=\\frac{2 \\tau}{\\ell} \\sin (\\theta / 2)\n$$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_879",
        "problem": "he schematic below shows the Hadley circulation in the Earth's tropical atmosphere around the spring equinox. Air rises from the equator and moves poleward in both hemispheres before descending in the subtropics at latitudes $\\pm \\varphi_{d}$ (where positive and negative latitudes refer to the northern and southern hemisphere respectively). The angular momentum about the Earth's spin axis is conserved for the upper branches of the circulation (enclosed by the dashed oval). Note that the schematic is not drawn to scale.\n\n[figure1]\n\n\nLet the temperature next to the surface and at the top of the atmosphere be $T_{H}$ and $T_{C}$ respectively. Given that the pressure difference between points $\\mathrm{A}$ and $\\mathrm{E}$ is $20 \\mathrm{hPa}$, calculate $T_{C}$ for $T_{H}=300 \\mathrm{~K}$.\n\nNote that the ratio of molar gas constant $(R)$ to molar heat capacity at constant pressure $\\left(c_{p}\\right)$ for air, $\\kappa$, is $2 / 7$.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nhe schematic below shows the Hadley circulation in the Earth's tropical atmosphere around the spring equinox. Air rises from the equator and moves poleward in both hemispheres before descending in the subtropics at latitudes $\\pm \\varphi_{d}$ (where positive and negative latitudes refer to the northern and southern hemisphere respectively). The angular momentum about the Earth's spin axis is conserved for the upper branches of the circulation (enclosed by the dashed oval). Note that the schematic is not drawn to scale.\n\n[figure1]\n\n\nLet the temperature next to the surface and at the top of the atmosphere be $T_{H}$ and $T_{C}$ respectively. Given that the pressure difference between points $\\mathrm{A}$ and $\\mathrm{E}$ is $20 \\mathrm{hPa}$, calculate $T_{C}$ for $T_{H}=300 \\mathrm{~K}$.\n\nNote that the ratio of molar gas constant $(R)$ to molar heat capacity at constant pressure $\\left(c_{p}\\right)$ for air, $\\kappa$, is $2 / 7$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of K, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_e7b9ba69a6dd20c5cfd1g-1.jpg?height=1148&width=1151&top_left_y=1096&top_left_x=384"
        ],
        "answer": [
            "195"
        ],
        "solution": "Since $p_{E}>p_{A}$ and $p_{A}=1000 \\mathrm{hPa}$, we have $p_{E}=1020 \\mathrm{hPa}$.\n\nFrom the adiabatic compression from $\\mathrm{D}$ to $\\mathrm{E}$, we have:\n\n$$\n\\begin{aligned}\n& p_{E}^{-\\kappa} T_{H}=p_{D}^{-\\kappa} T_{C} \\\\\n& T_{C}=\\left(\\frac{p_{D}}{p_{E}}\\right)^{\\kappa} \\times T_{H}=\\left(\\frac{225}{1020}\\right)^{2 / 7} \\times 300 K=195 K\n\\end{aligned}\n$$",
        "answer_type": "NV",
        "unit": [
            "K"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1104",
        "problem": "# A Three-body Problem and LISA \n\n[figure1]\n\nFIGURE 1 Coplanar orbits of three bodies.\n\nA third body of infinitesimal mass $\\mu$ is placed in a coplanar circular orbit about the same centre of mass so that $\\mu$ remains stationary relative to both $M$ and $m$ as shown in Figure 1. Assume that the infinitesimal mass is not collinear with $M$ and $m$. Find the values of the following parameters in terms of $R$ and $r$ :\n\ndistance from $\\mu$ to the centre of mass",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\n# A Three-body Problem and LISA \n\n[figure1]\n\nFIGURE 1 Coplanar orbits of three bodies.\n\nA third body of infinitesimal mass $\\mu$ is placed in a coplanar circular orbit about the same centre of mass so that $\\mu$ remains stationary relative to both $M$ and $m$ as shown in Figure 1. Assume that the infinitesimal mass is not collinear with $M$ and $m$. Find the values of the following parameters in terms of $R$ and $r$ :\n\ndistance from $\\mu$ to the centre of mass\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_27986d152c1a1dded222g-1.jpg?height=918&width=919&top_left_y=506&top_left_x=619"
        ],
        "answer": [
            "$\\sqrt{r^{2}+r R+R^{2}}$"
        ],
        "solution": "Since $\\mu$ is infinitesimal, it has no gravitational influences on the motion of neither $M$ nor $m$. For $\\mu$ to remain stationary relative to both $M$ and $m$ we must have:\n\n$$\n\\begin{aligned}\n\\frac{G M \\mu}{r_{1}^{2}} \\cos \\theta_{1}+\\frac{G m \\mu}{r_{2}^{2}} \\cos \\theta_{2} & =\\mu \\omega_{0}^{2} \\rho=\\frac{G(M+m) \\mu}{(R+r)^{3}} \\rho \\\\\n\\frac{G M \\mu}{r_{1}^{2}} \\sin \\theta_{1} & =\\frac{G m \\mu}{r_{2}^{2}} \\sin \\theta_{2}\n\\end{aligned}\n$$\n\nSubstituting $\\frac{G M}{r_{1}^{2}}$ from Eq. (5) into Eq. (4), and using the identity\n\n$\\sin \\theta_{1} \\cos \\theta_{2}+\\cos \\theta_{1} \\sin \\theta_{2}=\\sin \\left(\\theta_{1}+\\theta_{2}\\right)$, we get\n\n$$\nm \\frac{\\sin \\left(\\theta_{1}+\\theta_{2}\\right)}{r_{2}^{2}}=\\frac{(M+m)}{(R+r)^{3}} \\rho \\sin \\theta_{1}\n$$\n\nThe distances $r_{2}$ and $\\rho$, the angles $\\theta_{1}$ and $\\theta_{2}$ are related by two Sine Rule equations\n\n$$\n\\begin{aligned}\n& \\frac{\\sin \\psi_{1}}{\\rho}=\\frac{\\sin \\theta_{1}}{R} \\\\\n& \\frac{\\sin \\psi_{1}}{r_{2}}=\\frac{\\sin \\left(\\theta_{1}+\\theta_{2}\\right)}{R+r}\n\\end{aligned}\n$$\n\nSubstitute (7) into (6)\n\n$$\n\\frac{1}{r_{2}^{3}}=\\frac{R}{(R+r)^{4}} \\frac{(M+m)}{m}\n$$\n\nSince $\\frac{m}{M+m}=\\frac{R}{R+r}$,Eq. (10) gives\n\n$$\nr_{2}=R+r\n$$\n\nBy substituting $\\frac{G m}{r_{2}^{2}}$ from Eq. (5) into Eq. (4), and repeat a similar procedure, we get\n\n$$\nr_{1}=R+r\n$$\n\nHence, it is an equilateral triangle with\n\n$$\n\\begin{aligned}\n& \\psi_{1}=60^{\\circ} \\\\\n& \\psi_{2}=60^{\\circ}\n\\end{aligned}\n$$\n\nThe distance $\\rho$ is calculated from the Cosine Rule.\n\n$$\n\\begin{aligned}\n& \\rho^{2}=r^{2}+(R+r)^{2}-2 r(R+r) \\cos 60^{\\circ} \\\\\n& \\rho=\\sqrt{r^{2}+r R+R^{2}}\n\\end{aligned}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_825",
        "problem": "An electromagnetic wave can propagate in a transmission line in two opposite directions. For each direction of propagation, the characteristic impedance $Z_{0}$ can be used to relate the voltage $V_{0}$ and current $I_{0}$ amplitudes as in the Ohm's law, $Z_{0}=V_{0} / I_{0}$.\n\nConsider an interface between two transmission lines, with characteristic impedances $Z_{0}$ and $Z_{1}$. A schematic diagram of the circuit is shown below.\n\n[figure1]\n\nCircuit diagram of a transmission line of impedance $Z_{0}$ connected to a transmission line of impedance $Z_{1}$. The physical size of the interface is much smaller than the wavelength.\n\nWhen a signal $V_{\\mathrm{i}}$ sent into the transmission line with impedance $Z_{0}$ reaches the interface it is partially transmitted into the second transmission line, resulting in a signal $V_{\\mathrm{t}}$ in that line which propagates forward. Some of the signal may also be reflected, resulting in a backward propagating signal in the initial transmission line $V_{\\mathrm{r}}$.\n\nState the condition(s) for the signal $V_{\\mathrm{i}}$ to have gained a $\\pi$ phase change on re- $0.2 \\mathrm{pt}$ flection.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nAn electromagnetic wave can propagate in a transmission line in two opposite directions. For each direction of propagation, the characteristic impedance $Z_{0}$ can be used to relate the voltage $V_{0}$ and current $I_{0}$ amplitudes as in the Ohm's law, $Z_{0}=V_{0} / I_{0}$.\n\nConsider an interface between two transmission lines, with characteristic impedances $Z_{0}$ and $Z_{1}$. A schematic diagram of the circuit is shown below.\n\n[figure1]\n\nCircuit diagram of a transmission line of impedance $Z_{0}$ connected to a transmission line of impedance $Z_{1}$. The physical size of the interface is much smaller than the wavelength.\n\nWhen a signal $V_{\\mathrm{i}}$ sent into the transmission line with impedance $Z_{0}$ reaches the interface it is partially transmitted into the second transmission line, resulting in a signal $V_{\\mathrm{t}}$ in that line which propagates forward. Some of the signal may also be reflected, resulting in a backward propagating signal in the initial transmission line $V_{\\mathrm{r}}$.\n\nState the condition(s) for the signal $V_{\\mathrm{i}}$ to have gained a $\\pi$ phase change on re- $0.2 \\mathrm{pt}$ flection.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_c16db445016d4c0e9a0ag-3.jpg?height=166&width=782&top_left_y=842&top_left_x=637"
        ],
        "answer": [
            "$Z_{1}<Z_{0}$"
        ],
        "solution": "$\\pi$-shift implies opposite signs of $V_{\\mathrm{i}}$ and $V_{\\mathrm{r}}$ and hence requires $\\Gamma<0$. This implies $Z_{1}<Z_{0}$.",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1725",
        "problem": "一轻杆通过分别连在其两端的轻质弹簧 $\\mathrm{A}$ 和 B 悬挂在天花板下, 一物块 $\\mathrm{D}$ 通过轻质弹簧 C 连在轻杆上; A、B 和 C 的劲度系数分别为 $k_{1} 、 k_{2}$ 和 $k_{3}, \\mathrm{D}$ 的质量为 $m, \\mathrm{C}$ 与轻杆的连接点到 $\\mathrm{A}$ 和 $\\mathrm{B}$ 的水平距离分别为 $a$ 和 $b$; 整个系统的平衡时, 轻杆接近水平, 如图所示。假设物块 $\\mathrm{D}$ 在坚直方向做微小振动, $\\mathrm{A} 、 \\mathrm{~B}$ 始终可视为坚直, 忽略空气阻力。\n\n[图1]求物块 $\\mathrm{D}$ 上下微小振动的固有频率",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n这里是一些可能会帮助你解决问题的先验信息提示：\n一轻杆通过分别连在其两端的轻质弹簧 $\\mathrm{A}$ 和 B 悬挂在天花板下, 一物块 $\\mathrm{D}$ 通过轻质弹簧 C 连在轻杆上; A、B 和 C 的劲度系数分别为 $k_{1} 、 k_{2}$ 和 $k_{3}, \\mathrm{D}$ 的质量为 $m, \\mathrm{C}$ 与轻杆的连接点到 $\\mathrm{A}$ 和 $\\mathrm{B}$ 的水平距离分别为 $a$ 和 $b$; 整个系统的平衡时, 轻杆接近水平, 如图所示。假设物块 $\\mathrm{D}$ 在坚直方向做微小振动, $\\mathrm{A} 、 \\mathrm{~B}$ 始终可视为坚直, 忽略空气阻力。\n\n[图1]\n\n问题:\n求物块 $\\mathrm{D}$ 上下微小振动的固有频率\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_07aa406e17d01fd01b36g-01.jpg?height=428&width=457&top_left_y=937&top_left_x=1299"
        ],
        "answer": [
            "\\sqrt{\\frac{(a+b)^{2} k_{1} k_{2} k_{3}}{m\\left[a^{2} k_{1} k_{3}+b^{2} k_{2} k_{3}+(a+b)^{2} k_{1} k_{2}\\right]}}"
        ],
        "solution": "挂上物块 $\\mathrm{D}$ 前, 系统平衡时弹簧 $\\mathrm{C}$ 的末端 $E$ 点的坐标 $X_{E}$ 为\n\n$$\nx_{E}=l_{1}+l_{3}+\\frac{a\\left(l_{2}-l_{1}\\right)}{a+b}\n$$\n\n挂上物块 $\\mathrm{D}$ 后, 系统重新处于平衡位置时, 弹簧 $\\mathrm{C}$ 的末端从 $E$ 点移动到 $E^{\\prime}$ 点。 $E^{\\prime}$ 点的坐标 $x_{E^{\\prime}}$ 为\n\n$$\nx_{E^{\\prime}}=l_{1}+\\Delta x_{1}+l_{3}+\\Delta x_{3}+\\frac{a\\left(l_{2}-l_{1}+\\Delta x_{2}-\\Delta x_{1}\\right)}{a+b}\n$$\n\n将系统视为一等效弹簧, 则系统处于静平衡时等效弹簧伸长为\n\n$$\n\\Delta x_{E}=x_{E^{\\prime}}-x_{E}=\\Delta x_{1}+\\Delta x_{3}+\\frac{a\\left(\\Delta x_{2}-\\Delta x_{1}\\right)}{a+b}\n$$\n\n将(2)(7)(8)式代入(11)式, 得系统处于静平衡时等效弹簧的伸长为\n\n$$\n\\Delta x_{E}=\\frac{m g\\left(a^{2} k_{1}+b^{2} k_{2}\\right)}{(a+b)^{2} k_{1} k_{2}}+\\frac{m g}{k_{3}}\n$$\n\n因此，等效弹簧的等效劲度系数为\n\n$$\nk_{\\mathrm{e}}=\\frac{m g}{\\Delta x_{E}}=\\frac{(a+b)^{2} k_{1} k_{2} k_{3}}{a^{2} k_{1} k_{3}+b^{2} k_{2} k_{3}+(a+b)^{2} k_{1} k_{2}}\n$$\n\n物块 $\\mathrm{D}$ 上下微小振动的固有频率为\n\n$$\n\\omega_{\\mathrm{e}}=\\sqrt{\\frac{k_{\\mathrm{e}}}{m}}=\\sqrt{\\frac{(a+b)^{2} k_{1} k_{2} k_{3}}{m\\left[a^{2} k_{1} k_{3}+b^{2} k_{2} k_{3}+(a+b)^{2} k_{1} k_{2}\\right]}}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1740",
        "problem": "一个用电阻丝绕成的线圈, 浸没在量热器所盛的油中, 油的温度为 $0^{\\circ} \\mathrm{C}$, 当线圈两端加上一定电压后, 油温渐渐上升, $0^{\\circ} \\mathrm{C}$ 时温度升高的速率为 $5.0 \\mathrm{~K} \\cdot \\mathrm{min}^{-1}$, 持续一段时间后, 油温上升到 $30^{\\circ} \\mathrm{C}$, 此时温度升高的速率为 $4.5 \\mathrm{~K} \\cdot \\mathrm{min}^{-1}$, 这是因为线圈的电阻与温度有关。设温度为 $\\theta^{\\circ} \\mathrm{C}$ 时线圈的电阻为 $R_{\\theta}$, 温度为 $0^{\\circ} \\mathrm{C}$ 时线圈的电阻为 $R_{0}$, 则有 $R_{\\theta}=R_{0}(1-\\alpha \\theta), \\alpha$ 称为电阻的温度系数。试求此线圈电阻的温度系数。假设量热器及其中的油以及线圈所构成的系统温度升高的速率与该系统吸收的热量的速率（即单位时间内吸收的热量）成正比; 对油加热过程中加在线圈两端的电压恒定不变; 系统损失的热量可忽略不计。",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个数值。\n\n问题:\n一个用电阻丝绕成的线圈, 浸没在量热器所盛的油中, 油的温度为 $0^{\\circ} \\mathrm{C}$, 当线圈两端加上一定电压后, 油温渐渐上升, $0^{\\circ} \\mathrm{C}$ 时温度升高的速率为 $5.0 \\mathrm{~K} \\cdot \\mathrm{min}^{-1}$, 持续一段时间后, 油温上升到 $30^{\\circ} \\mathrm{C}$, 此时温度升高的速率为 $4.5 \\mathrm{~K} \\cdot \\mathrm{min}^{-1}$, 这是因为线圈的电阻与温度有关。设温度为 $\\theta^{\\circ} \\mathrm{C}$ 时线圈的电阻为 $R_{\\theta}$, 温度为 $0^{\\circ} \\mathrm{C}$ 时线圈的电阻为 $R_{0}$, 则有 $R_{\\theta}=R_{0}(1-\\alpha \\theta), \\alpha$ 称为电阻的温度系数。试求此线圈电阻的温度系数。假设量热器及其中的油以及线圈所构成的系统温度升高的速率与该系统吸收的热量的速率（即单位时间内吸收的热量）成正比; 对油加热过程中加在线圈两端的电压恒定不变; 系统损失的热量可忽略不计。\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n请记住，你的答案应以$\\mathrm{~K}^{-1}$为单位计算，但在给出最终答案时，请不要包含单位。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是不包含任何单位的数值。",
        "figure_urls": null,
        "answer": [
            "3.7 \\times 10^{-3}"
        ],
        "solution": "量热器、油和线圈构成的系统在单位时间内吸收的热量等于通过线圈的电流的电功率。设加在线圈两端的电压为 $U$, 当线圈的电阻为 $R_{0}$ 时, 电流的功率\n\n$$\nP_{0}=\\frac{U^{2}}{R_{0}}\n$$\n\n$$\n\\text { 根据题意, 有 } v_{0}=k P_{0}\n$$\n\n式中 $v_{0}$ 表示 $0^{\\circ} \\mathrm{C}$ 时系统升温的速率, $k$ 为比例系数, 同理当油温为 $0^{\\circ} \\mathrm{C}$ 时, 有\n\n$$\n\\begin{aligned}\n& P_{30}=\\frac{U^{2}}{R_{30}} \\\\\n& v_{30}=k P_{30}\n\\end{aligned}\n$$\n\n式中 $v_{30}$ 表示 $30^{\\circ} \\mathrm{C}$ 时系统升温的速率，由（1）、 (2)\n\n(3)、\n\n（4）各式得\n\n$$\n\\frac{R_{30}}{R_{0}}=\\frac{v_{30}}{v_{0}}=1+30 \\alpha\n$$\n\n代入数据解得 $\\alpha=3.7 \\times 10^{-3} \\mathrm{~K}^{-1}$",
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~K}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Physics_713",
        "problem": "An aluminum wire of varying thickness is pulled at each end with a force of $30 \\mathrm{~N}$. Which of the following statements is true? Note: tension is a force that tends to stretch something, and stress is the force per unit area.\nA: The tension is $60 \\mathrm{~N}$ all along the wire, and stress is constant along the wire\nB: The tension is $60 \\mathrm{~N}$ all along the wire, and stress varies along the wire\nC: The tension is $30 \\mathrm{~N}$ all along the wire, and stress is constant along the wire\nD: The tension is $30 \\mathrm{~N}$ all along the wire, and stress varies along the wire\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn aluminum wire of varying thickness is pulled at each end with a force of $30 \\mathrm{~N}$. Which of the following statements is true? Note: tension is a force that tends to stretch something, and stress is the force per unit area.\n\nA: The tension is $60 \\mathrm{~N}$ all along the wire, and stress is constant along the wire\nB: The tension is $60 \\mathrm{~N}$ all along the wire, and stress varies along the wire\nC: The tension is $30 \\mathrm{~N}$ all along the wire, and stress is constant along the wire\nD: The tension is $30 \\mathrm{~N}$ all along the wire, and stress varies along the wire\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": [
            "D"
        ],
        "solution": "The tension is 30N all along the wire because of Newton's third law. Since the crosssection varies, the stress is different at different cross-sections.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_1025",
        "problem": "## Kelvin water dropper\n\nThe following facts about the surface tension may turn out to be useful for this problem. For the molecules of a liquid, the positions at the liquid-air interface are less favourable as compared with the positions in the bulk of the liquid. This interface is described by the so-called surface energy, $U=\\sigma S$, where $S$ is the surface area of the interface and $\\sigma$ is the surface tension coefficient of the liquid. Moreover, two fragments of the liquid surface pull each other with a force $F=\\sigma l$, where $l$ is the length of a straight line separating the fragments.\n\n[figure1]\n\nA long metallic pipe with internal diameter $d$ is pointing directly downwards. Water is slowly dripping from a nozzle at its lower end, see fig. Water can be considered to be electrically conducting; its surface tension is $\\sigma$ and its density is $\\rho$. A droplet of radius $r$ hangs below the nozzle. The radius grows slowly in time until the droplet separates from the nozzle due to the free fall acceleration $g$. Always assume that $d \\ll r$.\n\nFind the radius $r_{\\max }$ of a drop just before it separates from the nozzle.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\n## Kelvin water dropper\n\nThe following facts about the surface tension may turn out to be useful for this problem. For the molecules of a liquid, the positions at the liquid-air interface are less favourable as compared with the positions in the bulk of the liquid. This interface is described by the so-called surface energy, $U=\\sigma S$, where $S$ is the surface area of the interface and $\\sigma$ is the surface tension coefficient of the liquid. Moreover, two fragments of the liquid surface pull each other with a force $F=\\sigma l$, where $l$ is the length of a straight line separating the fragments.\n\n[figure1]\n\nA long metallic pipe with internal diameter $d$ is pointing directly downwards. Water is slowly dripping from a nozzle at its lower end, see fig. Water can be considered to be electrically conducting; its surface tension is $\\sigma$ and its density is $\\rho$. A droplet of radius $r$ hangs below the nozzle. The radius grows slowly in time until the droplet separates from the nozzle due to the free fall acceleration $g$. Always assume that $d \\ll r$.\n\nFind the radius $r_{\\max }$ of a drop just before it separates from the nozzle.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6b45dae650949cf89919g-1.jpg?height=657&width=406&top_left_y=939&top_left_x=334"
        ],
        "answer": [
            "$\\sqrt[3]{\\frac{3 \\sigma d}{4 \\rho g}}$"
        ],
        "solution": "Let us write the force balance for the droplet. Since $d \\ll r$, we can neglect the force $\\frac{\\pi}{4} \\Delta p d^{2}$ due to the excess pressure $\\Delta p$ inside the tube. So, the gravity force $\\frac{4}{3} \\pi r_{\\max }^{3} \\rho g$ is balanced by the capillary force. When the droplet separates from the tube, the water surface forms in the vicinity of the nozzle a \"neck\", which has vertical tangent. In the horizontal cross-section of that \"neck\", the capillary force is vertical and can be calculated as $\\pi \\sigma d$. So,\n\n$$\nr_{\\max }=\\sqrt[3]{\\frac{3 \\sigma d}{4 \\rho g}}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_38",
        "problem": "An increase in the translational motion of the molecules of a gas confined in a steel tank will be observed in one of the following ways. It will produce an increase in:\nA: the temperature of the gas only.\nB: the pressure of the gas only.\nC: both the temperature and pressure of the gas.\nD: the temperature of the gas and a decrease in its pressure.\nE: the volume of the gas.\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn increase in the translational motion of the molecules of a gas confined in a steel tank will be observed in one of the following ways. It will produce an increase in:\n\nA: the temperature of the gas only.\nB: the pressure of the gas only.\nC: both the temperature and pressure of the gas.\nD: the temperature of the gas and a decrease in its pressure.\nE: the volume of the gas.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "C"
        ],
        "solution": "$\\mathrm{PV}=\\mathrm{nRT}$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_804",
        "problem": "A box of mass $5 \\mathrm{~kg}$ sits at rest on a horizontal floor. A vertically upwards force of $20 \\mathrm{~N}$ is exerted on the box by someone trying to lift it. What is the magnitude of the normal force exerted by the floor on the box while the $20 \\mathrm{~N}$ upwards force is being exerted?\na. $0 \\mathrm{~N}$ (the box is no longer in contact with the floor)\nA: $20 \\mathrm{~N}$\nB: $30 \\mathrm{~N}$\nC: $50 \\mathrm{~N}$\nD: $70 \\mathrm{~N}$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA box of mass $5 \\mathrm{~kg}$ sits at rest on a horizontal floor. A vertically upwards force of $20 \\mathrm{~N}$ is exerted on the box by someone trying to lift it. What is the magnitude of the normal force exerted by the floor on the box while the $20 \\mathrm{~N}$ upwards force is being exerted?\na. $0 \\mathrm{~N}$ (the box is no longer in contact with the floor)\n\nA: $20 \\mathrm{~N}$\nB: $30 \\mathrm{~N}$\nC: $50 \\mathrm{~N}$\nD: $70 \\mathrm{~N}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": [
            "C"
        ],
        "solution": "The weight of the box is $W=m g=-5 \\mathrm{~kg} \\times 10 \\mathrm{~m} \\mathrm{~s}^{-2}=-50 \\mathrm{~N}$, where $m$ is the mass of the box, and the negative sign means that the force is directed downwards. As the box is at rest, not accelerating, the total force on the box is $0 \\mathrm{~N}$. Hence, $W+V+N=0$ where $V=20 \\mathrm{~N}$ is the upwards force exerted by someone trying to lift it, and $N$ is the normal force. Rearranging,\n\n$N=-W-N=(50-20) \\mathrm{N}=30 \\mathrm{~N}$.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_611",
        "problem": "[figure1]\n\nThe wall of a neuron is made from an elastic membrane, which resists compression in the same way as a spring. It has an effective spring constant $k$ and an equilibrium thickness $d_{0}$. Assume that the membrane has a very large area $A$ and negligible curvature.\n\nThe neuron has \"ion pumps\" that can move ions across the membrane. In the resulting charged state, positive and negative ionic charge is arranged uniformly along the outer and inner surfaces of the membrane, respectively. The permittivity of the membrane is $\\epsilon$.\n\nSuppose that, after some amount of work is done by the ion pumps, the charges on the outer and inner surfaces are $Q$ and $-Q$, respectively. What is the thickness $d$ of the membrane?",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\n[figure1]\n\nThe wall of a neuron is made from an elastic membrane, which resists compression in the same way as a spring. It has an effective spring constant $k$ and an equilibrium thickness $d_{0}$. Assume that the membrane has a very large area $A$ and negligible curvature.\n\nThe neuron has \"ion pumps\" that can move ions across the membrane. In the resulting charged state, positive and negative ionic charge is arranged uniformly along the outer and inner surfaces of the membrane, respectively. The permittivity of the membrane is $\\epsilon$.\n\nSuppose that, after some amount of work is done by the ion pumps, the charges on the outer and inner surfaces are $Q$ and $-Q$, respectively. What is the thickness $d$ of the membrane?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_fed953f9e38b72bf8bd7g-14.jpg?height=358&width=927&top_left_y=469&top_left_x=596"
        ],
        "answer": [
            "$d_{0}-\\frac{Q^{2}}{2 \\epsilon A k}$"
        ],
        "solution": "One charge layer by itself creates an electric field $E_{1}=Q /(2 \\epsilon A)$ in each direction. So the force between the two sides of the membrane $F_{E}=Q E_{1}=Q^{2} /(2 \\epsilon A)$.\n\nThis electric force is balanced by the spring force $F_{s}=k x$, where $x=d_{0}-d$. Equating these two forces and solving for $d$ gives gives\n\n$$\nd=d_{0}-\\frac{Q^{2}}{2 \\epsilon A k} .\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_357",
        "problem": "Eero Uustalu and Jaan Kalda. You are given the following tools: a retroreflective film an enlarged bottom view of which is shown in the first figure; stand, ruler, laser pointer, sreen, graph paper, protractor.\n\n[figure1]\n\nWhile the top surface of the film is flat, the bottom surface is a periodic array of slanted triangular faces. Six such faces are shown enlarged in the seond figure; the faces 1,3 a corner of a cube, and the faces 2, 4 nd 6 are also perpendicular to each other On the right of the second figure, a cross section of the film is shown. The film's material between the slanted faces and the flat surface form microprisms. The refracting angles of these microprisms are denoted by $\\alpha_{i}, i=1,2, \\ldots 6$ (the index numbers correspond to those of the faces). Among the an lges $\\alpha_{i}$, some may be equal to each other.\n\n[figure2]\n\ncross-section\n\nWhen light falls onto the flat surface close to perpendicular incidence, it undergoes total internal reflections on the slanted faces, and as a result, its direction of propagation is rotated by $180^{\\circ}$. However, the microprisms can also serve as prisms diverting a light beam by an angle $\\beta$. The angle $\\beta$ depends on the angle of incidence, and on the prism angle $\\alpha=\\alpha_{i}$. Let $\\beta_{i}$ denote the minimal deflection angle for a fixed prism angle $\\alpha_{i}$\n\nDetermine the refraction coefficient of the film's material.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nEero Uustalu and Jaan Kalda. You are given the following tools: a retroreflective film an enlarged bottom view of which is shown in the first figure; stand, ruler, laser pointer, sreen, graph paper, protractor.\n\n[figure1]\n\nWhile the top surface of the film is flat, the bottom surface is a periodic array of slanted triangular faces. Six such faces are shown enlarged in the seond figure; the faces 1,3 a corner of a cube, and the faces 2, 4 nd 6 are also perpendicular to each other On the right of the second figure, a cross section of the film is shown. The film's material between the slanted faces and the flat surface form microprisms. The refracting angles of these microprisms are denoted by $\\alpha_{i}, i=1,2, \\ldots 6$ (the index numbers correspond to those of the faces). Among the an lges $\\alpha_{i}$, some may be equal to each other.\n\n[figure2]\n\ncross-section\n\nWhen light falls onto the flat surface close to perpendicular incidence, it undergoes total internal reflections on the slanted faces, and as a result, its direction of propagation is rotated by $180^{\\circ}$. However, the microprisms can also serve as prisms diverting a light beam by an angle $\\beta$. The angle $\\beta$ depends on the angle of incidence, and on the prism angle $\\alpha=\\alpha_{i}$. Let $\\beta_{i}$ denote the minimal deflection angle for a fixed prism angle $\\alpha_{i}$\n\nDetermine the refraction coefficient of the film's material.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_706aca6df357b4c9a255g-1.jpg?height=246&width=418&top_left_y=1593&top_left_x=1611",
            "https://cdn.mathpix.com/cropped/2024_03_06_706aca6df357b4c9a255g-1.jpg?height=164&width=373&top_left_y=448&top_left_x=2197"
        ],
        "answer": [
            "1.5"
        ],
        "solution": "From Snell's law, $\\sin (\\alpha-\\gamma)=$ $n \\sin (\\alpha / 2)$ so $n=\\sin (\\beta+\\gamma) / \\sin (\\gamma+\\beta / 2) \\approx$ 1.5 .",
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1190",
        "problem": "Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \\approx N \\approx A / 2$, where $A$ is the total number of nucleons $(A \\gg 1)$.\n\n## Atomic nucleus as closely packed system of nucleons\n\nIn a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \\mathrm{fm}\\left(1 \\mathrm{fm}=10^{-15} \\mathrm{~m}\\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\\frac{4}{3} r_{N}^{3} \\pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter.\n\n[figure1]\n\nCalculate what would be the packing factor $f$ if nucleons were arranged in a \"simple cubic\" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)].\n\nImportant: In all subsequent tasks, assume that the actual packing factor for nuclei is equal to the one from Task 1a. If you are not able to calculate it, in subsequent tasks use $f=1 / 2$.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nAlthough atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \\approx N \\approx A / 2$, where $A$ is the total number of nucleons $(A \\gg 1)$.\n\n## Atomic nucleus as closely packed system of nucleons\n\nIn a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \\mathrm{fm}\\left(1 \\mathrm{fm}=10^{-15} \\mathrm{~m}\\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\\frac{4}{3} r_{N}^{3} \\pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter.\n\n[figure1]\n\nCalculate what would be the packing factor $f$ if nucleons were arranged in a \"simple cubic\" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)].\n\nImportant: In all subsequent tasks, assume that the actual packing factor for nuclei is equal to the one from Task 1a. If you are not able to calculate it, in subsequent tasks use $f=1 / 2$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_656c8e5e21a256bf8ee1g-1.jpg?height=580&width=1125&top_left_y=1429&top_left_x=471"
        ],
        "answer": [
            "0.52"
        ],
        "solution": "In the SC-system, in each of 8 corners of a given cube there is one unit (atom, nucleon, etc.), but it is shared by 8 neighboring cubes - this gives a total of one nucleon per cube. If nucleons are touching, as we assume in our simplified model, then $a=2 r_{N}$ is the cube edge length $a$. The volume of one nucleon is then\n\n$V_{N}=\\frac{4}{3} r_{N}^{3} \\pi=\\frac{4}{3}\\left(\\frac{a}{2}\\right)^{3} \\pi=\\frac{4 a^{3}}{3 \\cdot 8} \\pi=\\frac{\\pi}{6} a^{3}$\n\nfrom which we obtain\n\n$f=\\frac{V_{N}}{a^{3}}=\\frac{\\pi}{6} \\approx 0.52$",
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1570",
        "problem": "图 a 是基于全内反射原理制备的折射率阶跃型光纤及其耦合光路示意图. 光纤内芯直径 $50 \\mu \\mathrm{m}$,折射率 $\\mathrm{n}_{1}=1.46$; 它的玻璃外包层的外径为 $125 \\mu \\mathrm{m}$, 折射率 $\\mathrm{n}_{2}=1.45$. 氦氛激光器输出一圆柱形平行光束, 为了将该激光束有效地耦合进入光纤传输, 可以在光纤前端放置一微球透镜进行聚焦和耦合, 微球透镜的直径 $\\mathrm{D}=3.00 \\mathrm{~mm}$, 折射率 $\\mathrm{n}=1.50$. 已知激光束中心轴通过微球透镜中心, 且与光纤对称轴重合. 空气折射率 $\\mathrm{n}_{0}=1.00$.\n\n[图1]\n\n图 a. 激光束经微球透镜耦合进入光纤示意图为了使进入光纤的全部光束能在光纤内长距离传输, 平行入射激光束的直径最大不能超过多少?",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个数值。\n这里是一些可能会帮助你解决问题的先验信息提示：\n图 a 是基于全内反射原理制备的折射率阶跃型光纤及其耦合光路示意图. 光纤内芯直径 $50 \\mu \\mathrm{m}$,折射率 $\\mathrm{n}_{1}=1.46$; 它的玻璃外包层的外径为 $125 \\mu \\mathrm{m}$, 折射率 $\\mathrm{n}_{2}=1.45$. 氦氛激光器输出一圆柱形平行光束, 为了将该激光束有效地耦合进入光纤传输, 可以在光纤前端放置一微球透镜进行聚焦和耦合, 微球透镜的直径 $\\mathrm{D}=3.00 \\mathrm{~mm}$, 折射率 $\\mathrm{n}=1.50$. 已知激光束中心轴通过微球透镜中心, 且与光纤对称轴重合. 空气折射率 $\\mathrm{n}_{0}=1.00$.\n\n[图1]\n\n图 a. 激光束经微球透镜耦合进入光纤示意图\n\n问题:\n为了使进入光纤的全部光束能在光纤内长距离传输, 平行入射激光束的直径最大不能超过多少?\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n请记住，你的答案应以$\\mathrm{~mm}$为单位计算，但在给出最终答案时，请不要包含单位。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是不包含任何单位的数值。",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_0207e542b9f7c87dc04dg-05.jpg?height=411&width=1077&top_left_y=271&top_left_x=472"
        ],
        "answer": [
            "$0.76$"
        ],
        "solution": "微球透镜的等效主点与球心重合，而等效焦距为\n\n$f_{e}=\\frac{\\left(2 n_{0}-n\\right) D}{4\\left(n-n_{0}\\right)}+\\frac{D}{2}=\\frac{n D}{4\\left(n-n_{0}\\right)}=\\frac{1.50 \\times 3.00 \\mathrm{~mm}}{4(1.50-1.00)}=2.25 \\mathrm{~mm}$\n\n(10) 4 分\n\n为了使进入光纤的全部光束能在光纤内长距离传输, 平行入射激光束的直径 $\\mathrm{d}$ 最大时, 光束在光纤端面上的入射角应满足最大入射角条件：\n\n$\\frac{\\frac{d}{2}}{\\sqrt{\\left(\\frac{d}{2}\\right)^{2}+\\left(\\frac{n D}{4\\left(n-n_{0}\\right)}\\right)^{2}}}=\\frac{1}{\\sqrt{1+\\left(\\frac{n D}{2 d\\left(n-n_{0}\\right)}\\right)^{2}}}=\\sin \\theta$\n\n(11) 4 分\n\n注意到\n\n$\\frac{1}{\\sqrt{1+\\left(\\frac{n D}{2 d\\left(n-n_{0}\\right)}\\right)^{2}}}=\\frac{2\\left(n-n_{0}\\right)}{n} \\frac{d}{D} \\frac{1}{\\sqrt{1+\\left(\\frac{2\\left(n-n_{0}\\right)}{n} \\frac{d}{D}\\right)^{2}}}$\n\n$\\approx \\frac{2\\left(n-n_{0}\\right)}{n} \\frac{d}{D}\\left[1-\\frac{1}{2}\\left(\\frac{2\\left(n-n_{0}\\right)}{n} \\frac{d}{D}\\right)^{2}\\right]=\\frac{2\\left(n-n_{0}\\right)}{n} \\frac{d}{D}$\n\n上式已略去 $\\left(\\frac{d}{D}\\right)^{3}$ 项, 将上式结果代入 (11) 式, 结合 (3) 式和题给数据得\n$d=\\frac{n}{2\\left(n-n_{0}\\right)} D \\sin \\theta=0.76 \\mathrm{~mm}$\n\n平行入射激光束的直径最大不能超过 $0.76 \\mathrm{~mm}$.",
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mm}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_490",
        "problem": "Consider the electron in the first part, except now take into account radiation. Assume that the orbit remains circular and the orbital radius $r$ changes by an amount $|\\Delta r| \\ll r$.\n\nConsider a circular orbit for the electron where $r<R$. Determine the change in the orbital radius $\\Delta r$ during one orbit in terms of any or all of $r, R, Q, e$, and any necessary fundamental constants. Be very specific about the sign of $\\Delta r$.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nConsider the electron in the first part, except now take into account radiation. Assume that the orbit remains circular and the orbital radius $r$ changes by an amount $|\\Delta r| \\ll r$.\n\nConsider a circular orbit for the electron where $r<R$. Determine the change in the orbital radius $\\Delta r$ during one orbit in terms of any or all of $r, R, Q, e$, and any necessary fundamental constants. Be very specific about the sign of $\\Delta r$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": [
            "$-\\frac{2 \\pi}{3}\\left(\\frac{e^{2}}{4 \\pi \\epsilon_{0} R m c^{2}}\\right) \\sqrt{\\frac{e Q}{4 \\pi \\epsilon_{0} R m c^{2}}} r$"
        ],
        "solution": "The energy radiated away is given by\n\n$$\n\\Delta E=-P T\n$$\n\nwhere $T$ is determined in the previous sections.\n\nIt is possible to compute the actual energy of each orbit, and it is fairly trivial to do for regions $r>R$, but perhaps there is an easier, more entertaining way. Consider\n\n$$\n\\Delta E=\\Delta K+\\Delta U\n$$\n\nand for small changes in $r$,\n\n$$\n\\frac{\\Delta U}{\\Delta r} \\approx-F=\\frac{e Q}{4 \\pi \\epsilon_{0}} \\frac{r}{R^{3}}\n$$\n\nThis implies the potential energy increases with increasing $r$, as expected. Now\n\n$$\n\\frac{\\Delta K}{\\Delta r} \\approx \\frac{d}{d r}\\left(\\frac{1}{2} m v^{2}\\right)=\\frac{1}{2} \\frac{d}{d r}\\left|r \\frac{m v^{2}}{r}\\right|\n$$\n\nbut $m v^{2} / r=F$, so\n\n$$\n\\frac{\\Delta K}{\\Delta r} \\approx \\frac{1}{2} \\frac{d}{d r}|r F|=\\frac{e Q}{4 \\pi \\epsilon_{0}} \\frac{r}{R^{3}}\n$$\n\nThis implies the kinetic energy increases with increasing $r$, also as expected, as this region acts like a multidimensional simple harmonic oscillator. Combining,\n\n$$\n\\frac{\\Delta E}{\\Delta r} \\approx 2 \\frac{e Q}{4 \\pi \\epsilon_{0}} \\frac{r}{R^{3}}=2 m a\n$$\n\nFinally,\n\n$$\n\\Delta r=-\\left(\\frac{1}{6 \\pi} \\frac{a^{2}}{c^{3} \\epsilon_{0}} e^{2}\\right)\\left(2 \\pi \\sqrt{\\frac{4 \\pi \\epsilon_{0} m R^{3}}{e Q}}\\right)\\left(\\frac{1}{2 m a}\\right)\n$$\n\nPlugging in the value of $a$, this can be simplified to\n\n$$\n\\Delta r=-\\frac{1}{6} \\sqrt{\\frac{e^{5} Q}{4 \\pi \\epsilon_{0}^{3} R\\left(m c^{2}\\right)^{3}}} \\frac{r}{R}\n$$\n\nAlternatively, we can write the result in terms of dimensionless groups,\n\n$$\n\\Delta r=-\\frac{2 \\pi}{3}\\left(\\frac{e^{2}}{4 \\pi \\epsilon_{0} R m c^{2}}\\right) \\sqrt{\\frac{e Q}{4 \\pi \\epsilon_{0} R m c^{2}}} r\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_1596",
        "problem": "激光瞄准系统的设计需考虑空气折射率的变化。由于受到地表状况、海拔高度、气温、湿度和空气密度等多种因素的影响, 空气的折射率在大气层中的分布是不均匀的,因而激光的传播路径并不是直线。为简化起见, 假设某地的空气折射率随高度 $y$ 的变化如下式所示\n\n$$\nn^{2}=n_{0}^{2}+\\alpha^{2} y,\n$$\n\n式中 $n_{0}$ 是 $y=0$ 处 (地面) 空气的折射率, $n_{0}$\n\n[图1]\n和 $\\alpha$ 均为大于零的已知常量。激光本身的传播时间可忽略。激光发射器位于坐标原点 O，如图。激光毁伤目标需要一定的照射时间。若目标 $\\mathrm{A}$ 处在激光发射器的可攻击范围内, 其初始位置为 $\\left(x_{0}, y_{0}\\right)$, 该目标在同一高度上以匀速度 $v$ 接近激光发射器。为了使激光能始终照射该目标，激光出射角 $\\theta_{0}$ 应如何随时间 $t$ 而变化?",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个方程。\n这里是一些可能会帮助你解决问题的先验信息提示：\n激光瞄准系统的设计需考虑空气折射率的变化。由于受到地表状况、海拔高度、气温、湿度和空气密度等多种因素的影响, 空气的折射率在大气层中的分布是不均匀的,因而激光的传播路径并不是直线。为简化起见, 假设某地的空气折射率随高度 $y$ 的变化如下式所示\n\n$$\nn^{2}=n_{0}^{2}+\\alpha^{2} y,\n$$\n\n式中 $n_{0}$ 是 $y=0$ 处 (地面) 空气的折射率, $n_{0}$\n\n[图1]\n和 $\\alpha$ 均为大于零的已知常量。激光本身的传播时间可忽略。激光发射器位于坐标原点 O，如图。\n\n问题:\n激光毁伤目标需要一定的照射时间。若目标 $\\mathrm{A}$ 处在激光发射器的可攻击范围内, 其初始位置为 $\\left(x_{0}, y_{0}\\right)$, 该目标在同一高度上以匀速度 $v$ 接近激光发射器。为了使激光能始终照射该目标，激光出射角 $\\theta_{0}$ 应如何随时间 $t$ 而变化?\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个方程，例如ANSWER=\\frac{x^2}{4}+\\frac{y^2}{2}=1",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_07aa406e17d01fd01b36g-05.jpg?height=634&width=737&top_left_y=1105&top_left_x=1019"
        ],
        "answer": [
            "$$ \\theta_{0}=\\arctan \\left(\\frac{2 n_{0}^{2}\\left(x_{0}-v t\\right)}{4 n_{0}^{2} y_{0}-\\alpha^{2}\\left(x_{0}-v t\\right)^{2}}+\\sqrt{\\left(\\frac{2 n_{0}^{2}\\left(x_{0}-v t\\right)}{4 n_{0}^{2} y_{0}-\\alpha^{2}\\left(x_{0}-v t\\right)^{2}}\\right)^{2}+\\frac{\\alpha^{2}\\left(x_{0}-v t\\right)^{2}}{4 n_{0}^{2} y_{0}-\\alpha^{2}\\left(x_{0}-v t\\right)^{2}}}\\right) $$"
        ],
        "solution": "将目标 $\\mathrm{A}$ 的坐标 $\\left(x_{a}, y_{a}\\right)$ 代入光线轨迹方程(9)式得\n\n$$\ny_{a}=\\frac{\\alpha^{2}}{4 n_{0}^{2} \\sin ^{2} \\theta_{0}} x_{a}^{2}+\\frac{x_{a}}{\\tan \\theta_{0}}\n$$\n\n将三角函数关系 $\\sin ^{2} \\theta_{0}=\\frac{\\tan ^{2} \\theta_{0}}{1+\\tan ^{2} \\theta_{0}}$ 代入上式可得\n\n$$\ny_{a}=\\frac{\\alpha^{2} x_{a}^{2}}{4 n_{0}^{2} \\tan ^{2} \\theta_{0}}\\left(1+\\tan ^{2} \\theta_{0}\\right)+\\frac{x_{a}}{\\tan \\theta_{0}}\n$$\n\n进一步整理上式可得关于 $\\tan \\theta_{0}$ 的一元二次方程\n\n$$\n\\left(4 n_{0}^{2} y_{a}-\\alpha^{2} x_{a}^{2}\\right) \\tan ^{2} \\theta_{0}-4 n_{0}^{2} x_{a} \\tan \\theta_{0}-\\alpha^{2} x_{a}^{2}=0\n$$\n\n解得\n\n$$\n\\tan \\theta_{0}=\\frac{2 n_{0}^{2} x_{a}}{4 n_{0}^{2} y_{a}-\\alpha^{2} x_{a}^{2}} \\pm \\sqrt{\\left(\\frac{2 n_{0}^{2} x_{a}}{4 n_{0}^{2} y_{a}-\\alpha^{2} x_{a}^{2}}\\right)^{2}+\\frac{\\alpha^{2} x_{a}^{2}}{4 n_{0}^{2} y_{a}-\\alpha^{2} x_{a}^{2}}}\n$$\n\n因为目标 A 在第 I 象限且处于可攻击范围内, 所以由（2）中的结论可知\n\n$$\n0 \\leq x_{a} \\leq \\frac{2 n_{0}}{\\alpha} \\sqrt{y_{a}},\n$$\n\n同时 $0 \\leq \\theta_{0} \\leq \\frac{\\pi}{2}$ 。故\n\n$$\n\\tan \\theta_{0} \\geq 0\n$$\n\n于是，上述解的右端根号前应取正号，因而\n\n$$\n\\tan \\theta_{0}=\\frac{2 n_{0}^{2} x_{a}}{4 n_{0}^{2} y_{a}-\\alpha^{2} x_{a}^{2}}+\\sqrt{\\left(\\frac{2 n_{0}^{2} x_{a}}{4 n_{0}^{2} y_{a}-\\alpha^{2} x_{a}^{2}}\\right)^{2}+\\frac{\\alpha^{2} x_{a}^{2}}{4 n_{0}^{2} y_{a}-\\alpha^{2} x_{a}^{2}}}\n$$\n\n由题设可知目标 $\\mathrm{A}$ 的运动轨迹为\n\n$$\nx_{a}=x_{0}-v t, \\quad y_{a}=y_{0}\n$$\n\n将(15)式代入(14)式, 可得激光出射角度 $\\theta_{0}$ 随时间 $t$ 的变化规律为\n\n$$\n\\theta_{0}=\\arctan \\left(\\frac{2 n_{0}^{2}\\left(x_{0}-v t\\right)}{4 n_{0}^{2} y_{0}-\\alpha^{2}\\left(x_{0}-v t\\right)^{2}}+\\sqrt{\\left(\\frac{2 n_{0}^{2}\\left(x_{0}-v t\\right)}{4 n_{0}^{2} y_{0}-\\alpha^{2}\\left(x_{0}-v t\\right)^{2}}\\right)^{2}+\\frac{\\alpha^{2}\\left(x_{0}-v t\\right)^{2}}{4 n_{0}^{2} y_{0}-\\alpha^{2}\\left(x_{0}-v t\\right)^{2}}}\\right)\n$$",
        "answer_type": "EQ",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_685",
        "problem": "A car travels with constant speed on a circular road on level ground as shown below. $\\mathrm{F}_{\\text {air }}$ is the force of air resistance on the car. Which of the other force is the horizontal force of the road on the car's tires?\n\n[figure1]\nA: $F_{a}$\nB: $F_{b}$\nC: $F_{c}$\nD: $F_{d}$\nE: $F_{e}$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA car travels with constant speed on a circular road on level ground as shown below. $\\mathrm{F}_{\\text {air }}$ is the force of air resistance on the car. Which of the other force is the horizontal force of the road on the car's tires?\n\n[figure1]\n\nA: $F_{a}$\nB: $F_{b}$\nC: $F_{c}$\nD: $F_{d}$\nE: $F_{e}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_22e26a14ee6fdd9254b6g-06.jpg?height=309&width=629&top_left_y=1212&top_left_x=1171"
        ],
        "answer": [
            "B"
        ],
        "solution": "$\\mathrm{F}_{a}=$ Centripetal force $\\mathrm{F}_{c}=$ Frictional force of the road, equals to force exerted by tires exert in the backward direction so that the car moves in the forward direction (Newton's third law Action-Reaction). The horizontal force on the car's tires, $\\mathrm{F}_{a}+\\mathrm{F}_{c}=\\mathrm{F}_{b}$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_418",
        "problem": "The resultant amplitude of the waves from the slits arriving at the screen at a maxima (a bright fringe), where the incoming waves arrive in phase, is 2A.This question relates to ways in which light energy may be concentrated in an interference pattern.\n\nFigure 1 shows wave fronts at normal incidence on a Young's double slit arrangement, illuminating them so that they both radiate in phase, each with an amplitude $A$ at the screen. One such slit alone will cause an intensity of illumination of $I$, where $I \\propto A^{2}$, in the central region of the screen where the Young's fringes pattern forms.\n\n[figure1]\nFigure: Young's double slits.\n\nWhat is the intensity at the maxima?",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\nHere is some context information for this question, which might assist you in solving it:\nThe resultant amplitude of the waves from the slits arriving at the screen at a maxima (a bright fringe), where the incoming waves arrive in phase, is 2A.\n\nproblem:\nThis question relates to ways in which light energy may be concentrated in an interference pattern.\n\nFigure 1 shows wave fronts at normal incidence on a Young's double slit arrangement, illuminating them so that they both radiate in phase, each with an amplitude $A$ at the screen. One such slit alone will cause an intensity of illumination of $I$, where $I \\propto A^{2}$, in the central region of the screen where the Young's fringes pattern forms.\n\n[figure1]\nFigure: Young's double slits.\n\nWhat is the intensity at the maxima?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_8ea586ad37c37c010606g-3.jpg?height=522&width=859&top_left_y=584&top_left_x=610"
        ],
        "answer": [
            "$4 I$"
        ],
        "solution": "$4 I$ as $I \\propto$ (amplitude) $^{2}$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_838",
        "problem": "[figure1]\n\nFig. 1 A monopole $q_{\\mathrm{m}}$ appears at a distance $h$ from a conducting thin film of thickness $d$. The origin of the coordinates is located on the upper surface.\n\nWe first focus on the initial response of the conducting thin film when at time $t=0$ a north monopole $q_{\\mathrm{m}}$ appears suddenly at the position $\\vec{r}_{\\mathrm{mp}}=h \\hat{z}(h>0)$, as is shown in Fig. 1 . The monopole remains stationary in all later times $(t>0)$.\n\nOur interest here is the initial total magnetic field $\\vec{B}(\\vec{\\rho}, z)$ in regions $z \\geq 0$ and $z \\leq-d$, and the induced electric current density in the thin film. The total magnetic field $\\vec{B}=\\vec{B}_{\\mathrm{mp}}+\\vec{B}^{\\prime}$, where magnetic fields $\\vec{B}_{\\mathrm{mp}}$ and $\\vec{B}^{\\prime}$ are, respectively, due to the monopole and the induced current in the thin film. The initial $\\vec{B}(\\vec{\\rho}, z)$ we refer to is at the time $t_{0}$, which falls within the interval $h / c \\leq t_{0} \\ll \\tau_{\\mathrm{c}}$. Here $\\tau_{\\mathrm{c}}$ is a time constant characterizing the subsequent response of the thin film, and $c$ is the speed of light in vacuum. In this problem, we take the limit $h / c \\rightarrow 0$ and hence let $t_{0}=0$.\n\nThe calculation of the initial total magnetic field $\\vec{B}(\\vec{\\rho}, z)$ (at $t_{0}=0$ ) is facilitated by introducing an image monopole. For $\\vec{B}(\\vec{\\rho}, z)$ in the region $z \\geq 0$, the image monopole has a magnetic charge $q_{\\mathrm{m}}$ and is located at $z=-h$. On the other hand, for $\\vec{B}(\\vec{\\rho}, z)$ in the region $z \\leq-d$, the image monopole has a magnetic charge $-q_{\\mathrm{m}}$ and is located at $z=h$.\n\nObtain the initial total magnetic field $\\vec{B}(\\vec{\\rho}, z)$ in $z \\leq-d$ at $t_{0}=0.2 \\mathrm{pt}$",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\n[figure1]\n\nFig. 1 A monopole $q_{\\mathrm{m}}$ appears at a distance $h$ from a conducting thin film of thickness $d$. The origin of the coordinates is located on the upper surface.\n\nWe first focus on the initial response of the conducting thin film when at time $t=0$ a north monopole $q_{\\mathrm{m}}$ appears suddenly at the position $\\vec{r}_{\\mathrm{mp}}=h \\hat{z}(h>0)$, as is shown in Fig. 1 . The monopole remains stationary in all later times $(t>0)$.\n\nOur interest here is the initial total magnetic field $\\vec{B}(\\vec{\\rho}, z)$ in regions $z \\geq 0$ and $z \\leq-d$, and the induced electric current density in the thin film. The total magnetic field $\\vec{B}=\\vec{B}_{\\mathrm{mp}}+\\vec{B}^{\\prime}$, where magnetic fields $\\vec{B}_{\\mathrm{mp}}$ and $\\vec{B}^{\\prime}$ are, respectively, due to the monopole and the induced current in the thin film. The initial $\\vec{B}(\\vec{\\rho}, z)$ we refer to is at the time $t_{0}$, which falls within the interval $h / c \\leq t_{0} \\ll \\tau_{\\mathrm{c}}$. Here $\\tau_{\\mathrm{c}}$ is a time constant characterizing the subsequent response of the thin film, and $c$ is the speed of light in vacuum. In this problem, we take the limit $h / c \\rightarrow 0$ and hence let $t_{0}=0$.\n\nThe calculation of the initial total magnetic field $\\vec{B}(\\vec{\\rho}, z)$ (at $t_{0}=0$ ) is facilitated by introducing an image monopole. For $\\vec{B}(\\vec{\\rho}, z)$ in the region $z \\geq 0$, the image monopole has a magnetic charge $q_{\\mathrm{m}}$ and is located at $z=-h$. On the other hand, for $\\vec{B}(\\vec{\\rho}, z)$ in the region $z \\leq-d$, the image monopole has a magnetic charge $-q_{\\mathrm{m}}$ and is located at $z=h$.\n\nObtain the initial total magnetic field $\\vec{B}(\\vec{\\rho}, z)$ in $z \\leq-d$ at $t_{0}=0.2 \\mathrm{pt}$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d32b3b2f89cebe6f1c2ag-2.jpg?height=642&width=1244&top_left_y=296&top_left_x=194"
        ],
        "answer": [
            "0"
        ],
        "solution": "In the $z \\leq-d$ region, the magnetic field $\\vec{B}=\\vec{B}^{\\prime}+\\vec{B}_{\\mathrm{mp}}$ at $t=t_{0}=0$ is given by\n\n$$\n\\vec{B}=0\n$$",
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_4",
        "problem": "If a copper wire is stretched to make it $0.1 \\%$ longer, what is the percentage change in its resistance?\nA: $0.01 \\%$\nB: $0.1 \\%$\nC: $0.2 \\%$\nD: $0.4 \\%$\nE: $0.5 \\%$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf a copper wire is stretched to make it $0.1 \\%$ longer, what is the percentage change in its resistance?\n\nA: $0.01 \\%$\nB: $0.1 \\%$\nC: $0.2 \\%$\nD: $0.4 \\%$\nE: $0.5 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "C"
        ],
        "solution": "$\\quad R=\\frac{\\rho l}{A} ; \\frac{\\Delta R}{R}=2 \\frac{\\Delta l}{l}$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_458",
        "problem": "Part (i)\n\nBy symmetry, the fields above and below the sheet are equal in magnitude and directed away from the sheet. By Gauss's Law, using a cylinder of base area $A$,\n\n$$\n2 E A=\\frac{\\sigma A}{\\epsilon_{0}} \\Rightarrow E=\\frac{\\sigma}{2 \\epsilon_{0}}\n$$\n\npointing directly away from the sheet in the $z$ direction, or\n\n$$\n\\mathbf{E}=\\frac{\\sigma}{2 \\epsilon} \\times \\begin{cases}\\hat{\\mathbf{z}} & \\text { above the sheet } \\\\ -\\hat{\\mathbf{z}} & \\text { below the sheet. }\\end{cases}\n$$In this problem assume that velocities $v$ are much less than the speed of light $c$, and therefore ignore relativistic contraction of lengths or time dilation.\n\nAn infinite uniform sheet has a surface charge density $\\sigma$ and has an infinitesimal thickness. The sheet lies in the $x y$ plane.\n\nAssuming the sheet is moving with velocity $\\tilde{\\mathbf{v}}=v \\hat{\\mathbf{x}}$ (parallel to the sheet), determine the electric field $\\tilde{\\mathbf{E}}$ (magnitude and direction) above and below the sheet.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\nHere is some context information for this question, which might assist you in solving it:\nPart (i)\n\nBy symmetry, the fields above and below the sheet are equal in magnitude and directed away from the sheet. By Gauss's Law, using a cylinder of base area $A$,\n\n$$\n2 E A=\\frac{\\sigma A}{\\epsilon_{0}} \\Rightarrow E=\\frac{\\sigma}{2 \\epsilon_{0}}\n$$\n\npointing directly away from the sheet in the $z$ direction, or\n\n$$\n\\mathbf{E}=\\frac{\\sigma}{2 \\epsilon} \\times \\begin{cases}\\hat{\\mathbf{z}} & \\text { above the sheet } \\\\ -\\hat{\\mathbf{z}} & \\text { below the sheet. }\\end{cases}\n$$\n\nproblem:\nIn this problem assume that velocities $v$ are much less than the speed of light $c$, and therefore ignore relativistic contraction of lengths or time dilation.\n\nAn infinite uniform sheet has a surface charge density $\\sigma$ and has an infinitesimal thickness. The sheet lies in the $x y$ plane.\n\nAssuming the sheet is moving with velocity $\\tilde{\\mathbf{v}}=v \\hat{\\mathbf{x}}$ (parallel to the sheet), determine the electric field $\\tilde{\\mathbf{E}}$ (magnitude and direction) above and below the sheet.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [the electric field $\\tilde{\\mathbf{E}}$ above the sheet, the electric field $\\tilde{\\mathbf{E}}$ below the sheet].\nTheir answer types are, in order, [expression, expression].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": [
            "$\\frac{\\sigma}{2 \\epsilon} \\times \\hat{\\mathbf{z}}$",
            "$\\frac{\\sigma}{2 \\epsilon} \\times -\\hat{\\mathbf{z}}$"
        ],
        "solution": "The motion does not affect the electric field, so the answer is the same as that of part (i).",
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "the electric field $\\tilde{\\mathbf{E}}$ above the sheet",
            "the electric field $\\tilde{\\mathbf{E}}$ below the sheet"
        ],
        "type_sequence": [
            "EX",
            "EX"
        ],
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_707",
        "problem": "A ball of mass $m=200 \\mathrm{~g}$ is hung in the middle of horizontal elastic cord of length $L=1.5 \\mathrm{~m}$, attached to the ceiling, as shown in figure below.\n\nThe elastic constant of the cord is $k=300 \\mathrm{~N} / \\mathrm{m}$.\n\n[figure1]\n\nCalculate the period of very small oscillations from the equilibrium point.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nA ball of mass $m=200 \\mathrm{~g}$ is hung in the middle of horizontal elastic cord of length $L=1.5 \\mathrm{~m}$, attached to the ceiling, as shown in figure below.\n\nThe elastic constant of the cord is $k=300 \\mathrm{~N} / \\mathrm{m}$.\n\n[figure1]\n\nCalculate the period of very small oscillations from the equilibrium point.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of s, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_e69eaad2d8db15283465g-09.jpg?height=263&width=1657&top_left_y=855&top_left_x=234"
        ],
        "answer": [
            "0.408"
        ],
        "solution": "\\begin{aligned}\nm \\ddot{y} & =-m g+2 K\\left(2 \\sqrt{(d-y)^{2}+w^{2}}-2 w\\right)\\left(\\frac{d-y}{w}\\right) \\\\\n& =-m g+2 K\\left(2 \\sqrt{d^{2}+w^{2}-2 y d}-2 w\\right)\\left(\\frac{d-y}{w}\\right) \\\\\n& =-m g+2 K\\left(2 \\sqrt{d^{2}+w^{2}}\\left(1-\\frac{y d}{d^{2}+w^{2}}\\right)-2 w\\right)\\left(\\frac{d-y}{w}\\right) \\\\\n& =-m g+2 K\\left(2 \\sqrt{d^{2}+w^{2}}-2 w\\right)-2 \\frac{y d}{\\sqrt{d^{2}+w^{2}}}\\left(\\frac{d}{w}-\\frac{y}{w}\\right)\n\\end{aligned}\n\nWe defined, $2 K\\left(2 \\sqrt{d^{2}+w^{2}}-2 w\\right)\\left(\\frac{d}{w}\\right)=m g$, so,\n\n$m \\ddot{y}=\\frac{-4 K y d}{\\sqrt{d^{2}+w^{2}}} \\frac{d}{w}-2 K\\left(2 \\sqrt{d^{2}+w^{2}}-2 w\\right) \\frac{y}{w}$\n\n$2 K\\left(2 \\sqrt{d^{2}+w^{2}}-2 w\\right) \\frac{y}{w}=m g \\frac{y}{d}$\n\n$m \\ddot{y}=-4 k \\frac{d^{2}}{w \\sqrt{d^{2}+w^{2}}} y-m g d \\frac{y}{d}$\n\n$\\ddot{y}=-\\left(\\frac{4 k d^{2}}{m w \\sqrt{d^{2}+w^{2}}}+\\frac{g}{d}\\right) y$\n\nso, $T=\\frac{2 \\pi}{\\sqrt{\\frac{4 k d^{2}}{m w \\sqrt{d^{2}+w^{2}}}+\\frac{g}{d}}}=0.408 \\mathrm{~s}$",
        "answer_type": "NV",
        "unit": [
            "s"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_621",
        "problem": "This is a parallax effect, as shown in the below diagram:\n\n[figure1]\n\nThe angle Earth makes with MAR-Kappa is the angular shift it makes in the night sky.\n\nThen we have that $D=R_{E S} /(\\theta / 2)=1.9 \\times 10^{17} \\mathrm{~m}$.Scientists have recently detected a new star, the MAR-Kappa. The star is almost a perfect blackbody, and its measured light spectrum is shown below.\n\n[figure2]\n\nThe total measured light intensity from MAR-Kappa is $I=1.12 \\times 10^{-8} \\mathrm{~W} / \\mathrm{m}^{2}$. The mass of MAR-Kappa is estimated to be $3.5 \\times 10^{30} \\mathrm{~kg}$. It is stationary relative to the sun. You may find the Stefan-Boltzmann law useful, which states the power emitted by a blackbody with area $A$ is $\\sigma A T^{4}$.\n\nThe spectrum of wavelengths $\\lambda$ emitted from a blackbody only depends on $h, c, k_{B}, \\lambda$, and $T$. \n\nThe \"lines\" in the spectrum result from atoms in the star absorbing specific wavelengths of the emitted light. One contribution to the width of the spectral lines is the Doppler shift associated with the thermal motion of the atoms in the star. The spectral line at $\\lambda=389 \\mathrm{~nm}$ is due to helium. \n\nOver the course of a year, MAR-Kappa appears to oscillate between two positions in the background night sky, which are an angular distance of $1.6 \\times 10^{-6} \\mathrm{rad}$ apart. What is the radius of MAR-Kappa?\n\n Assume that MAR-Kappa lies in the same plane as the Earth's orbit, which is circular with radius $1.5 \\times 10^{11} \\mathrm{~m}$.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThis is a parallax effect, as shown in the below diagram:\n\n[figure1]\n\nThe angle Earth makes with MAR-Kappa is the angular shift it makes in the night sky.\n\nThen we have that $D=R_{E S} /(\\theta / 2)=1.9 \\times 10^{17} \\mathrm{~m}$.\n\nproblem:\nScientists have recently detected a new star, the MAR-Kappa. The star is almost a perfect blackbody, and its measured light spectrum is shown below.\n\n[figure2]\n\nThe total measured light intensity from MAR-Kappa is $I=1.12 \\times 10^{-8} \\mathrm{~W} / \\mathrm{m}^{2}$. The mass of MAR-Kappa is estimated to be $3.5 \\times 10^{30} \\mathrm{~kg}$. It is stationary relative to the sun. You may find the Stefan-Boltzmann law useful, which states the power emitted by a blackbody with area $A$ is $\\sigma A T^{4}$.\n\nThe spectrum of wavelengths $\\lambda$ emitted from a blackbody only depends on $h, c, k_{B}, \\lambda$, and $T$. \n\nThe \"lines\" in the spectrum result from atoms in the star absorbing specific wavelengths of the emitted light. One contribution to the width of the spectral lines is the Doppler shift associated with the thermal motion of the atoms in the star. The spectral line at $\\lambda=389 \\mathrm{~nm}$ is due to helium. \n\nOver the course of a year, MAR-Kappa appears to oscillate between two positions in the background night sky, which are an angular distance of $1.6 \\times 10^{-6} \\mathrm{rad}$ apart. What is the radius of MAR-Kappa?\n\n Assume that MAR-Kappa lies in the same plane as the Earth's orbit, which is circular with radius $1.5 \\times 10^{11} \\mathrm{~m}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~m}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_fed953f9e38b72bf8bd7g-18.jpg?height=152&width=1352&top_left_y=970&top_left_x=408",
            "https://cdn.mathpix.com/cropped/2024_03_06_fed953f9e38b72bf8bd7g-17.jpg?height=640&width=1051&top_left_y=507&top_left_x=537"
        ],
        "answer": [
            "$1.3 \\times 10^{9}$"
        ],
        "solution": "The luminosity (total power given off by the star) is given by\n\n$$\n4 \\pi D^{2} I=L=4 \\pi\\left(1.9 \\times 10^{17} \\mathrm{~m}\\right)^{2} \\cdot\\left(1.12 \\times 10^{-8} \\mathrm{~W} / \\mathrm{m}^{2}\\right)=5.1 \\times 10^{27} \\mathrm{~W}\n$$\n\nwith $D$ the distance from Earth to MAR-Kappa. The radius of the star is given by $\\sqrt{L /\\left(4 \\pi \\sigma T^{4}\\right)}=1.3 \\times 10^{9} \\mathrm{~m}$.",
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~m}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_533",
        "problem": "In this problem we consider a simplified model of the electromagnetic radiation inside a cubical box of side length $L$. In this model, the electric field has spatial dependence\n\n$$\nE(x, y, z)=E_{0} \\sin \\left(k_{x} x\\right) \\sin \\left(k_{y} y\\right) \\sin \\left(k_{z} z\\right)\n$$\n\nwhere one corner of the box lies at the origin and the box is aligned with the $x, y$, and $z$ axes. Let $h$ be Planck's constant, $k_{B}$ be Boltzmann's constant, and $c$ be the speed of light.\n\nIn the model, each permitted value of the triple $\\left(k_{x}, k_{y}, k_{z}\\right)$ corresponds to a quantum state. These states can be visualized in a state space, which is a notional three-dimensional space with axes corresponding to $k_{x}, k_{y}$, and $k_{z}$. How many states occupy a volume $s$ of state space, if $s$ is large enough that the discreteness of the states can be ignored?",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nIn this problem we consider a simplified model of the electromagnetic radiation inside a cubical box of side length $L$. In this model, the electric field has spatial dependence\n\n$$\nE(x, y, z)=E_{0} \\sin \\left(k_{x} x\\right) \\sin \\left(k_{y} y\\right) \\sin \\left(k_{z} z\\right)\n$$\n\nwhere one corner of the box lies at the origin and the box is aligned with the $x, y$, and $z$ axes. Let $h$ be Planck's constant, $k_{B}$ be Boltzmann's constant, and $c$ be the speed of light.\n\nIn the model, each permitted value of the triple $\\left(k_{x}, k_{y}, k_{z}\\right)$ corresponds to a quantum state. These states can be visualized in a state space, which is a notional three-dimensional space with axes corresponding to $k_{x}, k_{y}$, and $k_{z}$. How many states occupy a volume $s$ of state space, if $s$ is large enough that the discreteness of the states can be ignored?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": [
            "$\\frac{L^{3} s}{\\pi^{3}}$"
        ],
        "solution": "In the abstract state space, the states are spaced a distance $\\pi / L$ apart. Each can therefore be thought of as occupying volume $\\pi^{3} / L^{3}$, and the number of states in the volume $s$ is\n\n$$\nN=\\frac{L^{3} s}{\\pi^{3}}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_1556",
        "problem": "具有一定能量、动量的光子还具有角动量。圆偏振光的光子的角动量大小为 $\\hbar$ 。光子被物体吸收后, 光子的能量、动量和角动量就全部传给物体。物体吸收光子获得的角动量可以使物体转动。科学家利用这一原理, 在连续的圆偏振激光照射下, 实现了纳米颗粒的高速转动, 获得了迄今为止液体环境中转速最高的微尺度转子。\n\n如图, 一金纳米球颗粒放置在两片水平光滑玻璃平板之间，并整体（包括玻璃平板）浸在水中。一束圆偏振激光从上往下照射到金纳米颗粒上。已知该束入射激光在真空中的波长 $\\lambda=830 \\mathrm{~nm}$, 经显微物镜聚焦后（仍假设为平面波, 每个光子具有沿传播方向的角动量 $\\hbar$ ) 光斑直径 $d=1.20 \\times 10^{-6} \\mathrm{~m}$, 功率 $P=20.0 \\mathrm{~mW}$ 。金纳米球颗粒的半径 $R=100 \\mathrm{~nm}$, 金的密度 $\\rho=19.32 \\times 10^{3} \\mathrm{~kg} / \\mathrm{m}^{3}$ 。忽略光在介质\n\n圆偏振激光\n[图1]\n界面上的反射以及玻璃、水对光的吸收等损失, 仅从金纳米颗粒吸收光子获得角动量驱动其转动的角度分析下列问题（计算结果取 3 位有效数字）：求该束激光的频率 $V$ 和光强 $I$ （在传播方向上单位横截面积所传输的功率）。",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题包含多个待求解的量。\n这里是一些可能会帮助你解决问题的先验信息提示：\n具有一定能量、动量的光子还具有角动量。圆偏振光的光子的角动量大小为 $\\hbar$ 。光子被物体吸收后, 光子的能量、动量和角动量就全部传给物体。物体吸收光子获得的角动量可以使物体转动。科学家利用这一原理, 在连续的圆偏振激光照射下, 实现了纳米颗粒的高速转动, 获得了迄今为止液体环境中转速最高的微尺度转子。\n\n如图, 一金纳米球颗粒放置在两片水平光滑玻璃平板之间，并整体（包括玻璃平板）浸在水中。一束圆偏振激光从上往下照射到金纳米颗粒上。已知该束入射激光在真空中的波长 $\\lambda=830 \\mathrm{~nm}$, 经显微物镜聚焦后（仍假设为平面波, 每个光子具有沿传播方向的角动量 $\\hbar$ ) 光斑直径 $d=1.20 \\times 10^{-6} \\mathrm{~m}$, 功率 $P=20.0 \\mathrm{~mW}$ 。金纳米球颗粒的半径 $R=100 \\mathrm{~nm}$, 金的密度 $\\rho=19.32 \\times 10^{3} \\mathrm{~kg} / \\mathrm{m}^{3}$ 。忽略光在介质\n\n圆偏振激光\n[图1]\n界面上的反射以及玻璃、水对光的吸收等损失, 仅从金纳米颗粒吸收光子获得角动量驱动其转动的角度分析下列问题（计算结果取 3 位有效数字）：\n\n问题:\n求该束激光的频率 $V$ 和光强 $I$ （在传播方向上单位横截面积所传输的功率）。\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你的最终解答的量应该按以下顺序输出：[该束激光的频率 $V$, 该束激光的光强 $I$]\n它们的单位依次是[$\\mathrm{~Hz}$, $\\mathrm{~W} / \\mathrm{m}^{2}$]，但在你给出最终答案时不应包含单位。\n它们的答案类型依次是[数值, 数值]\n你需要在输出的最后用以下格式总结答案：“最终答案是\\boxed{ANSWER}”，其中ANSWER应为你的最终答案序列，用英文逗号分隔，例如：5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_bc7c55a7ed04447daac3g-02.jpg?height=450&width=450&top_left_y=1883&top_left_x=1292"
        ],
        "answer": [
            "$3.61 \\times 10^{14}$",
            "$1.77 \\times 10^{10}$"
        ],
        "solution": "该束激光的频率\n\n$$\nv=\\frac{c}{\\lambda}=3.61 \\times 10^{14} \\mathrm{~Hz}\n$$\n\n入射激光的强度\n\n$$\nI=\\frac{P}{\\pi\\left(\\frac{d}{2}\\right)^{2}}=1.77 \\times 10^{10} \\mathrm{~W} / \\mathrm{m}^{2}\n$$",
        "answer_type": "MPV",
        "unit": [
            "$\\mathrm{~Hz}$",
            "$\\mathrm{~W} / \\mathrm{m}^{2}$"
        ],
        "answer_sequence": [
            "该束激光的频率 $V$",
            "该束激光的光强 $I$"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_338",
        "problem": "[figure1]\n\nA circuit is made out of a battery, a switch, res istors and capacitors as shown in the image. The resistors all have a resistance of $R$, the capacitor all have a capacitance of $C$ and the battery has a voltage of $U$. The point $\\mathrm{A}$ is connected to the ground and so it has a potential of $0 \\mathrm{~V}$. In the be ginning the switch is open and all the capacitor have no charge.\n\nWhat is the potential at points B and $\\mathrm{C}$ after we have closed the switch and waited fo all the potentials to stabilize?",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\n[figure1]\n\nA circuit is made out of a battery, a switch, res istors and capacitors as shown in the image. The resistors all have a resistance of $R$, the capacitor all have a capacitance of $C$ and the battery has a voltage of $U$. The point $\\mathrm{A}$ is connected to the ground and so it has a potential of $0 \\mathrm{~V}$. In the be ginning the switch is open and all the capacitor have no charge.\n\nWhat is the potential at points B and $\\mathrm{C}$ after we have closed the switch and waited fo all the potentials to stabilize?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of U, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_e6c1582ce7f1c05fa0a6g-1.jpg?height=657&width=642&top_left_y=257&top_left_x=794",
            "https://cdn.mathpix.com/cropped/2024_03_14_83cc90b265723e4db811g-2.jpg?height=307&width=666&top_left_y=708&top_left_x=775"
        ],
        "answer": [
            "\\frac{11}{13}"
        ],
        "solution": "Once the potentials have stabilized, there is no current through any capacitor, therefore we can then analyze the resistor network by effectively cutting away all the capacitors. We get the following equivalent circuit\n\n[figure2]\n\nWe can analyze it using the laws of serie and parallel connections. The resistance of the three resistors between the points $\\mathrm{X}$ and $\\mathrm{C}$ i $R_{X C}=\\frac{1}{R^{-1}+(2 R)^{-1}}=\\frac{2}{3} R$. The resistance of the network between $\\mathrm{X}$ and $\\mathrm{B}$ is then $R_{X B}=$ $\\frac{1}{\\left(R_{X C}+R\\right)^{-1}+R^{-1}}=\\frac{5}{8} R$. Therefore the potentials $\\varphi_{B}=\\frac{R}{R_{X B}+R} U=\\frac{8}{13} U$ and $\\varphi_{C}=\\varphi_{B}+$ $\\frac{R}{R_{X C}+R}\\left(U-\\varphi_{B}\\right)=\\frac{11}{13} U$.",
        "answer_type": "NV",
        "unit": [
            "U"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_495",
        "problem": "A uniform solid spherical ball starts from rest on a loop-the-loop track. It rolls without slipping along the track. However, it does not have enough speed to make it to the top of the loop. From what height $h$ would the ball need to start in order to land at point $\\mathrm{P}$ directly underneath the top of the loop? Express your answer in terms of $R$, the radius of the loop. Assume that the radius of the ball is very small compared to the radius of the loop, and that there are no energy losses due to friction.\n\n[figure1]",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nA uniform solid spherical ball starts from rest on a loop-the-loop track. It rolls without slipping along the track. However, it does not have enough speed to make it to the top of the loop. From what height $h$ would the ball need to start in order to land at point $\\mathrm{P}$ directly underneath the top of the loop? Express your answer in terms of $R$, the radius of the loop. Assume that the radius of the ball is very small compared to the radius of the loop, and that there are no energy losses due to friction.\n\n[figure1]\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_33ea9fa74c8b34628eedg-14.jpg?height=602&width=879&top_left_y=710&top_left_x=623"
        ],
        "answer": [
            "$\\frac{37}{20} R$"
        ],
        "solution": "We fix the origin at $P$. Assume the ball leaves at an angle $\\theta$ away from the vertical. At this point, the $x$ and $y$ coordinates are\n\n$$\nx=R \\sin \\theta, \\quad y=R(1+\\cos \\theta)\n$$\n\nBy energy conservation, we have\n\n$$\nm g(h-y)=\\frac{1}{2} m v^{2}+\\frac{1}{2} I \\omega^{2}=\\frac{1}{2} m(1+\\beta) v^{2}\n$$\n\nwhere $\\beta=2 / 5$, and we used the fact that the ball rolls without slipping.\n\nLet $v$ be the speed of the ball when it leaves the loop. Then its velocity components at that moment are\n\n$$\nv_{x}=-v \\cos \\theta, \\quad v_{y}=v \\sin \\theta .\n$$\n\nAssuming the ball impacts $P$ at time $t$,\n\n$$\ny=\\frac{1}{2} g t^{2}-v_{y} t, \\quad x=-v_{x} t\n$$\n\nThe second equation yields\n\n$$\nt=\\frac{R}{v} \\frac{\\sin \\theta}{\\cos \\theta}\n$$\nand plugging this into the first equation gives\n\n$$\nR+R \\cos \\theta=\\frac{1}{2} g\\left(\\frac{R}{v} \\frac{\\sin \\theta}{\\cos \\theta}\\right)^{2}-v \\sin \\theta \\frac{R}{v} \\frac{\\sin \\theta}{\\cos \\theta}\n$$\n\nwhich simplifies to\n\n$$\n1+\\cos \\theta=\\frac{g R}{2 v^{2}} \\frac{\\sin ^{2} \\theta}{\\cos \\theta}\n$$\n\nNow, the ball leaves the surface when the normal component of the force of the loop on the ball just drops to zero. This happens when\n\n$$\nm g \\cos \\theta=m \\frac{v^{2}}{R} \\Rightarrow \\frac{v^{2}}{g R}=\\cos \\theta\n$$\n\nand plugging this into the previous equation gives\n\n$$\n1+\\cos \\theta=\\frac{1}{2} \\frac{1-\\cos ^{2} \\theta}{\\cos ^{2} \\theta} \\Rightarrow 2 \\cos ^{2} \\theta=1-\\cos \\theta\n$$\n\nThis is a quadratic equation with solutions\n\n$$\n\\cos \\theta=\\frac{-1 \\pm \\sqrt{1+8}}{4}=-\\frac{1}{4} \\pm \\frac{3}{4}\n$$\n\nOnly the positive answer of $\\cos \\theta=1 / 2$ is relevant here, though the negative answer is still physical!\n\nNow that we know $\\theta$, getting the final answer is straightforward. We combine the energy conservation equation and the condition\n\n$$\nm g \\cos \\theta=m \\frac{v^{2}}{R}\n$$\n\nto find\n\n$$\nh=1+\\left(\\frac{1}{2}(1+\\beta)+1\\right) R \\cos \\theta=\\frac{37}{20} R\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_81",
        "problem": "On December 21, 1968, NASA launched Apollo 8. This was the first manned mission to the moon, and after 68 hours, went into orbit around the moon but never landed on the surface. That was reserved for Apollo 11. During one five-minute portion of the launch, Apollo 8 increased its speed from 7,600 to 10,800 meters per second. What was the acceleration of Apollo 8 during this time period?\nA: $8.92 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}$\nB: $53.3 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}$\nC: $64.0 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}$\nD: $10.7 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}$\nE: $18.7 \\frac{m}{s^{2}}$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOn December 21, 1968, NASA launched Apollo 8. This was the first manned mission to the moon, and after 68 hours, went into orbit around the moon but never landed on the surface. That was reserved for Apollo 11. During one five-minute portion of the launch, Apollo 8 increased its speed from 7,600 to 10,800 meters per second. What was the acceleration of Apollo 8 during this time period?\n\nA: $8.92 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}$\nB: $53.3 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}$\nC: $64.0 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}$\nD: $10.7 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}$\nE: $18.7 \\frac{m}{s^{2}}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "D"
        ],
        "solution": "$v=v_{0}+a t$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_430",
        "problem": "This question is about stopping distances for motor vehicles.\n\nMany of the students participating in this Physics Challenge in the UK will also be learning to drive. So, the extract from the Highway Code reproduced below in Fig. 1 may well be familiar to those who have worked for the theory test.\n\nWe will explore the physics of this table together with some associated phenomena.\n\n[figure1]\n\nFigure 1: A table of stopping distances taken from the Highway Code.\n\nCalculate the deceleration of a car when it brakes from a speed of $80 \\mathrm{~km} \\mathrm{~h}^{-1}$.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about stopping distances for motor vehicles.\n\nMany of the students participating in this Physics Challenge in the UK will also be learning to drive. So, the extract from the Highway Code reproduced below in Fig. 1 may well be familiar to those who have worked for the theory test.\n\nWe will explore the physics of this table together with some associated phenomena.\n\n[figure1]\n\nFigure 1: A table of stopping distances taken from the Highway Code.\n\nCalculate the deceleration of a car when it brakes from a speed of $80 \\mathrm{~km} \\mathrm{~h}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $$\\mathrm{~m} \\mathrm{~s}^{-2}$$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_22bb4cfada5ea54fde10g-2.jpg?height=428&width=1579&top_left_y=2102&top_left_x=244"
        ],
        "answer": [
            "6.5"
        ],
        "solution": "With $v^{2}=0$,\n\n$$\n\\begin{aligned}\ndeceleration &=\\frac{u^{2}}{2 s} \\\\\n& =(80 \\mathrm{~km} / \\mathrm{h})^{2} /[2 \\times 38 \\mathrm{~m}] \\\\\n& =(80000)^{2} /\\left[2 \\times(3600)^{2} \\times 38\\right] \\quad \\text { unit conversion } \\\\\n& =6.5 \\mathrm{~m} \\mathrm{~s}^{-2}\n\\end{aligned}\n$$",
        "answer_type": "NV",
        "unit": [
            "$$\\mathrm{~m} \\mathrm{~s}^{-2}$$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_467",
        "problem": "Two large parallel plates of area $A$ are placed at $x=0$ and $x=d \\ll \\sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium.\n\nFor large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that\n\n$$\nv=\\mu E\n$$\n\nwhere $E$ is the local electric field and $\\mu$ is the charge mobility.\n\nIn the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\\rho(x)$. Use charge conservation to find a relationship between $\\rho(x), v(x)$, and their derivatives.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an equation.\n\nproblem:\nTwo large parallel plates of area $A$ are placed at $x=0$ and $x=d \\ll \\sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium.\n\nFor large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that\n\n$$\nv=\\mu E\n$$\n\nwhere $E$ is the local electric field and $\\mu$ is the charge mobility.\n\nIn the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\\rho(x)$. Use charge conservation to find a relationship between $\\rho(x), v(x)$, and their derivatives.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an equation, e.g. ANSWER=\\frac{x^2}{4}+\\frac{y^2}{2}=1",
        "figure_urls": null,
        "answer": [
            "$v \\frac{\\mathrm{d} \\rho}{\\mathrm{d} x}+\\rho \\frac{\\mathrm{d} v}{\\mathrm{~d} x}=0$"
        ],
        "solution": "In the steady state, the current is the same everywhere. Consider the region $(x, x+d x)$. The time it takes for the charge in the second region to leave is $\\frac{\\mathrm{d} x}{v(x)}$. The amount of charge that leaves is $\\rho A \\mathrm{~d} x$. The current is thus given by $\\rho A v$, so $\\rho v$ is constant. Alternatively, one can write this as\n\n$$\nv \\frac{\\mathrm{d} \\rho}{\\mathrm{d} x}+\\rho \\frac{\\mathrm{d} v}{\\mathrm{~d} x}=0\n$$\n\nBoth forms were accepted.",
        "answer_type": "EQ",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_907",
        "problem": "Calculation in the previous part shows that in order to build the space elevator, it is neccessary to have light materials with very high tensile strength. Carbon nanotubes are materials that meet such requirements because of strong chemical bondings between very light atoms. Two natural polymorphs of carbon are diamond and graphite. In diamond every carbon atom is surrounded by four nearest neighbor (NN) atoms to form a tetrahedron. Graphite has a layer structure. In each layer, carbon atoms are arranged in a hexagonal plane lattice with three NNs. Although diamond is known as the hardest materials, covalent bondings between carbon atoms in hexagonal layers of graphite is stronger than those between carbon atoms in diamond tetrahedra. Graphite is much softer than diamond because of the van der Waals bonding between carbon atoms of different layers, which is much weaker than covalent bonding.\n\n[figure1]\n\nFigure 2. Graphite structure\n(a)\n[figure2]\n\nFigure 3. Graphene (a) and carbon nanotube (b).\n\nA monatomic layer in graphite is called graphene and has monoatomic thickness. Isolated graphene sheet is not stable and has a tendency to roll up to form carbon spheres or carbon nanotubes. The hexagonal crystal lattice of graphene is depicted in Fig. 4. The distance between two NN carbon atoms is $a=0.142 \\mathrm{~nm}$ and the distance between two closest parallel bondings is $b=0.246 \\mathrm{~nm}$. Because the covalent bondings between carbon atoms in graphene are very strong, mechanical properties of carbon nanotubes are very special. They have an extremely large Young's modulus and tensile strength, as well as a very light density. Young's modulus is defined as the ratio of the stress along an axis to the strain (ratio of deformation over initial length) along that axis in the range of stress in which Hooke's law holds.\n\n[figure3]\n\nFigure 4. Graphene.\n\n## Theory\n\n[figure4]\n\nFigure 5. An illustration of a carbon nanotube with 9 carbon-carbon parallel bondings. Note: In this problem, there are 27 carbon-carbon parallel bondings. (1) parallel bond; (2) slanted bond; (3) tube axis.\n\nNow we examine some mechanical properties of a carbon nanotube having 27 carbon-carbon bondings parallel to the tube axis (for an illustration, see Figure 5). The bonding between two carbon atoms can be described by the Morse potential $V(x)=V_{0}\\left(e^{-4 \\frac{x}{a}}-2 e^{-2 \\frac{x}{a}}\\right)$. Here $a=0.142 \\mathrm{~nm}$ is the equilibrium distance between two $\\mathrm{NN}$ carbon atoms, $V_{0}=4.93 \\mathrm{eV}$ is the bonding energy, and $x$ is the displacement of the atom from the equilibrium position. Hereafter, we approximate the Morse potential by a quadratic potential $V(x)=P+Q x^{2}$. All non-nearest-neighbor interactions are neglected. In this approximation, one can propose that carbon atoms are bonded through \"springs\" with the spring constant $k$. Changes in angles between bonds are neglected.\n\nFind coefficients $P$ and $Q$ in term of $a$ and $V_{0}$",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an equation.\n\nproblem:\nCalculation in the previous part shows that in order to build the space elevator, it is neccessary to have light materials with very high tensile strength. Carbon nanotubes are materials that meet such requirements because of strong chemical bondings between very light atoms. Two natural polymorphs of carbon are diamond and graphite. In diamond every carbon atom is surrounded by four nearest neighbor (NN) atoms to form a tetrahedron. Graphite has a layer structure. In each layer, carbon atoms are arranged in a hexagonal plane lattice with three NNs. Although diamond is known as the hardest materials, covalent bondings between carbon atoms in hexagonal layers of graphite is stronger than those between carbon atoms in diamond tetrahedra. Graphite is much softer than diamond because of the van der Waals bonding between carbon atoms of different layers, which is much weaker than covalent bonding.\n\n[figure1]\n\nFigure 2. Graphite structure\n(a)\n[figure2]\n\nFigure 3. Graphene (a) and carbon nanotube (b).\n\nA monatomic layer in graphite is called graphene and has monoatomic thickness. Isolated graphene sheet is not stable and has a tendency to roll up to form carbon spheres or carbon nanotubes. The hexagonal crystal lattice of graphene is depicted in Fig. 4. The distance between two NN carbon atoms is $a=0.142 \\mathrm{~nm}$ and the distance between two closest parallel bondings is $b=0.246 \\mathrm{~nm}$. Because the covalent bondings between carbon atoms in graphene are very strong, mechanical properties of carbon nanotubes are very special. They have an extremely large Young's modulus and tensile strength, as well as a very light density. Young's modulus is defined as the ratio of the stress along an axis to the strain (ratio of deformation over initial length) along that axis in the range of stress in which Hooke's law holds.\n\n[figure3]\n\nFigure 4. Graphene.\n\n## Theory\n\n[figure4]\n\nFigure 5. An illustration of a carbon nanotube with 9 carbon-carbon parallel bondings. Note: In this problem, there are 27 carbon-carbon parallel bondings. (1) parallel bond; (2) slanted bond; (3) tube axis.\n\nNow we examine some mechanical properties of a carbon nanotube having 27 carbon-carbon bondings parallel to the tube axis (for an illustration, see Figure 5). The bonding between two carbon atoms can be described by the Morse potential $V(x)=V_{0}\\left(e^{-4 \\frac{x}{a}}-2 e^{-2 \\frac{x}{a}}\\right)$. Here $a=0.142 \\mathrm{~nm}$ is the equilibrium distance between two $\\mathrm{NN}$ carbon atoms, $V_{0}=4.93 \\mathrm{eV}$ is the bonding energy, and $x$ is the displacement of the atom from the equilibrium position. Hereafter, we approximate the Morse potential by a quadratic potential $V(x)=P+Q x^{2}$. All non-nearest-neighbor interactions are neglected. In this approximation, one can propose that carbon atoms are bonded through \"springs\" with the spring constant $k$. Changes in angles between bonds are neglected.\n\nFind coefficients $P$ and $Q$ in term of $a$ and $V_{0}$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an equation, e.g. ANSWER=\\frac{x^2}{4}+\\frac{y^2}{2}=1",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9fac1528de308fa347bbg-2.jpg?height=549&width=525&top_left_y=1850&top_left_x=731",
            "https://cdn.mathpix.com/cropped/2024_03_14_9fac1528de308fa347bbg-3.jpg?height=430&width=900&top_left_y=710&top_left_x=518",
            "https://cdn.mathpix.com/cropped/2024_03_14_9fac1528de308fa347bbg-3.jpg?height=451&width=691&top_left_y=1856&top_left_x=660",
            "https://cdn.mathpix.com/cropped/2024_03_14_9fac1528de308fa347bbg-4.jpg?height=944&width=1144&top_left_y=172&top_left_x=184"
        ],
        "answer": [
            "$Q=\\frac{4 V_{0}}{a^{2}}$"
        ],
        "solution": "Expand exponential function in series, and limit to the lowest power of $x$, one has $V=V_{0}\\left(-1+\\frac{4 x^{2}}{a^{2}}\\right)$ and gets $P=-V_{0}$ and  $Q=\\frac{4 V_{0}}{a^{2}}$.",
        "answer_type": "EQ",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1395",
        "problem": "把沿 $x$ 方向通有电流 ( $x$ 方向的电场强度为 $E_{x}$ ) 的长方体形的半导体材料, 放在沿 $z$ 方向匀强磁场中, 磁感应强度大小为 $B_{z}$ 。在垂直于电场和磁场的 $+y$ 或 $-y$ 方向将产生一个横向电场 $E_{y}$, 这个现象称为霍尔效应, 由霍尔效应产生的电场称为霍尔电场。实验表明霍尔电场 $E_{y}$与电流的电流密度 $J_{x}$ 和\n\n磁感应强度 $B_{Z}$ 的乘积成正比, 即 $E_{y}=R_{H} J_{x} B_{Z}$ 比例系数 $R_{H}$ 称为霍尔系数。\n\n某半导体材料样品中有两种载流子: 空穴和电子; 空穴和电子在单位电场下的平均速度 (即载流子的平均速度与电场成正比的比例系数) 分别为 $\\mu_{\\mathrm{p}}$ 和 $-\\mu_{\\mathrm{n}}$, 空穴和电子的数密度分别为 $p$和 $n$, 电荷分别为 $e$ 和 $-e$ 。试确定该半导体材料的霍尔系数。",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n\n问题:\n把沿 $x$ 方向通有电流 ( $x$ 方向的电场强度为 $E_{x}$ ) 的长方体形的半导体材料, 放在沿 $z$ 方向匀强磁场中, 磁感应强度大小为 $B_{z}$ 。在垂直于电场和磁场的 $+y$ 或 $-y$ 方向将产生一个横向电场 $E_{y}$, 这个现象称为霍尔效应, 由霍尔效应产生的电场称为霍尔电场。实验表明霍尔电场 $E_{y}$与电流的电流密度 $J_{x}$ 和\n\n磁感应强度 $B_{Z}$ 的乘积成正比, 即 $E_{y}=R_{H} J_{x} B_{Z}$ 比例系数 $R_{H}$ 称为霍尔系数。\n\n某半导体材料样品中有两种载流子: 空穴和电子; 空穴和电子在单位电场下的平均速度 (即载流子的平均速度与电场成正比的比例系数) 分别为 $\\mu_{\\mathrm{p}}$ 和 $-\\mu_{\\mathrm{n}}$, 空穴和电子的数密度分别为 $p$和 $n$, 电荷分别为 $e$ 和 $-e$ 。试确定该半导体材料的霍尔系数。\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d716ce15f03757bb482eg-10.jpg?height=251&width=394&top_left_y=740&top_left_x=1348"
        ],
        "answer": [
            "$\\frac{1}{e} \\frac{p \\mu_{\\mathrm{p}}^{2}-n \\mu_{\\mathrm{n}}^{2}}{\\left(p \\mu_{\\mathrm{p}}+n \\mu_{\\mathrm{n}}\\right)^{2}}$"
        ],
        "solution": "为确定起见, 取坐标系如图所示, 磁场沿 $\\mathrm{z}$ 方向, 通电电流 $J_{x}$ 沿 $\\mathrm{x}$ 方向。设半导体材料中的载流子空穴和电子沿 $x$ 方向的平均速率分别为 $v_{\\mathrm{px}}$ 和 $v_{\\mathrm{nx}}$, 沿 $x$ 方向的电流密度为\n\n$$\nJ_{\\mathrm{x}}=e p\\left(v_{\\mathrm{p}}\\right)_{\\mathrm{x}}+(-e) n\\left(-v_{\\mathrm{n}}\\right)_{\\mathrm{x}}\n$$\n\n[图1]\n\n式中\n\n$$\n\\begin{gathered}\n\\left(v_{p}\\right)_{x}=\\mu_{p} E_{x} \\\\\n\\left(-v_{n}\\right)_{x}=-\\mu_{n} E_{x}\n\\end{gathered}\n$$\n\n如果沿 $x$ 方向的电流中只有一种载流子, 则当作用于载流子的洛仑兹力与霍尔电场的作用力平衡时, 霍尔电场达到稳定, 如金属导体。在半导体中, 存在两种载流子, 同一磁场对两种载流子受到的洛仑兹力方向相同, 所受到的霍尔电场力方向相反, 两种载流子受到的洛仑兹力不可能同时与霍尔电场力平衡, 所以在半导体样品存在载流子的横向流动, 当任何时刻流向样品同一侧面的空穴数与电子数相等时, 霍尔电场便达到稳定。设两种载流子的横向平均速率分别为 $\\left(v_{\\mathrm{p}}\\right)_{y}$ 和 $\\left(v_{\\mathrm{n}}\\right)_{y}$ 则横向电流密度为\n\n$$\nJ_{\\mathrm{y}}=e p\\left(-v_{\\mathrm{p}}\\right)_{\\mathrm{y}}+(-e) n\\left(-v_{\\mathrm{n}}\\right)_{\\mathrm{y}}\n$$\n\n这时, 空穴在横向受到的作用力的大小为\n\n$$\nF_{\\mathrm{py}}=e\\left[E_{y}-\\left(v_{\\mathrm{p}}\\right)_{x} B_{z}\\right]\n$$\n\n电子在横向受到的作用力的大小为\n\n$$\nF_{\\mathrm{ny}}=(-e)\\left[E_{y}-\\left(-v_{\\mathrm{n}}\\right)_{x} B_{z}\\right]\n$$\n\n故两种载流子的横向平均速度为\n\n$$\n\\begin{aligned}\n& \\left(-v_{\\mathrm{p}}\\right)_{\\mathrm{y}}=\\mu_{\\mathrm{p}}\\left[E_{\\mathrm{y}}-\\left(v_{\\mathrm{p}}\\right)_{\\mathrm{x}} B_{\\mathrm{z}}\\right] \\\\\n& \\left(-v_{\\mathrm{n}}\\right)_{\\mathrm{y}}=-\\mu_{\\mathrm{p}}\\left[E_{\\mathrm{y}}+\\left(v_{\\mathrm{n}}\\right)_{\\mathrm{x}} B_{\\mathrm{z}}\\right]\n\\end{aligned}\n$$\n\n霍尔电场达到稳定时有\n\n$$\nJ_{y}=0\n$$\n\n由(4)(7)(8)及)及(2)(3)式得\n\n$$\nE_{\\mathrm{y}}=\\frac{p \\mu_{\\mathrm{p}}^{2}-n \\mu_{\\mathrm{n}}^{2}}{\\left(p \\mu_{\\mathrm{p}}+n \\mu_{\\mathrm{n}}\\right)} E_{\\mathrm{x}} B_{\\mathrm{z}}\n$$\n\n根据霍尔系数的定义以及(4)(10)两式得\n\n$$\nR_{\\mathrm{H}}=\\frac{1}{e} \\frac{p \\mu_{\\mathrm{p}}^{2}-n \\mu_{\\mathrm{n}}^{2}}{\\left(p \\mu_{\\mathrm{p}}+n \\mu_{\\mathrm{n}}\\right)^{2}}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Physics_965",
        "problem": "The terminal voltage of a dc power supply is measured as $5.00 \\mathrm{~V}$ when it is on open circuit. A $2.00 \\Omega$ resistor is connected across the terminals and the voltage drops by $0.100 \\mathrm{~V}$.\n\nIf the supply is treated as a simple emf and internal resistance, what would be the value of the internal resistance?",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe terminal voltage of a dc power supply is measured as $5.00 \\mathrm{~V}$ when it is on open circuit. A $2.00 \\Omega$ resistor is connected across the terminals and the voltage drops by $0.100 \\mathrm{~V}$.\n\nIf the supply is treated as a simple emf and internal resistance, what would be the value of the internal resistance?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of \\Omega, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_38b9374d4ba914cc7d40g-4.jpg?height=285&width=486&top_left_y=1077&top_left_x=271"
        ],
        "answer": [
            "0.041"
        ],
        "solution": "[figure1]\n\nIdentify potentials\n\n$$\nI=\\frac{4.9}{2}=\\frac{0.1}{r} \\rightarrow r=\\frac{0.2}{4.9}=\\frac{2}{49} \\Omega=0.041 \\Omega\n$$",
        "answer_type": "NV",
        "unit": [
            "\\Omega"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_700",
        "problem": "he temperature of a gas is increased by the same amount in two difference processes.\n\nProcess A: The volume is kept constant.\n\nProcess B: The pressure is kept constant.\n\nWhich of the following is correct?\nA: The heat required in process $A$ is greater.\nB: The heat required in process $B$ is greater.\nC: The same heat is required in both cases.\nD: The change in entropy is the same for both processes.\nE: The change in internal energy is the same for both processes.\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nhe temperature of a gas is increased by the same amount in two difference processes.\n\nProcess A: The volume is kept constant.\n\nProcess B: The pressure is kept constant.\n\nWhich of the following is correct?\n\nA: The heat required in process $A$ is greater.\nB: The heat required in process $B$ is greater.\nC: The same heat is required in both cases.\nD: The change in entropy is the same for both processes.\nE: The change in internal energy is the same for both processes.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "B"
        ],
        "solution": "$$\n\\Delta Q=\\Delta U+\\Delta W\n$$\n\nWhen the volume is constant, no work is done by the gas. So the amount of heat needed is less.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_37",
        "problem": "An electron ( $\\left.e^{-}\\right)$is moving and has kinetic energy of $1.00 \\mathrm{MeV}$. It makes a head-on collision with a positron $\\left(e^{+}\\right)$that is at rest. In the collision the two particles annihilate each other and are replaced by two photons $(\\gamma)$ of equal energy, each traveling at angles $\\theta$ to the electron's original direction of motion. The reaction is $\\mathrm{e}^{-}+\\mathrm{e}^{+} \\rightarrow 2 \\gamma$. Determine the angle of emission $\\theta$ of each photon.\nA: $15.3^{0}$\nB: $25.3^{\\circ}$\nC: $35.3^{0}$\nD: $45.3^{\\circ}$\nE: $55.3^{\\circ}$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn electron ( $\\left.e^{-}\\right)$is moving and has kinetic energy of $1.00 \\mathrm{MeV}$. It makes a head-on collision with a positron $\\left(e^{+}\\right)$that is at rest. In the collision the two particles annihilate each other and are replaced by two photons $(\\gamma)$ of equal energy, each traveling at angles $\\theta$ to the electron's original direction of motion. The reaction is $\\mathrm{e}^{-}+\\mathrm{e}^{+} \\rightarrow 2 \\gamma$. Determine the angle of emission $\\theta$ of each photon.\n\nA: $15.3^{0}$\nB: $25.3^{\\circ}$\nC: $35.3^{0}$\nD: $45.3^{\\circ}$\nE: $55.3^{\\circ}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "D"
        ],
        "solution": "The moving electron has momentum, $p$, along the positive $x$-axis and kinetic energy, $K$. The total energy, $E$, of the electron and the stationary positron before the collision is $\\mathrm{E}=\\mathrm{K}+2 \\mathrm{mc}^{2}=2.022 \\mathrm{MeV}$. The two photons emerge from the collision each with energy $\\mathrm{E}_{\\gamma}=1 / 2 \\mathrm{E}=1.011 \\mathrm{MeV}$ as given by conservation of energy, and each with magnitude of momentum $\\mathrm{p}_{\\gamma}=\\mathrm{E}_{\\gamma} / \\mathrm{c}=1.011 \\mathrm{MeV} / \\mathrm{c}$. The momentum vectors of the photons make angles $\\pm \\theta$ with the $\\mathrm{x}$-axis. Conservation of momentum in the $\\mathrm{x}$-direction is $\\mathrm{p}=2 \\mathrm{p} \\gamma \\cos \\theta$ from which $\\theta=45.3^{\\circ}$.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_877",
        "problem": "In the jets from AGN, we have populations of highly energetic electrons in regions with strong magnetic fields. This creates the conditions for the emission of high fluxes of synchrotron radiation. The electrons are often so highly energetic, that they can be described as ultra relativistic with $\\gamma \\gg 1$.\n \nAs the electron is accelerated due to the magnetic field it emits electromagnetic radiation. In a frame at which the electron is momentarily at rest, there is no preferred direction for the emission of the radiation. Half is emitted in the forward direction, and half in the backward direction. However, in the frame of the observer, for an electron moving at an ultra relativistic speed, with $\\gamma \\gg 1$, the radiation is concentrated in a forward cone with $\\theta \\lesssim 1 / \\gamma$ (so the total angle of cone is $2 / \\gamma$ ). As the electron is gyrating around the magnetic field, any observer will only see pulses of radiation as the forward cone sweeps through the line of sight.\n[figure1]\n\nFigure 3: The diagram on the left shows the distribution of power in radiation from an electron accelerating up the page in the frame at which the electron in momentarily at rest. The diagram on the right shows the distribution of power in radiation for the same electron in the observer's frame, where most radiation is emitted in the forward cone. In the observers frame, the direction of the electron's acceleration is shown by a vector labelled $\\mathbf{a}$ and the direction of its velocity is shown by a vector labelled $\\mathbf{v}$.\n\n Find the duration of a pulse, $\\Delta t$, of synchrotron radiation observed from an electron with lorentz factor $\\gamma$, travelling at an angle $\\phi$ to the magnetic field.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an equation.\n\nproblem:\nIn the jets from AGN, we have populations of highly energetic electrons in regions with strong magnetic fields. This creates the conditions for the emission of high fluxes of synchrotron radiation. The electrons are often so highly energetic, that they can be described as ultra relativistic with $\\gamma \\gg 1$.\n \nAs the electron is accelerated due to the magnetic field it emits electromagnetic radiation. In a frame at which the electron is momentarily at rest, there is no preferred direction for the emission of the radiation. Half is emitted in the forward direction, and half in the backward direction. However, in the frame of the observer, for an electron moving at an ultra relativistic speed, with $\\gamma \\gg 1$, the radiation is concentrated in a forward cone with $\\theta \\lesssim 1 / \\gamma$ (so the total angle of cone is $2 / \\gamma$ ). As the electron is gyrating around the magnetic field, any observer will only see pulses of radiation as the forward cone sweeps through the line of sight.\n[figure1]\n\nFigure 3: The diagram on the left shows the distribution of power in radiation from an electron accelerating up the page in the frame at which the electron in momentarily at rest. The diagram on the right shows the distribution of power in radiation for the same electron in the observer's frame, where most radiation is emitted in the forward cone. In the observers frame, the direction of the electron's acceleration is shown by a vector labelled $\\mathbf{a}$ and the direction of its velocity is shown by a vector labelled $\\mathbf{v}$.\n\n Find the duration of a pulse, $\\Delta t$, of synchrotron radiation observed from an electron with lorentz factor $\\gamma$, travelling at an angle $\\phi$ to the magnetic field.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an equation, e.g. ANSWER=\\frac{x^2}{4}+\\frac{y^2}{2}=1",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_2416d49d47cb88c0a72bg-4.jpg?height=166&width=842&top_left_y=1502&top_left_x=618"
        ],
        "answer": [
            "$$ \\Delta t_{\\mathrm{a}}=\\frac{m_{e}}{\\gamma^{2} e B} . $$"
        ],
        "solution": "The observer only sees the synchrotron emission when they are within the forward light cone. As the electron is gyrating around the magnetic field, this direction is changing. The observer is in this light cone for time $\\Delta t=\\frac{2 \\theta}{\\Omega}=\\frac{2 m}{e B}$. However, the emitting electron is moving directly toward the observer over this time, so although the light emitted at the start of the pulse is ahead of the light at the end of the pulse, it is only ahead by $c \\Delta t\\left(1-\\frac{v}{c}\\right)$. The pulse then has an apparent duration of\n\n$$\n\\Delta t_{a}=\\Delta t\\left(1-\\frac{v}{c}\\right) .\n$$\n\nSince $\\left(1-\\frac{v}{c}\\right)\\left(1+\\frac{v}{c}\\right)=1-\\frac{v^{2}}{c^{2}}=\\frac{1}{\\gamma^{2}}$, we can write $\\left(1-\\frac{v}{c}\\right)=\\frac{1}{\\gamma^{2}\\left(1+\\frac{v}{c}\\right)}$. As the electrons are ultrarelativistic, $\\left(1+\\frac{v}{c}\\right)=2$, and\n\n$$\n\\Delta t_{\\mathrm{a}}=\\frac{m_{e}}{\\gamma^{2} e B} .\n$$",
        "answer_type": "EQ",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_735",
        "problem": "A very smart bird is sitting on a roof holding a shell in its beak. It wants to throw the shell so that it hits the ground with the highest speed possible. Assuming air resistance is negligible, and initial speed is constant in all scenarios $\\left(v_{A}=v_{B}=v_{c}\\right)$, how should the bird throw the shell so that its final speed before hitting the ground is greatest?\n\n[figure1]\nA: Throw it upwards $\\left(v_{A}\\right)$\nB: Throw it horizontally $\\left(v_{B}\\right)$\nC: Throw it downwards $\\left(v_{C}\\right)$\nD: $\\mathrm{A}$ and $\\mathrm{C}$ produce a faster final speed than $\\mathrm{B}$\nE: A, B, and C all produce the same final speed\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA very smart bird is sitting on a roof holding a shell in its beak. It wants to throw the shell so that it hits the ground with the highest speed possible. Assuming air resistance is negligible, and initial speed is constant in all scenarios $\\left(v_{A}=v_{B}=v_{c}\\right)$, how should the bird throw the shell so that its final speed before hitting the ground is greatest?\n\n[figure1]\n\nA: Throw it upwards $\\left(v_{A}\\right)$\nB: Throw it horizontally $\\left(v_{B}\\right)$\nC: Throw it downwards $\\left(v_{C}\\right)$\nD: $\\mathrm{A}$ and $\\mathrm{C}$ produce a faster final speed than $\\mathrm{B}$\nE: A, B, and C all produce the same final speed\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_a0d0ab960474d1b0609cg-03.jpg?height=398&width=678&top_left_y=156&top_left_x=1168"
        ],
        "answer": [
            "E"
        ],
        "solution": "Conservation of energy:\n\nSince the ball starts at the same height and moves at the same initial speed, we can use conservation of energy to find the speed of the shell when it hits the ground.\n\n$\\frac{1}{2} m v^{2}=\\frac{1}{2} m v_{0}^{2}+m g h$\n\n$\\therefore v=\\sqrt{v_{0}^{2}+2 g h}$ in all cases.\n\nAlternate solution:\n\n$v_{A}=v_{B}=v_{C}=v_{0}$\n\nThrow upwards: $v_{A, f i n}=\\sqrt{v_{0}^{2}+2 g h}$\n\nThrow downwards: $v_{C, f i n}=\\sqrt{\\left(-v_{0}\\right)^{2}+2 g h}=\\sqrt{v_{0}^{2}+2 g h}$\n\nThrow horizontally:\n\nInitial vertical velocity is 0 , constant horizontal velocity is $v_{B x}=v_{B}=v_{0}$\n\nFinal vertical: $v_{B y}=-\\sqrt{2 g h}$\n\nTotal: $v_{B, f i n}=\\sqrt{v_{B x}^{2}+v_{B y}^{2}}=\\sqrt{v_{0}^{2}+2 g h}$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_618",
        "problem": "Beloit College has a \"homemade\" $500 \\mathrm{kV}$ VanDeGraff proton accelerator, designed and constructed by the students and faculty.\n[figure1]\n\nAccelerator dome (assume it is a sphere); accelerating column; bending electromagnet\n\nThe accelerator dome, an aluminum sphere of radius $a=0.50$ meters, is charged by a rubber belt with width $w=10 \\mathrm{~cm}$ that moves with speed $v_{b}=20 \\mathrm{~m} / \\mathrm{s}$. The accelerating column consists of 20 metal rings separated by glass rings; the rings are connected in series with $500 \\mathrm{M} \\Omega$ resistors. The proton beam has a current of $25 \\mu \\mathrm{A}$ and is accelerated through $500 \\mathrm{kV}$ and then passes through a tuning electromagnet. The electromagnet consists of wound copper pipe as a conductor. The electromagnet effectively creates a uniform field $B$ inside a circular region of radius $b=10 \\mathrm{~cm}$ and zero outside that region.\n\n[figure2]\n\nOnly six of the 20 metals rings and resistors are shown in the figure. The fuzzy grey path is the path taken by the protons as they are accelerated from the dome, through the electromagnet, into the target.\n\nThe electromagnet is composed of layers of spiral wound copper pipe; the pipe has inner diameter $d_{i}=0.40 \\mathrm{~cm}$ and outer diameter $d_{o}=0.50 \\mathrm{~cm}$. The copper pipe is wound into this flat spiral that has an inner diameter $D_{i}=20 \\mathrm{~cm}$ and outer diameter $D_{o}=50 \\mathrm{~cm}$. Assuming the pipe almost touches in the spiral winding, determine the length $L$ in one spiral.\n\n[figure3]",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBeloit College has a \"homemade\" $500 \\mathrm{kV}$ VanDeGraff proton accelerator, designed and constructed by the students and faculty.\n[figure1]\n\nAccelerator dome (assume it is a sphere); accelerating column; bending electromagnet\n\nThe accelerator dome, an aluminum sphere of radius $a=0.50$ meters, is charged by a rubber belt with width $w=10 \\mathrm{~cm}$ that moves with speed $v_{b}=20 \\mathrm{~m} / \\mathrm{s}$. The accelerating column consists of 20 metal rings separated by glass rings; the rings are connected in series with $500 \\mathrm{M} \\Omega$ resistors. The proton beam has a current of $25 \\mu \\mathrm{A}$ and is accelerated through $500 \\mathrm{kV}$ and then passes through a tuning electromagnet. The electromagnet consists of wound copper pipe as a conductor. The electromagnet effectively creates a uniform field $B$ inside a circular region of radius $b=10 \\mathrm{~cm}$ and zero outside that region.\n\n[figure2]\n\nOnly six of the 20 metals rings and resistors are shown in the figure. The fuzzy grey path is the path taken by the protons as they are accelerated from the dome, through the electromagnet, into the target.\n\nThe electromagnet is composed of layers of spiral wound copper pipe; the pipe has inner diameter $d_{i}=0.40 \\mathrm{~cm}$ and outer diameter $d_{o}=0.50 \\mathrm{~cm}$. The copper pipe is wound into this flat spiral that has an inner diameter $D_{i}=20 \\mathrm{~cm}$ and outer diameter $D_{o}=50 \\mathrm{~cm}$. Assuming the pipe almost touches in the spiral winding, determine the length $L$ in one spiral.\n\n[figure3]\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~m}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_8c5c0b6efdeaae46cc88g-19.jpg?height=468&width=1592&top_left_y=438&top_left_x=259",
            "https://cdn.mathpix.com/cropped/2024_03_06_8c5c0b6efdeaae46cc88g-19.jpg?height=493&width=1268&top_left_y=1339&top_left_x=426",
            "https://cdn.mathpix.com/cropped/2024_03_06_8c5c0b6efdeaae46cc88g-22.jpg?height=377&width=293&top_left_y=221&top_left_x=965"
        ],
        "answer": [
            "33"
        ],
        "solution": "Treat the problem as two dimensional. The area of the spiral is\n\n$$\nA=\\frac{\\pi}{4}\\left(D_{o}^{2}-D_{i}^{2}\\right)\n$$\n\nThe area of the pipe is\n\n$$\nA=L d_{o}\n$$\n\nEquating and solving,\n\n$$\nL=\\frac{\\pi\\left(D_{o}^{2}-D_{i}^{2}\\right)}{4 d_{o}}=33 \\mathrm{~m}\n$$",
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~m}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1320",
        "problem": "如图, 半径为 $R$ 、质量为 $M$ 的半球静置于光滑水平桌面上, 在半球顶点上有一质量为 $m$ 、半径为 $r$ 的匀质小球。某时刻, 小球受到微小的扰动后由静止开始沿半球表面运动。在运动过程中, 小球相对于半球的位置由角位置 $\\theta$ 描述, $\\theta$ 为两球心的连线与坚直方向之间的夹角。已知小球绕其对称轴的转动惯量为 $\\frac{2}{5} m r^{2}$, 小球与半球之间的动摩擦因数为 $\\mu$, 假定最大静摩擦力等于滑动摩擦力。重力加速度大小为 $g$ 。\n\n[图1]当小球刚好运动到角位置 $\\theta_{3}$ 时脱离半球，求此时小球质心相对于半球运动速度的大小 $v_{m}\\left(\\theta_{3}\\right)$ 。",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n这里是一些可能会帮助你解决问题的先验信息提示：\n如图, 半径为 $R$ 、质量为 $M$ 的半球静置于光滑水平桌面上, 在半球顶点上有一质量为 $m$ 、半径为 $r$ 的匀质小球。某时刻, 小球受到微小的扰动后由静止开始沿半球表面运动。在运动过程中, 小球相对于半球的位置由角位置 $\\theta$ 描述, $\\theta$ 为两球心的连线与坚直方向之间的夹角。已知小球绕其对称轴的转动惯量为 $\\frac{2}{5} m r^{2}$, 小球与半球之间的动摩擦因数为 $\\mu$, 假定最大静摩擦力等于滑动摩擦力。重力加速度大小为 $g$ 。\n\n[图1]\n\n问题:\n当小球刚好运动到角位置 $\\theta_{3}$ 时脱离半球，求此时小球质心相对于半球运动速度的大小 $v_{m}\\left(\\theta_{3}\\right)$ 。\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_35bc41298eef336dfdafg-01.jpg?height=337&width=486&top_left_y=351&top_left_x=1276"
        ],
        "answer": [
            "$$\\sqrt{(R+r) g \\cos \\theta_{3}}$$"
        ],
        "solution": "在小球刚好运动到角位置 $\\theta_{3}$ 处脱离半球的瞬间,\n\n$$\nN=0\n$$\n\n此时半球的加速度为零。因此, 在小球脱离半球的瞬间, 小球质心相对于半球运动速度的大小 $v_{m}\\left(\\theta_{3}\\right)$ 满足\n\n$$\nm g \\cos \\theta_{3}=m \\frac{v_{\\mathrm{C}}^{\\prime 2}}{R+r}\n$$\n\n由此得\n\n$$\nv_{\\mathrm{C}}^{\\prime}=\\sqrt{(R+r) g \\cos \\theta_{3}}\n$$\n$$\n\\frac{2}{7} g \\sin \\theta_{2}-\\mu g \\cos \\theta_{2}+a_{M}\\left(\\theta_{2}\\right)\\left(\\frac{2}{7} \\cos \\theta_{2}+\\mu \\sin \\theta_{2}\\right)+\\frac{\\mu(M+m)^{2} V_{M}^{2}\\left(\\theta_{2}\\right)}{(R+r) m^{2} \\cos ^{2} \\theta_{2}}=0\n$$\n\n式中 $V_{M}\\left(\\theta_{2}\\right)$ 和 $a_{M}\\left(\\theta_{2}\\right)$ 如(4)(5)式 $\\left(\\theta_{1} \\rightarrow \\theta_{2}\\right)$ 所示。",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_351",
        "problem": "Consider two absolutely elastic dielec tric balls of radius $r$ and mass $m$ one of which carries isotropically distributed charge $-q$, and the other $-+q$. There is so strong homogen eous magnetic field $B$, parallel to the axis $z$, tha electrostatic interaction of the two charges can be neglected; neglect also gravity and friction forces. The first ball (negatively charged) moves with speed $v$ and collides with the second ball which had been resting at the origin. The collision is central, and immediately before the im pact, the velocity of the first ball was parallel to the $x$-axis.\n\nWhat is the speed of the second ball immediately after the collision?",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nConsider two absolutely elastic dielec tric balls of radius $r$ and mass $m$ one of which carries isotropically distributed charge $-q$, and the other $-+q$. There is so strong homogen eous magnetic field $B$, parallel to the axis $z$, tha electrostatic interaction of the two charges can be neglected; neglect also gravity and friction forces. The first ball (negatively charged) moves with speed $v$ and collides with the second ball which had been resting at the origin. The collision is central, and immediately before the im pact, the velocity of the first ball was parallel to the $x$-axis.\n\nWhat is the speed of the second ball immediately after the collision?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": [
            "v"
        ],
        "solution": "After the first collision, let the velocities of the first and second ball be $v_{1}$ and $v_{2}$ respectively. Applying the conservation of energy gives $\\frac{m v^{2}}{2}=\\frac{m v_{1}^{2}}{2}+\\frac{m v_{2}^{2}}{2}$ or $v^{2}=v_{1}^{2}+v_{2}^{2}$. Conservation of momentum yields $m v=m v_{1}+m v_{2}$ or $v=v_{1}+v_{2}$. Combining the two equations gives $v^{2}=v_{2}^{2}+\\left(v-v_{2}\\right)^{2}=v^{2}-2 v v_{2}+2 v_{2}^{2}$ and $v_{2}=0 ; v$.\nSince the first solution corresponds to the case when the collision doesn't happen, the speed of he second ball must be $v_{2}=v$.",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_96",
        "problem": "A machine gun fires $100 \\mathrm{~g}$ bullets at a speed of $1000 \\mathrm{~m} / \\mathrm{s}$. The person holding the machine gun in their hands can exert an average force of $150 \\mathrm{~N}$ against the gun. If the gun is to remain stationary, what is the maximum number of bullets that can be fired per minute?\nA: 10\nB: 15\nC: 30\nD: 60\nE: 90\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA machine gun fires $100 \\mathrm{~g}$ bullets at a speed of $1000 \\mathrm{~m} / \\mathrm{s}$. The person holding the machine gun in their hands can exert an average force of $150 \\mathrm{~N}$ against the gun. If the gun is to remain stationary, what is the maximum number of bullets that can be fired per minute?\n\nA: 10\nB: 15\nC: 30\nD: 60\nE: 90\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "E"
        ],
        "solution": "$\\mathbf{E} \\quad n=\\frac{F}{m v}$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_66",
        "problem": "What does a Venturi meter measure?\nA: fluid pressure\nB: fluid density\nC: fluid speed\nD: fluid flow\nE: fluid temperature\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat does a Venturi meter measure?\n\nA: fluid pressure\nB: fluid density\nC: fluid speed\nD: fluid flow\nE: fluid temperature\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "D"
        ],
        "solution": "See: https://forumautomation.com/t/what-is-a-venturi-meter-and-how-does-itwork/8455. A Venturi meter measures fluid flow.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_961",
        "problem": "The idea of centre of mass is an important concept and is more appreciated with examples of its use.\n\nA $34 \\mathrm{~cm}$ long uniform straight rod lies on a smooth horizontal surface and it is seen to be spinning round whilst also moving across the surface (translating). At one particular moment in time it is observed that the velocities of the ends of the rod are normal to the rod and have values, $2.6 \\mathrm{~m} \\mathrm{~s}^{-1}$ and $4.2 \\mathrm{~m} \\mathrm{~s}^{-1}$ as illustrated in Fig. 2.\n\n[figure1]\n\nFigure 2: A uniform rod which is rotating and translating across a smooth horizontal surface.\n\n At what frequency does it rotate?",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe idea of centre of mass is an important concept and is more appreciated with examples of its use.\n\nA $34 \\mathrm{~cm}$ long uniform straight rod lies on a smooth horizontal surface and it is seen to be spinning round whilst also moving across the surface (translating). At one particular moment in time it is observed that the velocities of the ends of the rod are normal to the rod and have values, $2.6 \\mathrm{~m} \\mathrm{~s}^{-1}$ and $4.2 \\mathrm{~m} \\mathrm{~s}^{-1}$ as illustrated in Fig. 2.\n\n[figure1]\n\nFigure 2: A uniform rod which is rotating and translating across a smooth horizontal surface.\n\n At what frequency does it rotate?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of Hz, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_9de0d5715d0b2f377364g-04.jpg?height=560&width=355&top_left_y=1496&top_left_x=859"
        ],
        "answer": [
            "3.2"
        ],
        "solution": "$$\nf=\\frac{1}{T}=\\frac{\\text { speed }}{\\text { distance }}=\\frac{3.4}{2 \\pi \\times 0.17}\n$$\n\n$3.2 \\mathrm{~Hz}$",
        "answer_type": "NV",
        "unit": [
            "Hz"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_566",
        "problem": "A newly discovered subatomic particle, the $S$ meson, has a mass $M$. When at rest, it lives for exactly $\\tau=3 \\times 10^{-8}$ seconds before decaying into two identical particles called $P$ mesons (peons?) that each have a mass of $\\alpha M$.\n\nIn a reference frame where the $\\mathrm{S}$ meson travels 9 meters between creation and decay, determine the velocity. Write the answers in terms of $M$, the speed of light $c$, and any numerical constants.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nA newly discovered subatomic particle, the $S$ meson, has a mass $M$. When at rest, it lives for exactly $\\tau=3 \\times 10^{-8}$ seconds before decaying into two identical particles called $P$ mesons (peons?) that each have a mass of $\\alpha M$.\n\nIn a reference frame where the $\\mathrm{S}$ meson travels 9 meters between creation and decay, determine the velocity. Write the answers in terms of $M$, the speed of light $c$, and any numerical constants.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": [
            "$\\frac{c}{\\sqrt{1+\\alpha^{-2}}}$"
        ],
        "solution": "By ordinary kinematics, we have\n\n$$\nd=v t=v \\gamma \\tau, \\quad \\gamma=1 / \\sqrt{1-v^{2} / c^{2}}\n$$\n\nwhere $\\gamma$ is the time dilation factor. Now we have to solve for $v$. Defining $\\alpha=d / c \\tau$,\n\n$$\n\\alpha=\\frac{v}{\\sqrt{c^{2}-v^{2}}} \\Rightarrow \\alpha^{2}=\\frac{v^{2}}{c^{2}-v^{2}}=\\frac{1}{1-v^{2} / c^{2}}-1\n$$\n\nSolving for $v$ gives\n\n$$\nv=\\frac{c}{\\sqrt{1+\\alpha^{-2}}}\n$$\n\nPlugging in the numbers, we find $\\alpha=1$, so $v=c / \\sqrt{2}$.",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_40",
        "problem": "A combination of two thin convex lenses is placed as shown at right. An object is placed $5 \\mathrm{~cm}$ in front of $L_{1}$ which has a focal length of $10 \\mathrm{~cm}$. $L_{2}$ is $10 \\mathrm{~cm}$ behind $L_{1}$ and has a focal length of $12 \\mathrm{~cm}$. How far from $L_{2}$ is the final image for this lens combination?\n\n[figure1]\nA: $8 \\mathrm{~cm}$\nB: $15 \\mathrm{~cm}$\nC: $22 \\mathrm{~cm}$\nD: $24 \\mathrm{~cm}$\nE: $30 \\mathrm{~cm}$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA combination of two thin convex lenses is placed as shown at right. An object is placed $5 \\mathrm{~cm}$ in front of $L_{1}$ which has a focal length of $10 \\mathrm{~cm}$. $L_{2}$ is $10 \\mathrm{~cm}$ behind $L_{1}$ and has a focal length of $12 \\mathrm{~cm}$. How far from $L_{2}$ is the final image for this lens combination?\n\n[figure1]\n\nA: $8 \\mathrm{~cm}$\nB: $15 \\mathrm{~cm}$\nC: $22 \\mathrm{~cm}$\nD: $24 \\mathrm{~cm}$\nE: $30 \\mathrm{~cm}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_bd21c863b16c0f40a895g-07.jpg?height=293&width=683&top_left_y=1705&top_left_x=1187"
        ],
        "answer": [
            "E"
        ],
        "solution": "$\\mathbf{E} \\quad \\frac{1}{f}=\\frac{1}{d_{0}}+\\frac{1}{d_{i}}$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1613",
        "problem": "中性粒子分析器 (Neutral-Particle Analyser) 是核聚变研究中测量快离子温度及其能量分布的重要设备. 其基本原理如图所示, 通过对高能量 (200eV 30KeV) 中性原子（它们容易穿透探测区中的电磁区域) 的能量和动量的测量, 可诊断曾与这些中性原子充分碰撞过的粒子的性质. 为了测量中性原子的能量分布, 首先让中性原子电离然后让离子束以 $\\theta$ 角入射到间距为 $d$ 、电压为 $V$ 的平行板电极组成的区域, 经电场偏转后离开电场区域, 在保证所测量离子不碰到上极板的前提下, 通过测量入射孔 $\\mathrm{A}$ 和出射孔 $\\mathrm{B}$ 间平行于极板方向的距离 1 来决定离子的能量. 设 $\\mathrm{A}$ 与下极板的距离为 $h_{1}, \\mathrm{~B}$ 与下极板的距离为 $h_{2}$, 已知离子所带电荷为 $q$.\n\n[图1]推导离子能量 $E$ 与 $I$ 的关系, 并给出离子在极板内垂直于极板方向的最大飞行距离",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个方程。\n这里是一些可能会帮助你解决问题的先验信息提示：\n中性粒子分析器 (Neutral-Particle Analyser) 是核聚变研究中测量快离子温度及其能量分布的重要设备. 其基本原理如图所示, 通过对高能量 (200eV 30KeV) 中性原子（它们容易穿透探测区中的电磁区域) 的能量和动量的测量, 可诊断曾与这些中性原子充分碰撞过的粒子的性质. 为了测量中性原子的能量分布, 首先让中性原子电离然后让离子束以 $\\theta$ 角入射到间距为 $d$ 、电压为 $V$ 的平行板电极组成的区域, 经电场偏转后离开电场区域, 在保证所测量离子不碰到上极板的前提下, 通过测量入射孔 $\\mathrm{A}$ 和出射孔 $\\mathrm{B}$ 间平行于极板方向的距离 1 来决定离子的能量. 设 $\\mathrm{A}$ 与下极板的距离为 $h_{1}, \\mathrm{~B}$ 与下极板的距离为 $h_{2}$, 已知离子所带电荷为 $q$.\n\n[图1]\n\n问题:\n推导离子能量 $E$ 与 $I$ 的关系, 并给出离子在极板内垂直于极板方向的最大飞行距离\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个方程，例如ANSWER=\\frac{x^2}{4}+\\frac{y^2}{2}=1",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_bc0dbcee918772e93d28g-04.jpg?height=471&width=829&top_left_y=1706&top_left_x=996"
        ],
        "answer": [
            "$$ H=E d \\sin ^{2} \\theta /(q V) $$"
        ],
        "solution": "对于微观粒子可以忽略地球重力的作用, 因此离子在极板外做匀速直线运动, 以入射孔所在处为坐标原点, 以垂直于极板的方向为 $\\mathrm{Y}$ 轴方向, 平行于极板方向为 $\\mathrm{X}$ 轴方向, 则离子在坐标 $\\left(h_{1} \\cot \\theta, h_{1}\\right)$ 处进入极板区, 进入极板区后离子做斜抛运动, 初速度为\n\n$$\nv_{0}=\\sqrt{2 E / m}\n$$\n\n加速度大小为 $a_{y}=-q V /(m d)$\n\n$\\mathrm{Y}$ 轴方向速度为零时, 对应的坐标为\n\n$$\n\\left(h_{1} \\cot \\theta+E d \\sin 2 \\theta /(q V), \\quad h_{1}+E d \\sin ^{2} \\theta /(q V)\\right)\n$$\n离子离开极板时, 速度大小仍为 $v_{0}=\\sqrt{2 E / m}$, 后作匀速直线运动, 因此出射孔在 $\\mathrm{X}$ 轴方向的坐标为\n\n$$\nl=\\left(h_{1}+h_{2}\\right) \\cot \\theta+2 E d \\sin (2 \\theta) /(q V)\n$$\n\n离子的能量与 $l$ 的关系为\n\n$$\nE=q V\\left[l-\\left(h_{1}+h_{2}\\right) \\cot \\theta\\right] /(2 d \\sin 2 \\theta)\n$$\n\n能量为 $E$ 的离子在极板内垂直于极板方向的最大飞行距离\n\n$$\nH=E d \\sin ^{2} \\theta /(q V)\n$$",
        "answer_type": "EQ",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1477",
        "problem": "2005 年诺贝尔奖颁发给了梶田隆章（Takaaki KaJita）和阿瑟 B 麦克唐纳（Arthur B.McDonald），奖励他们分别身为各自团队中的核心研究者, 和同事一起发现了中微子振荡, 在粒子物理领域开辟了新的疆土。一种常见的探测中微子的方法是在氢核（即质子）上俘获中微子, 生成一个正电子和一个中子, 称为反贝塔衰变反应（IBD）。下面说法中正确的有\nA: 反应方程式可以写为 $p+\\bar{v}_{e} \\rightarrow n+\\mathrm{e}^{+}$, 其中 $\\bar{v}_{e}$ 为反电子中微子\nB: 中子和正电子的静质量之和大于质子静质量, 中微子的静质量趋于 0\nC: 自由的中子也可以进行衰变, 产生中微子, 反应方程式为 $n \\rightarrow p+\\bar{v}_{e}+\\mathrm{e}^{+}$\nD: 如果被反应前质量是静止的，则产生的正电子和中子的动量之和不等于 0\n",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n2005 年诺贝尔奖颁发给了梶田隆章（Takaaki KaJita）和阿瑟 B 麦克唐纳（Arthur B.McDonald），奖励他们分别身为各自团队中的核心研究者, 和同事一起发现了中微子振荡, 在粒子物理领域开辟了新的疆土。一种常见的探测中微子的方法是在氢核（即质子）上俘获中微子, 生成一个正电子和一个中子, 称为反贝塔衰变反应（IBD）。下面说法中正确的有\n\nA: 反应方程式可以写为 $p+\\bar{v}_{e} \\rightarrow n+\\mathrm{e}^{+}$, 其中 $\\bar{v}_{e}$ 为反电子中微子\nB: 中子和正电子的静质量之和大于质子静质量, 中微子的静质量趋于 0\nC: 自由的中子也可以进行衰变, 产生中微子, 反应方程式为 $n \\rightarrow p+\\bar{v}_{e}+\\mathrm{e}^{+}$\nD: 如果被反应前质量是静止的，则产生的正电子和中子的动量之和不等于 0\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "A",
            "B",
            "D"
        ],
        "solution": "根据题意可知, 质子在俘获中微子后, 生成一个正电子和一个中子, 所以其反应方程式就如选项 A 所说的那样, 所以选项 A 正确; 中微子是俘获的, 说明它是运动的, 根据爱因斯坦的质能方程,运动的能量可以转弯为质量, 所以反应后的质量之和大于反应前质子的质量, 选项 B 正确; 中子进行衰变,应该产生质子和电子, 而不是正电子, 故选项 C 错误; 若反应前质子是静止的, 但中微子是运动的, 所以反应前的总动量不等于零, 所以反应后的总动量也不会等于零, 选项 D 正确。",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Physics_1111",
        "problem": "Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \\approx N \\approx A / 2$, where $A$ is the total number of nucleons $(A \\gg 1)$.\n\n## Transfer reactions\n\n[figure1]\n\nThe ${ }^{58} \\mathrm{Ni}$ nucleus produced in the excited state discussed in the part a), deexcites into its ground state by emitting a gamma-photon in the direction of its motion. Consider this decay in the frame of reference in which ${ }^{58} \\mathrm{Ni}$ is at rest to find the recoil energy of ${ }^{58} \\mathrm{Ni}$ (i.e. kinetic energy which ${ }^{58} \\mathrm{Ni}$ acquires after the emission of the photon). What is the photon energy in that system? What is the photon energy in the lab system of reference (i.e. what would be the energy of the photon measured in the detector which is positioned in the direction in which the ${ }^{58} \\mathrm{Ni}$ nucleus moves)?",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nAlthough atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \\approx N \\approx A / 2$, where $A$ is the total number of nucleons $(A \\gg 1)$.\n\n## Transfer reactions\n\n[figure1]\n\nThe ${ }^{58} \\mathrm{Ni}$ nucleus produced in the excited state discussed in the part a), deexcites into its ground state by emitting a gamma-photon in the direction of its motion. Consider this decay in the frame of reference in which ${ }^{58} \\mathrm{Ni}$ is at rest to find the recoil energy of ${ }^{58} \\mathrm{Ni}$ (i.e. kinetic energy which ${ }^{58} \\mathrm{Ni}$ acquires after the emission of the photon). What is the photon energy in that system? What is the photon energy in the lab system of reference (i.e. what would be the energy of the photon measured in the detector which is positioned in the direction in which the ${ }^{58} \\mathrm{Ni}$ nucleus moves)?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of MeV, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_656c8e5e21a256bf8ee1g-3.jpg?height=716&width=1450&top_left_y=1815&top_left_x=306"
        ],
        "answer": [
            "10.925"
        ],
        "solution": "For gamma-emission from the static nucleus, laws of conservation of energy and momentum give:\n\n$$\n\\begin{array}{r}\nE_{x}\\left({ }^{58} \\mathrm{Ni}\\right)=E_{\\gamma}+E_{\\text {recoil }} \\\\\np_{\\gamma}=p_{\\text {recoil }}\n\\end{array}\n$$\n\nGamma-ray and recoiled nucleus have, of course, opposite directions. For gamma-ray (photon), energy and momentum are related as:\n\n$$\nE_{\\gamma}=p_{\\gamma} \\cdot c\n$$\n\nIn part a) we have seen that the nucleus motion in this energy range is not relativistic, so we have:\n\n$$\nE_{\\text {recoil }}=\\frac{p_{\\text {recoil }}^{2}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right)}=\\frac{p_{\\gamma}^{2}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right)}=\\frac{E_{\\gamma}^{2}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right) \\cdot c^{2}}\n$$\n\nInserting this into law of energy conservation Eq. (36), we get:\n\n$$\nE_{x}\\left({ }^{58} \\mathrm{Ni}\\right)=E_{\\gamma}+E_{\\text {recoil }}=E_{\\gamma}+\\frac{E_{\\gamma}^{2}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right) \\cdot c^{2}}\n$$\n\nThis reduces to the quadratic equation:\n\n$$\nE_{\\gamma}^{2}+2 m\\left({ }^{58} \\mathrm{Ni}\\right) c^{2} \\cdot E_{\\gamma}+2 m\\left({ }^{58} \\mathrm{Ni}\\right) c^{2} E_{x}\\left({ }^{58} \\mathrm{Ni}\\right)=0\n$$\n\nwhich gives the following solution:\n\n$$\n\\begin{aligned}\nE_{\\gamma} & =\\frac{-2 m\\left({ }^{58} \\mathrm{Ni}\\right) c^{2}+\\sqrt{4\\left(m\\left({ }^{58} \\mathrm{Ni}\\right) c^{2}\\right)^{2}+8 m\\left({ }^{58} \\mathrm{Ni}\\right) c^{2} E_{x}\\left({ }^{58} \\mathrm{Ni}\\right)}}{2}= \\\\\n& =\\sqrt{\\left(m\\left({ }^{58} \\mathrm{Ni}\\right) c^{2}\\right)^{2}+2 m\\left({ }^{58} \\mathrm{Ni}\\right) c^{2} E_{x}\\left({ }^{58} \\mathrm{Ni}\\right)}-m\\left({ }^{58} \\mathrm{Ni}\\right) c^{2}\n\\end{aligned}\n$$\n\nInserting numbers gives:\n\n$$\nE_{\\gamma}=10.8633 \\mathrm{MeV}\n$$\n\nThe equation (37) can also be reduced to an approximate equation before inserting numbers:\n\n$$\nE_{\\gamma}=E_{x}\\left(1-\\frac{E_{x}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right) c^{2}}\\right)=10.8633 \\mathrm{MeV}\n$$\n\nThe recoil energy is now easily found as:\n\n$$\nE_{\\text {recoil }}=E_{x}\\left({ }^{58} \\mathrm{Ni}\\right)-E_{\\gamma}=1.1 \\mathrm{keV}\n$$\n\nDue to the fact that nucleus emitting gamma-ray $\\left({ }^{58} \\mathrm{Ni}\\right)$ is moving with the high velocity, the energy of gamma ray will be changed because of the Doppler effect. The relativistic Doppler effect (when source is moving towards observer/detector) is given with this formula:\n\n$$\nf_{\\text {detector }}=f_{\\gamma, \\text { emitted }} \\sqrt{\\frac{1+\\beta}{1-\\beta}}\n$$\n\nand since there is a simple relation between photon energy and frequency $(E=h f)$, we get the similar expression for energy:\n\n$$\nE_{\\text {detector }}=E_{\\gamma, \\text { emitted }} \\sqrt{\\frac{1+\\beta}{1-\\beta}}\n$$\n\nwhere $\\beta=\\mathrm{v} / \\mathrm{c}$ and $\\mathrm{v}$ is the velocity of emitter (the ${ }^{58} \\mathrm{Ni}$ nucleus). Taking the calculated value of the ${ }^{58} \\mathrm{Ni}$ velocity (equation 29 ) we get:\n\n$$\nE_{\\text {detector }}=E_{\\gamma, \\text { emitted }} \\sqrt{\\frac{1+\\beta}{1-\\beta}}=10.863 \\sqrt{\\frac{1+0.00565}{1-0.00565}}=10.925 \\mathrm{MeV}\n$$",
        "answer_type": "NV",
        "unit": [
            "MeV"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_145",
        "problem": "Raindrops with a number density of $n$ drops per cubic meter and radius $r_{0}$ hit the ground with a speed $v_{0}$. The resulting pressure on the ground from the rain is $P_{0}$. If the number density is doubled, the drop radius is halved, and the speed is halved, the new pressure will be\nA: $P_{0}$\nB: $P_{0} / 2$\nC: $P_{0} / 4$\nD: $P_{0} / 8$\nE: $P_{0} / 16 $ \n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRaindrops with a number density of $n$ drops per cubic meter and radius $r_{0}$ hit the ground with a speed $v_{0}$. The resulting pressure on the ground from the rain is $P_{0}$. If the number density is doubled, the drop radius is halved, and the speed is halved, the new pressure will be\n\nA: $P_{0}$\nB: $P_{0} / 2$\nC: $P_{0} / 4$\nD: $P_{0} / 8$\nE: $P_{0} / 16 $ \n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "E"
        ],
        "solution": "Each drop has $1 / 8$ the mass and $1 / 2$ the speed, so $1 / 16$ the momentum. The rate at which they hit the ground is the same, since the number density is doubled but the speed is halved. Thus the pressure is $P_{0} / 16$.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_892",
        "problem": "The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and H. Stormer at Bell Labs in 1981. In the experiment electrons were confined in two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the two-dimensional electron layer). A strong uniform magnetic field $B$ was applied perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when a current $I$ was passing through the sample, the voltage $V_{\\mathrm{H}}$ across the current path exhibited an unexpected quantized plateau (corresponding to a Hall resistance $R_{\\mathrm{H}}=$ $3 h / e^{2}$ ) at sufficiently low temperatures. The appearance of the plateau would imply the presence of fractionally charged quasiparticles in the system, which we analyze below. For simplicity, we neglect the scattering of the electrons by random potential, as well as the electron spin.\n\nIn a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions.\n\nWe know that electrons move in circular orbits in the magnetic field. In the quantum mechanical picture, the impinging magnetic field $B$ could be viewed as creating tiny whirlpools, so-called vortices, in the sea of electrons - one whirlpool for each flux quantum $h / e$ of the magnetic field, where $h$ is the Planck's constant and $e$ the elementary charge of an electron. For the case of $R_{\\mathrm{H}}=3 h / e^{2}$, which was discovered by Tsui and Stormer, derive the ratio of the number of the electrons $N$ to the number of the flux quanta $N_{\\phi}$,\nknown as the filling factor $v$.\n\n(a)\n\n[figure1]\n\n(b)\n\n[figure2]\n\nFigure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\\mathrm{H}}=3 h / e^{2}$ are due to the FQHE.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and H. Stormer at Bell Labs in 1981. In the experiment electrons were confined in two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the two-dimensional electron layer). A strong uniform magnetic field $B$ was applied perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when a current $I$ was passing through the sample, the voltage $V_{\\mathrm{H}}$ across the current path exhibited an unexpected quantized plateau (corresponding to a Hall resistance $R_{\\mathrm{H}}=$ $3 h / e^{2}$ ) at sufficiently low temperatures. The appearance of the plateau would imply the presence of fractionally charged quasiparticles in the system, which we analyze below. For simplicity, we neglect the scattering of the electrons by random potential, as well as the electron spin.\n\nIn a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions.\n\nWe know that electrons move in circular orbits in the magnetic field. In the quantum mechanical picture, the impinging magnetic field $B$ could be viewed as creating tiny whirlpools, so-called vortices, in the sea of electrons - one whirlpool for each flux quantum $h / e$ of the magnetic field, where $h$ is the Planck's constant and $e$ the elementary charge of an electron. For the case of $R_{\\mathrm{H}}=3 h / e^{2}$, which was discovered by Tsui and Stormer, derive the ratio of the number of the electrons $N$ to the number of the flux quanta $N_{\\phi}$,\nknown as the filling factor $v$.\n\n(a)\n\n[figure1]\n\n(b)\n\n[figure2]\n\nFigure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\\mathrm{H}}=3 h / e^{2}$ are due to the FQHE.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_10f4032f500207d762acg-2.jpg?height=386&width=963&top_left_y=441&top_left_x=455",
            "https://cdn.mathpix.com/cropped/2024_03_14_10f4032f500207d762acg-2.jpg?height=954&width=1302&top_left_y=959&top_left_x=320"
        ],
        "answer": [
            "1/3"
        ],
        "solution": "The Hall resistance can be rewritten as\n\n$$\nR_{H}=\\frac{1}{e} \\frac{\\phi}{N}=\\frac{h}{e^{2}} \\frac{\\phi /(h / e)}{N}=\\frac{h}{e^{2}} \\frac{N_{\\phi}}{N}\n$$\n\nAt the plateau, $\\nu=N / N_{\\phi}=1 / 3$.",
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_564",
        "problem": "Consider the circuit shown below. $I_{s}$ is a constant current source, meaning that no matter what device is connected between points $\\mathrm{A}$ and $\\mathrm{B}$, the current provided by the constant current source is the same.\n\n[figure1]\n\nConnect instead an ideal ammeter between $A$ and $B$. Determine the current in terms of any or all of $R$ and $I_{s}$.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nConsider the circuit shown below. $I_{s}$ is a constant current source, meaning that no matter what device is connected between points $\\mathrm{A}$ and $\\mathrm{B}$, the current provided by the constant current source is the same.\n\n[figure1]\n\nConnect instead an ideal ammeter between $A$ and $B$. Determine the current in terms of any or all of $R$ and $I_{s}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d2de61e197238a85a597g-05.jpg?height=347&width=838&top_left_y=474&top_left_x=619"
        ],
        "answer": [
            "$\\frac{1}{9} I_{s}$"
        ],
        "solution": "An ideal ammeter has zero resistance, so we just need to find the current through the effective $6 R$ resistor that connects the two vertical branches. This current will flow to the left.\n\nBy symmetry, the current through each vertical resistance of $2 R$ must be the same, as well as the currents through each vertical resistance of $4 R$. This gives the system of equations\n\n$$\n\\begin{aligned}\nI_{s} & =I_{2}+I_{4} \\\\\nI_{2} & =I_{6}+I_{4} \\\\\nI_{4}(4 R) & =I_{2}(2 R)+I_{6}(6 R)\n\\end{aligned}\n$$\n\nCopyright (C)2015 American Association of Physics Teachers\n\nEliminating $I_{2}$ gives\n\n$$\n\\begin{aligned}\nI_{s} & =I_{6}+2 I_{4} \\\\\n4 I_{4} & =2\\left(I_{6}+I_{4}\\right)+6 I_{6}\n\\end{aligned}\n$$\n\nFinally, eliminating $I_{4}$ gives $I_{4}=4 I_{6}$ and\n\n$$\nI_{6}=\\frac{1}{9} I_{s}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1544",
        "problem": "电子感应加速器利用变化的磁场来加速电子。电子绕平均半径为 $R$ 的环形轨道（轨道位于真空管道内）运动, 磁感应强度方向与环形轨道平面垂直。电子被感应电场加速, 感应电场的方向与环形轨道相切。电子电荷量为 $e$ 。\n\n[图1]为了使电子在不断增强的磁场中沿着半径不变的圆轨道加速运动, 求 $\\frac{\\Delta B}{\\Delta t}$ 和 $\\frac{\\Delta \\bar{B}}{\\Delta t}$ 之间必须满足的定量关系",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个方程。\n这里是一些可能会帮助你解决问题的先验信息提示：\n电子感应加速器利用变化的磁场来加速电子。电子绕平均半径为 $R$ 的环形轨道（轨道位于真空管道内）运动, 磁感应强度方向与环形轨道平面垂直。电子被感应电场加速, 感应电场的方向与环形轨道相切。电子电荷量为 $e$ 。\n\n[图1]\n\n问题:\n为了使电子在不断增强的磁场中沿着半径不变的圆轨道加速运动, 求 $\\frac{\\Delta B}{\\Delta t}$ 和 $\\frac{\\Delta \\bar{B}}{\\Delta t}$ 之间必须满足的定量关系\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个方程，例如ANSWER=\\frac{x^2}{4}+\\frac{y^2}{2}=1",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_7973e92ea1a5d86ba5a6g-05.jpg?height=371&width=417&top_left_y=1893&top_left_x=1362"
        ],
        "answer": [
            "$$ \\frac{\\Delta B}{\\Delta t}=\\frac{1}{2} \\frac{\\Delta \\bar{B}}{\\Delta t} $$"
        ],
        "solution": "(3)和(9)式所表示的是同样的力的大小。联立(3)(9)式得\n\n$$\n\\frac{\\Delta B}{\\Delta t}=\\frac{1}{2} \\frac{\\Delta \\bar{B}}{\\Delta t}\n$$\n\n这就是为了使电子在不断增强的磁场中沿着半径不变的圆轨道加速运动, $\\frac{\\Delta B}{\\Delta t}$ 和 $\\frac{\\Delta \\bar{B}}{\\Delta t}$ 之间必须满足的定量关系。",
        "answer_type": "EQ",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_624",
        "problem": "Suppose a domino stands upright on a table. It has height $h$, thickness $t$, width $w$ (as shown below), and mass $m$. The domino is free to rotate about its edges, but will not slide across the table.\n[figure1]\n\nSuppose we give the domino a sharp, horizontal impulsive push with total momentum $p$.\n\nAt what height $H$ above the table is the impulse $p$ required to topple the domino smallest?",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nSuppose a domino stands upright on a table. It has height $h$, thickness $t$, width $w$ (as shown below), and mass $m$. The domino is free to rotate about its edges, but will not slide across the table.\n[figure1]\n\nSuppose we give the domino a sharp, horizontal impulsive push with total momentum $p$.\n\nAt what height $H$ above the table is the impulse $p$ required to topple the domino smallest?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_8c5c0b6efdeaae46cc88g-15.jpg?height=456&width=832&top_left_y=528&top_left_x=644"
        ],
        "answer": [
            "$h$"
        ],
        "solution": "A convenient method is to look at the angular momentum in the domino because this is easy to calculate and describes rotational motion. Because the push is horizontal, the moment arm of the push (about the domino's rotational axis) is purely vertical. That means the angular momentum of the push is $p H$, where $p$ is the momentum imparted and $H$ is the height of the push. There is some minimum angular momentum $L_{\\min }$ to topple the domino, so we set $L_{\\min }=p H$. The bigger $H$, the smaller $p$, so we should choose the largest possible $H$. In other words, we should push at the very top of the domino, $H=h$. While we're on this part, note that if the push weren't constrained to be horizontal, the push could be a little bit smaller since the moment arm could be the entire diagonal of the thin edge of the domino.",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_76",
        "problem": "A $3.00 \\mathrm{~kg}$ bucket of water is raised with an upward acceleration of $2.20 \\mathrm{~m} / \\mathrm{s}^{2}$ from a well by means of an attached rope. What is the tension in the rope?\nA: $30.2 \\mathrm{~N}$\nB: $33.3 \\mathrm{~N}$\nC: $36.6 \\mathrm{~N}$\nD: $39.0 \\mathrm{~N}$\nE: $43.2 \\mathrm{~N}$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA $3.00 \\mathrm{~kg}$ bucket of water is raised with an upward acceleration of $2.20 \\mathrm{~m} / \\mathrm{s}^{2}$ from a well by means of an attached rope. What is the tension in the rope?\n\nA: $30.2 \\mathrm{~N}$\nB: $33.3 \\mathrm{~N}$\nC: $36.6 \\mathrm{~N}$\nD: $39.0 \\mathrm{~N}$\nE: $43.2 \\mathrm{~N}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "C"
        ],
        "solution": "$T=\\sum F=m a+m g$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_614",
        "problem": "In this problem, use a particle-like model of photons: they propagate in straight lines and obey the law of reflection, but are subject to the quantum uncertainty principle. You may use small-angle approximations throughout the problem.\n\nA photon with wavelength $\\lambda$ has traveled from a distant star to a telescope mirror, which has a circular cross-section with radius $R$ and a focal length $f \\gg R$. The path of the photon is nearly aligned to the axis of the mirror, but has some slight uncertainty $\\Delta \\theta$. The photon reflects off the mirror and travels to a detector, where it is absorbed by a particular pixel on a charge-coupled device (CCD).\n\nSuppose the telescope mirror is manufactured so that photons coming in parallel to each other are focused to the same pixel on the CCD, regardless of where they hit the mirror. Then all small cross-sectional areas of the mirror are equally likely to include the point of reflection for a photon.\n\nFind the standard deviation $\\Delta r$ of the distribution for $r$, the distance from the center of the telescope mirror to the point of reflection of the photon.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nIn this problem, use a particle-like model of photons: they propagate in straight lines and obey the law of reflection, but are subject to the quantum uncertainty principle. You may use small-angle approximations throughout the problem.\n\nA photon with wavelength $\\lambda$ has traveled from a distant star to a telescope mirror, which has a circular cross-section with radius $R$ and a focal length $f \\gg R$. The path of the photon is nearly aligned to the axis of the mirror, but has some slight uncertainty $\\Delta \\theta$. The photon reflects off the mirror and travels to a detector, where it is absorbed by a particular pixel on a charge-coupled device (CCD).\n\nSuppose the telescope mirror is manufactured so that photons coming in parallel to each other are focused to the same pixel on the CCD, regardless of where they hit the mirror. Then all small cross-sectional areas of the mirror are equally likely to include the point of reflection for a photon.\n\nFind the standard deviation $\\Delta r$ of the distribution for $r$, the distance from the center of the telescope mirror to the point of reflection of the photon.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": [
            "$\\frac{R}{\\sqrt{18}}$"
        ],
        "solution": "The square of the standard deviation is the variance, so\n\n$$\n(\\Delta r)^{2}=\\left\\langle r^{2}\\right\\rangle-\\langle r\\rangle^{2}\n$$\n\nComputing the average value of $r^{2}$ is mathematically the exact same thing as computing the moment of inertia of a uniform disk; we have\n\n$$\n\\left\\langle r^{2}\\right\\rangle=\\frac{1}{\\pi R^{2}} \\int_{0}^{R} r^{2}(2 \\pi r d r)=\\frac{1}{2} R^{2}\n$$\n\nSimilarly, the average value of $r$ is\n\n$$\n\\langle r\\rangle=\\frac{1}{\\pi R^{2}} \\int_{0}^{R} r(2 \\pi r d r)=\\frac{2}{3} R\n$$\n\nThen we have\n\n$$\n\\Delta r=\\frac{R}{\\sqrt{18}}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_65",
        "problem": "Earth rotates once every $23 \\mathrm{~h} 56 \\mathrm{~min}$ around an axis that runs from the North to South Pole. What is the angular velocity of the Earth?\nA: $7.29 \\times 10^{-5} \\mathrm{rad} / \\mathrm{s}$\nB: $7.59 \\times 10^{-5} \\mathrm{rad} / \\mathrm{s}$\nC: $7.09 \\times 10^{-5} \\mathrm{rad} / \\mathrm{s}$\nD: $4.37 \\times 10^{-3} \\mathrm{rad} / \\mathrm{s}$\nE: $4.37 \\times 10^{-5} \\mathrm{rad} / \\mathrm{s}$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nEarth rotates once every $23 \\mathrm{~h} 56 \\mathrm{~min}$ around an axis that runs from the North to South Pole. What is the angular velocity of the Earth?\n\nA: $7.29 \\times 10^{-5} \\mathrm{rad} / \\mathrm{s}$\nB: $7.59 \\times 10^{-5} \\mathrm{rad} / \\mathrm{s}$\nC: $7.09 \\times 10^{-5} \\mathrm{rad} / \\mathrm{s}$\nD: $4.37 \\times 10^{-3} \\mathrm{rad} / \\mathrm{s}$\nE: $4.37 \\times 10^{-5} \\mathrm{rad} / \\mathrm{s}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "A"
        ],
        "solution": "$\\mathrm{T}=23 \\mathrm{~h} 56 \\mathrm{~min}=86,160 \\mathrm{~s} ; f=\\frac{1}{T} ; \\omega=2 \\pi f$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_376",
        "problem": "Eero Uustalu and Jaan Kalda. You are given the following tools: a retroreflective film an enlarged bottom view of which is shown in the first figure; stand, ruler, laser pointer, sreen, graph paper, protractor.\n\n[figure1]\n\nWhile the top surface of the film is flat, the bottom surface is a periodic array of slanted triangular faces. Six such faces are shown enlarged in the seond figure; the faces 1,3 a corner of a cube, and the faces 2, 4 nd 6 are also perpendicular to each other On the right of the second figure, a cross section of the film is shown. The film's material between the slanted faces and the flat surface form microprisms. The refracting angles of these microprisms are denoted by $\\alpha_{i}, i=1,2, \\ldots 6$ (the index numbers correspond to those of the faces). Among the an lges $\\alpha_{i}$, some may be equal to each other.\n\n[figure2]\n\ncross-section\n\nWhen light falls onto the flat surface close to perpendicular incidence, it undergoes total internal reflections on the slanted faces, and as a result, its direction of propagation is rotated by $180^{\\circ}$. However, the microprisms can also serve as prisms diverting a light beam by an angle $\\beta$. The angle $\\beta$ depends on the angle of incidence, and on the prism angle $\\alpha=\\alpha_{i}$. Let $\\beta_{i}$ denote the minimal deflection angle for a fixed prism angle $\\alpha_{i}$\n\nDetermine the prism angles $\\alpha_{i}$, $i=1,2, \\ldots 6$",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nEero Uustalu and Jaan Kalda. You are given the following tools: a retroreflective film an enlarged bottom view of which is shown in the first figure; stand, ruler, laser pointer, sreen, graph paper, protractor.\n\n[figure1]\n\nWhile the top surface of the film is flat, the bottom surface is a periodic array of slanted triangular faces. Six such faces are shown enlarged in the seond figure; the faces 1,3 a corner of a cube, and the faces 2, 4 nd 6 are also perpendicular to each other On the right of the second figure, a cross section of the film is shown. The film's material between the slanted faces and the flat surface form microprisms. The refracting angles of these microprisms are denoted by $\\alpha_{i}, i=1,2, \\ldots 6$ (the index numbers correspond to those of the faces). Among the an lges $\\alpha_{i}$, some may be equal to each other.\n\n[figure2]\n\ncross-section\n\nWhen light falls onto the flat surface close to perpendicular incidence, it undergoes total internal reflections on the slanted faces, and as a result, its direction of propagation is rotated by $180^{\\circ}$. However, the microprisms can also serve as prisms diverting a light beam by an angle $\\beta$. The angle $\\beta$ depends on the angle of incidence, and on the prism angle $\\alpha=\\alpha_{i}$. Let $\\beta_{i}$ denote the minimal deflection angle for a fixed prism angle $\\alpha_{i}$\n\nDetermine the prism angles $\\alpha_{i}$, $i=1,2, \\ldots 6$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [$\\alpha_{1}$, $\\alpha_{2}$].\nTheir units are, in order, [^{\\circ}, ^{\\circ}], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_706aca6df357b4c9a255g-1.jpg?height=246&width=418&top_left_y=1593&top_left_x=1611",
            "https://cdn.mathpix.com/cropped/2024_03_06_706aca6df357b4c9a255g-1.jpg?height=164&width=373&top_left_y=448&top_left_x=2197"
        ],
        "answer": [
            "50",
            "58"
        ],
        "solution": "Minimal deflection angle corresponds to the rays traversing the prisms symmetrically. This allows us to conveniently find the minimal deflection angle in terms of $\\alpha$ and $\\gamma$, where $\\gamma$ is the angle of the film with respect to the laser beam. Further measurements show $\\gamma_{1}=11^{\\circ}$ and $\\gamma_{2}=9^{\\circ}$. From geometry, we get $\\beta / 2=\\alpha / 2-\\gamma$ so $\\alpha=2 \\gamma+\\beta$. This gives $\\alpha_{1}=50^{\\circ}, \\alpha_{2}=58^{\\circ}$.",
        "answer_type": "MPV",
        "unit": [
            "^{\\circ}",
            "^{\\circ}"
        ],
        "answer_sequence": [
            "$\\alpha_{1}$",
            "$\\alpha_{2}$"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_152",
        "problem": "Two blocks are suspended by two massless elastic strings to the ceiling as shown in the figure. The masses of the upper and lower block are $m_{1}=2 \\mathrm{~kg}$ and $m_{2}=4 \\mathrm{~kg}$ respectively. If the upper string is suddenly cut just above the top block what are the accelerations of the two blocks at the moment when the top block begins to fall?\n\n[figure1]\nA: upper: $10 \\mathrm{~m} / \\mathrm{s}^{2}$; lower: 0\nB: upper: $10 \\mathrm{~m} / \\mathrm{s}^{2}$; lower: $10 \\mathrm{~m} / \\mathrm{s}^{2}$\nC: upper: $20 \\mathrm{~m} / \\mathrm{s}^{2}$; lower: $10 \\mathrm{~m} / \\mathrm{s}^{2}$\nD: upper: $30 \\mathrm{~m} / \\mathrm{s}^{2}$; lower: $0 $ \nE: upper: $30 \\mathrm{~m} / \\mathrm{s}^{2}$; lower: $10 \\mathrm{~m} / \\mathrm{s}^{2}$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTwo blocks are suspended by two massless elastic strings to the ceiling as shown in the figure. The masses of the upper and lower block are $m_{1}=2 \\mathrm{~kg}$ and $m_{2}=4 \\mathrm{~kg}$ respectively. If the upper string is suddenly cut just above the top block what are the accelerations of the two blocks at the moment when the top block begins to fall?\n\n[figure1]\n\nA: upper: $10 \\mathrm{~m} / \\mathrm{s}^{2}$; lower: 0\nB: upper: $10 \\mathrm{~m} / \\mathrm{s}^{2}$; lower: $10 \\mathrm{~m} / \\mathrm{s}^{2}$\nC: upper: $20 \\mathrm{~m} / \\mathrm{s}^{2}$; lower: $10 \\mathrm{~m} / \\mathrm{s}^{2}$\nD: upper: $30 \\mathrm{~m} / \\mathrm{s}^{2}$; lower: $0 $ \nE: upper: $30 \\mathrm{~m} / \\mathrm{s}^{2}$; lower: $10 \\mathrm{~m} / \\mathrm{s}^{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_06de5165d0cf42eb6dbcg-11.jpg?height=366&width=265&top_left_y=476&top_left_x=930"
        ],
        "answer": [
            "D"
        ],
        "solution": "Draw free body diagrams for object 1 and 2 before the string is cut. These diagrams still apply immediately after the string is cut, except that the cut string no longer contributes a force in the free body diagram for object 1. Use the resulting free body diagrams to find acceleration of the two blocks.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1257",
        "problem": "A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q$ ' placed inside the sphere (you do not have to prove it). **Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface).**\n\n[figure1]\n\nA point charge $q$ with mass $m$ is suspended on a thread of length $L$ which is attached to a wall, in the vicinity of the grounded metallic sphere. In your considerations, ignore all electrostatic effects of the wall. The point charge makes a mathematical pendulum (see Fig. 3). The point at which the thread is attached to the wall is at a distancel from the center of the sphere. Assume that the effects of gravity are negligible.\n\n[figure2]\n\nFind the frequency for small oscillations of the pendulum.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nA point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q$ ' placed inside the sphere (you do not have to prove it). **Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface).**\n\n[figure1]\n\nA point charge $q$ with mass $m$ is suspended on a thread of length $L$ which is attached to a wall, in the vicinity of the grounded metallic sphere. In your considerations, ignore all electrostatic effects of the wall. The point charge makes a mathematical pendulum (see Fig. 3). The point at which the thread is attached to the wall is at a distancel from the center of the sphere. Assume that the effects of gravity are negligible.\n\n[figure2]\n\nFind the frequency for small oscillations of the pendulum.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_843c51f18af1a9802d4bg-1.jpg?height=774&width=1627&top_left_y=1046&top_left_x=220",
            "https://i.postimg.cc/rmvjsmdm/image.png"
        ],
        "answer": [
            "$\\frac{q}{(l-L)^{2}-R^{2}} \\sqrt{\\frac{R l}{4 \\pi \\varepsilon_{0}} \\frac{1}{m L}}$"
        ],
        "solution": "The equation of motion of the mathematical pendulum is\n\n$m L \\ddot{\\alpha}=-F_{\\perp}$\n\nAs we are interested in small oscillations, the angle $\\alpha$ is small, i.e. for its value in radians we have $\\alpha$ much smaller than 1 . For a small value of argument of trigonometric functions we have approximate relations $\\sin x \\approx x$ and $\\cos x \\approx 1-x^{2} / 2$. So for small oscillations of the pendulum we have $\\beta \\approx \\alpha L /(l-L)$ and $\\gamma \\approx l \\alpha /(l-L)$.\n\nCombining these relations with (13) we obtain\n\n$m L \\frac{d^{2} \\alpha}{d t^{2}}+\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R d}{\\left(d^{2}-R^{2}\\right)^{2}}\\left(1+\\frac{L}{d}\\right) \\alpha=0$\n\nWhere $d=l-L$ what leads to\n\n$$\n\\begin{aligned}\n& \\omega=\\frac{q}{d^{2}-R^{2}} \\sqrt{\\frac{R d}{4 \\pi \\varepsilon_{0}} \\frac{1}{m L}\\left(1+\\frac{L}{d}\\right)}= \\\\\n& =\\frac{q}{(l-L)^{2}-R^{2}} \\sqrt{\\frac{R l}{4 \\pi \\varepsilon_{0}} \\frac{1}{m L}}\n\\end{aligned}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_203",
        "problem": "A satellite is following an elliptical orbit around the Earth. Its engines are capable of providing a one-time impulse of a fixed magnitude. In order to maximize the energy of the satellite, the impulse should be\n\n[figure1]\nA: directed along the satellite's velocity and applied when the satellite is in its perigee. \nB: directed along the satellite's velocity and applied when the satellite is in apogee.\nC: directed toward the Earth and applied when the satellite is in perigee.\nD: directed toward the Earth and applied when the satellite is in apogee.\nE: directed away from the Earth and applied when the satellite is in apogee.\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA satellite is following an elliptical orbit around the Earth. Its engines are capable of providing a one-time impulse of a fixed magnitude. In order to maximize the energy of the satellite, the impulse should be\n\n[figure1]\n\nA: directed along the satellite's velocity and applied when the satellite is in its perigee. \nB: directed along the satellite's velocity and applied when the satellite is in apogee.\nC: directed toward the Earth and applied when the satellite is in perigee.\nD: directed toward the Earth and applied when the satellite is in apogee.\nE: directed away from the Earth and applied when the satellite is in apogee.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_6e7e781e6599f99f638bg-04.jpg?height=387&width=957&top_left_y=354&top_left_x=584"
        ],
        "answer": [
            "A"
        ],
        "solution": "The change in the velocity $\\Delta \\mathbf{v}$ has fixed magnitude. Since the kinetic energy is proportional to $v^{2}$, the greatest change is attained when the impulse is parallel to the velocity, when the speed is as large as possible, which occurs at perigee.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_475",
        "problem": "The ship starts out in a circular orbit around the sun very near the Earth and has a goal of moving to a circular orbit around the Sun that is very close to Mars. It will make this transfer in an elliptical orbit as shown in bold in the diagram below. This is accomplished with an initial velocity boost near the Earth $\\Delta v_{1}$ and then a second velocity boost near Mars $\\Delta v_{2}$. Assume that both of these boosts are from instantaneous impulses, and ignore mass changes in the rocket as well as gravitational attraction to either Earth or Mars. Don't ignore the\n\nSun! Assume that the Earth and Mars are both in circular orbits around the Sun of radii $R_{E}$ and $R_{M}=R_{E} / \\alpha$ respectively. The orbital speeds are $v_{E}$ and $v_{M}$ respectively.\n\n[figure1]\n\nWhat is the angular separation between Earth and Mars, as measured from the Sun, at the time of launch so that the rocket will start from Earth and arrive at Mars when it reaches the orbit of Mars? Express your answer in terms of $\\alpha$.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThe ship starts out in a circular orbit around the sun very near the Earth and has a goal of moving to a circular orbit around the Sun that is very close to Mars. It will make this transfer in an elliptical orbit as shown in bold in the diagram below. This is accomplished with an initial velocity boost near the Earth $\\Delta v_{1}$ and then a second velocity boost near Mars $\\Delta v_{2}$. Assume that both of these boosts are from instantaneous impulses, and ignore mass changes in the rocket as well as gravitational attraction to either Earth or Mars. Don't ignore the\n\nSun! Assume that the Earth and Mars are both in circular orbits around the Sun of radii $R_{E}$ and $R_{M}=R_{E} / \\alpha$ respectively. The orbital speeds are $v_{E}$ and $v_{M}$ respectively.\n\n[figure1]\n\nWhat is the angular separation between Earth and Mars, as measured from the Sun, at the time of launch so that the rocket will start from Earth and arrive at Mars when it reaches the orbit of Mars? Express your answer in terms of $\\alpha$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d2de61e197238a85a597g-16.jpg?height=658&width=656&top_left_y=365&top_left_x=778"
        ],
        "answer": [
            "$\\pi\\left(1-\\left(\\frac{\\alpha+1}{2}\\right)^{3 / 2}\\right)$"
        ],
        "solution": "Kepler's third law gives the time for the orbital transfer,\n\n$$\n\\frac{T}{T_{M}}=\\frac{1}{2}\\left(\\frac{\\left(R_{E}+R_{M}\\right) / 2}{R_{M}}\\right)^{3 / 2}=\\frac{1}{2}\\left(\\frac{\\alpha+1}{2}\\right)^{3 / 2}\n$$\n\nDuring this time Mars moves through an angle of\n\n$$\n2 \\pi \\frac{T}{T_{M}}=\\pi\\left(\\frac{\\alpha+1}{2}\\right)^{3 / 2}\n$$\n\nwhile the rocket moves through an angle of $\\pi$, so the angular separation from Earth will be\n\n$$\n\\theta=\\pi\\left(1-\\left(\\frac{\\alpha+1}{2}\\right)^{3 / 2}\\right)\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_397",
        "problem": "Consider a hypothetical interstellar travel with a photon-propelled spaceship of initial rest mass $M=1 \\times 10^{5} \\mathrm{~kg}$. The on-board fuel (antimatter) is annihilated with an equal mass of matter to create photons yielding a reactive force. The matter required for annihilation is collected from the very sparse plasma of the interstellar space (assume that the velocity of the interstellar plasma is zero in the Earth's frame of reference). The speed of light is $c=3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$.\n\nWhat should be the initial rate $\\mu(\\mathrm{kg} / \\mathrm{s})$ at which the antimatter should be burned for the acceleration to be equal to the free fall acceleration on Earth $\\left(g=9.81 \\mathrm{~m} / \\mathrm{s}^{2}\\right)$",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nConsider a hypothetical interstellar travel with a photon-propelled spaceship of initial rest mass $M=1 \\times 10^{5} \\mathrm{~kg}$. The on-board fuel (antimatter) is annihilated with an equal mass of matter to create photons yielding a reactive force. The matter required for annihilation is collected from the very sparse plasma of the interstellar space (assume that the velocity of the interstellar plasma is zero in the Earth's frame of reference). The speed of light is $c=3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$.\n\nWhat should be the initial rate $\\mu(\\mathrm{kg} / \\mathrm{s})$ at which the antimatter should be burned for the acceleration to be equal to the free fall acceleration on Earth $\\left(g=9.81 \\mathrm{~m} / \\mathrm{s}^{2}\\right)$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of \\mathrm{~kg} / \\mathrm{s}, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": [
            "$1.635 \\times 10^{-3}$"
        ],
        "solution": "At non-relativistic speeds, we can apply classical momentum and energy conservation to find the acceleration in terms of the antimatter burning rate $\\mu$. In a time interval $\\Delta t$, a mass of $\\Delta m=\\mu \\Delta t$ antimatter annihilates with an equal mass of matter. The resulting photons have an energy equal to the annihilated rest energy $\\Delta E=2 \\Delta m c^{2}$. For maximal acceleration, the photons have to be all emitted in the same direction (can be achieved, for example, using mirrors). The resulting photon cloud will then have a momentum of $\\Delta p=\\Delta E / c$. From the conservation of momentum, the space ship must get a momentum boost in the opposite direction, equal to $\\Delta p=M \\Delta v=M g \\Delta t$. Combining everything, we get $\\mu=M g /(2 c)=$ $1.635 \\times 10^{-3} \\mathrm{~kg} / \\mathrm{s}$",
        "answer_type": "NV",
        "unit": [
            "\\mathrm{~kg} / \\mathrm{s}"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_916",
        "problem": "Consider a gas of ultra relativistic electrons $(\\gamma \\gg 1)$, with an isotropic distribution of velocities (does not depend on direction). The proper number density of particles with energies between $\\epsilon$ and $\\epsilon+d \\epsilon$ is given by $f(\\epsilon) d \\epsilon$, where $\\epsilon$ is the energy per particle. Consider also a wall of area $\\Delta A$, which is in contact with the gas.\n\nDerive a relationship between the pressure and volume of an ultra relativistic $0.6 \\mathrm{pt}$ electron gas undergoing an adiabatic expansion.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an equation.\n\nproblem:\nConsider a gas of ultra relativistic electrons $(\\gamma \\gg 1)$, with an isotropic distribution of velocities (does not depend on direction). The proper number density of particles with energies between $\\epsilon$ and $\\epsilon+d \\epsilon$ is given by $f(\\epsilon) d \\epsilon$, where $\\epsilon$ is the energy per particle. Consider also a wall of area $\\Delta A$, which is in contact with the gas.\n\nDerive a relationship between the pressure and volume of an ultra relativistic $0.6 \\mathrm{pt}$ electron gas undergoing an adiabatic expansion.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an equation, e.g. ANSWER=\\frac{x^2}{4}+\\frac{y^2}{2}=1",
        "figure_urls": null,
        "answer": [
            "$\\frac{P V^{4 / 3}}{P_{0} V_{0}^{4 / 3}}  =1$"
        ],
        "solution": "For an adiabatic process $d Q=0$ so $d E=d W=-P d V . d E=d(3 P V)=3 P d V+3 V d P$, so equating these expressions gives\n\n$$\n\\begin{aligned}\n3 P d V+3 V d P & =-p d V \\\\\n4 P d V & =-3 V d P \\\\\n4 \\frac{d V}{V} & =-3 \\frac{d P}{P} \\\\\n4 \\int_{V_{0}}^{V} \\frac{d V^{\\prime}}{V^{\\prime}} & =-3 \\int_{P_{0}}^{P} \\frac{d P^{\\prime}}{P} \\\\\n4 \\ln \\left(\\frac{V}{V_{0}}\\right) & =-3 \\ln \\left(\\frac{P}{P_{0}}\\right) \\\\\n\\frac{P V^{4 / 3}}{P_{0} V_{0}^{4 / 3}} & =1\n\\end{aligned}\n$$",
        "answer_type": "EQ",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_940",
        "problem": "The terminal voltage of a dc power supply is measured as $5.00 \\mathrm{~V}$ when it is on open circuit. A $2.00 \\Omega$ resistor is connected across the terminals and the voltage drops by $0.100 \\mathrm{~V}$.\n\nIf the load resistor is reduced to $0.400 \\Omega$, what would be the terminal voltage now?",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe terminal voltage of a dc power supply is measured as $5.00 \\mathrm{~V}$ when it is on open circuit. A $2.00 \\Omega$ resistor is connected across the terminals and the voltage drops by $0.100 \\mathrm{~V}$.\n\nIf the load resistor is reduced to $0.400 \\Omega$, what would be the terminal voltage now?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of V, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_38b9374d4ba914cc7d40g-4.jpg?height=323&width=469&top_left_y=1455&top_left_x=311"
        ],
        "answer": [
            "4.54"
        ],
        "solution": "[figure1]\n\n$$\n\\begin{array}{ll}\nI=\\frac{V}{0.4}=\\frac{5}{0.4+\\frac{2}{49}} \\quad \\text { or similar idea } \\\\\nV=\\frac{2 \\times 49}{0.4 \\times 49+2}=\\frac{98}{21.6}=4.54 \\mathrm{~V}\n\\end{array}\n$$",
        "answer_type": "NV",
        "unit": [
            "V"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_1346",
        "problem": "用薄膜制备技术在某均质硅基片上沉积一层均匀等厚氮化镓薄膜,制备出一个硅基氮化镓样品, 如图 I 所示。\n\n[图1]当用波长范围为 $450 \\sim 1200 \\mathrm{~nm}$ 的光垂直均匀照射该\n\n样品氮化镓表面, 观察到其反射光谱仅有两种波长的光获得最大相干加强, 其中之一波长为 $600 \\mathrm{~nm}$; 氮化镓的折射率 $n$ 与入射光在真空中波长 $\\lambda$ (单位 $\\mathrm{nm}$ ) 之间的关系（色散关系）为\n\n$$\nn^{2}=2.26^{2}+\\frac{330.1^{2}}{\\lambda^{2}-265.7^{2}}\n$$\n\n在该样品硅基片的另一面左、右对称的两个半面上分别均匀沉积一光谱选择性材料涂层,如图 II 所示;对某种特定波长的光, 左半面涂层全吸收, 右半面全反射。用两根长均为 $a$ 的轻细线坚直悬挂该样品, 样品长为 $a$ 、宽为 $b$, 可绕过其中心的光滑坚直固定轴 $\\mathrm{OO}^{\\prime}$ 转动, 也可上下移动, 如图 III 所示。开始时,样品静止, 用上述特定波长的强激光持续均匀垂直照射该样品涂层表面。此后保持激光方向始终不变, 样品绕 $\\mathrm{OO}^{\\prime}$ 轴转动直至稳定。涂层表面始终被激光完全照射。不计激光对样品侧面的照射。设硅基片厚度为 $d^{\\prime}$ 、密度为 $\\rho^{\\prime}$, 氮化镓薄膜的厚度为 $d$ 、密度为 $\\rho$, 涂层质量可忽略, 真空的介电常量 $\\varepsilon_{0}$, 重力加速度大小为 $g$ 。若样品稳定后相对于光照前原位形的转角为 $\\alpha$, 求所用强激光的电场强度有效值 $E$ 。\n\n[图2]",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n这里是一些可能会帮助你解决问题的先验信息提示：\n用薄膜制备技术在某均质硅基片上沉积一层均匀等厚氮化镓薄膜,制备出一个硅基氮化镓样品, 如图 I 所示。\n\n[图1]\n\n问题:\n当用波长范围为 $450 \\sim 1200 \\mathrm{~nm}$ 的光垂直均匀照射该\n\n样品氮化镓表面, 观察到其反射光谱仅有两种波长的光获得最大相干加强, 其中之一波长为 $600 \\mathrm{~nm}$; 氮化镓的折射率 $n$ 与入射光在真空中波长 $\\lambda$ (单位 $\\mathrm{nm}$ ) 之间的关系（色散关系）为\n\n$$\nn^{2}=2.26^{2}+\\frac{330.1^{2}}{\\lambda^{2}-265.7^{2}}\n$$\n\n在该样品硅基片的另一面左、右对称的两个半面上分别均匀沉积一光谱选择性材料涂层,如图 II 所示;对某种特定波长的光, 左半面涂层全吸收, 右半面全反射。用两根长均为 $a$ 的轻细线坚直悬挂该样品, 样品长为 $a$ 、宽为 $b$, 可绕过其中心的光滑坚直固定轴 $\\mathrm{OO}^{\\prime}$ 转动, 也可上下移动, 如图 III 所示。开始时,样品静止, 用上述特定波长的强激光持续均匀垂直照射该样品涂层表面。此后保持激光方向始终不变, 样品绕 $\\mathrm{OO}^{\\prime}$ 轴转动直至稳定。涂层表面始终被激光完全照射。不计激光对样品侧面的照射。设硅基片厚度为 $d^{\\prime}$ 、密度为 $\\rho^{\\prime}$, 氮化镓薄膜的厚度为 $d$ 、密度为 $\\rho$, 涂层质量可忽略, 真空的介电常量 $\\varepsilon_{0}$, 重力加速度大小为 $g$ 。若样品稳定后相对于光照前原位形的转角为 $\\alpha$, 求所用强激光的电场强度有效值 $E$ 。\n\n[图2]\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a47de6806e8da0a0f86dg-04.jpg?height=71&width=423&top_left_y=1238&top_left_x=1225",
            "https://cdn.mathpix.com/cropped/2024_03_31_a47de6806e8da0a0f86dg-04.jpg?height=886&width=488&top_left_y=1675&top_left_x=1166",
            "https://cdn.mathpix.com/cropped/2024_03_31_a47de6806e8da0a0f86dg-27.jpg?height=608&width=623&top_left_y=1492&top_left_x=1139"
        ],
        "answer": [
            "$2\\left[\\frac{\\left(\\rho^{\\prime} d^{\\prime}+\\rho d\\right) g}{\\varepsilon_{0}} \\frac{\\sin \\frac{\\alpha}{2}}{\\cos ^{2} \\alpha}\\right]^{1 / 2}$"
        ],
        "solution": "设频率为 $v$ 的单色光, 每秒垂直入射到物体表面每平方米上的光子数为 $N$ 。每个光子传给物体的动量为 $\\frac{h v}{c}$ ( $h$ 为普朗克常量, $c$ 是真空中的光速), 若入射光子全被物体吸收,则该表面每平方米在每秒内所获得的动量即光压 $P_{0}$ 为\n\n$$\nP_{0}=N \\frac{h v}{c}\n$$\n\n每个光子传给物体的能量为 $h v$, 每秒垂直入射到全吸收物体表面每平方米上的能量为\n\n$$\nN h v=\\bar{w} C\n$$\n\n式中 $\\bar{w}$ 为电磁波平均能量密度\n\n$$\n\\bar{w}=\\varepsilon_{0} E^{2}\n$$\n\n由以上三式得, 频率为 $v$ 的单色光正入射到全吸收表面时对该表面的光压 $P_{0}$ 为\n\n$$\nP_{0}=\\bar{w}=\\varepsilon_{0} E^{2}\n$$\n\n辐射压强对样品的合力矩为\n\n$$\n\\begin{aligned}\nL_{p} & =\\left(2 P_{0} \\cdot \\frac{a b}{2} \\cdot \\cos \\alpha\\right) \\cdot \\frac{a}{4} \\cdot \\cos \\alpha-\\left(P_{0} \\cdot \\frac{a b}{2} \\cdot \\cos \\alpha\\right) \\cdot \\frac{a}{4} \\cdot \\cos \\alpha \\\\\n& =P_{0} \\cdot \\frac{a^{2} b}{8} \\cdot \\cos ^{2} \\alpha\n\\end{aligned}\n$$\n\n如图, 由几何关系可知\n\n$$\nA C=a \\sin \\frac{\\alpha}{2}\n$$\n\n由此得\n\n$$\n\\beta=\\frac{\\alpha}{2}\n$$\n\n样品在坚直方向受力平衡\n\n$$\n2 N \\cos \\beta=M g\n$$\n\n式中 $N$ 为细绳的拉力, $M$ 为基片质量 $m^{\\prime}$ 和氮化镓薄膜质量 $m$ 的和\n\n[图3]\n\n$$\nM=m^{\\prime}+m=\\left(\\rho^{\\prime} d^{\\prime}+\\rho d\\right) a b\n$$\n\n轻细线拉力 $N$ 提供给样品转矩为\n\n$$\nL_{N}=2(N \\sin \\beta) \\cdot\\left(\\frac{a}{2} \\cos \\frac{\\alpha}{2}\\right)\n$$\n\n根据样品所受力矩平衡条件\n\n$$\nL_{N}=L_{p}\n$$\n\n将(13)(14)(15)(16)式代入上式得\n\n$$\n\\left(\\rho^{\\prime} d^{\\prime}+\\rho d\\right) a b \\cdot g \\cdot \\frac{a}{2} \\sin \\frac{\\alpha}{2}=P_{0} \\cdot \\frac{a^{2} b}{8} \\cdot \\cos ^{2} \\alpha\n$$\n\n由(11)(17)式得\n\n$$\n\\frac{\\sin \\frac{\\alpha}{2}}{\\cos ^{2} \\alpha}=\\frac{\\varepsilon_{0} E^{2}}{4\\left(\\rho^{\\prime} d^{\\prime}+\\rho d\\right) g}\n$$\n\n即\n\n$$\nE=2\\left[\\frac{\\left(\\rho^{\\prime} d^{\\prime}+\\rho d\\right) g}{\\varepsilon_{0}} \\frac{\\sin \\frac{\\alpha}{2}}{\\cos ^{2} \\alpha}\\right]^{1 / 2}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_656",
        "problem": "[figure1]\n\nA block accelerates uniformly along a straight line from $A$ to $C$ as shown above. Its speed at point $A$ is $v_{A}$ and that at point $C$ is $v_{C}$. What is its speed at point $B$, which is the mid-point between point $A$ and $C$ ?\nA: $\\frac{2 v_{A} v_{C}}{v_{A}+v_{C}}$\nB: $\\frac{v_{A}+v_{C}}{2}$\nC: $\\sqrt{\\frac{v_{A}^{2}+v_{C}^{2}}{2}}$\nD: $\\sqrt{v_{A} v_{C}}$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nA block accelerates uniformly along a straight line from $A$ to $C$ as shown above. Its speed at point $A$ is $v_{A}$ and that at point $C$ is $v_{C}$. What is its speed at point $B$, which is the mid-point between point $A$ and $C$ ?\n\nA: $\\frac{2 v_{A} v_{C}}{v_{A}+v_{C}}$\nB: $\\frac{v_{A}+v_{C}}{2}$\nC: $\\sqrt{\\frac{v_{A}^{2}+v_{C}^{2}}{2}}$\nD: $\\sqrt{v_{A} v_{C}}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_a0d0ab960474d1b0609cg-04.jpg?height=87&width=618&top_left_y=447&top_left_x=298"
        ],
        "answer": [
            "C"
        ],
        "solution": "The speeds at point $\\mathrm{B}$ and $\\mathrm{C}$ are given by\n$$\n\\begin{aligned}\n& v_B^2=v_A^2+2 a s \\\\\n& v_C^2=v_B^2+2 a s\n\\end{aligned}\n$$\nwhere $a$ is the acceleration and $s$ is the distance travelled between the neighboring points.\nEliminating 2 as from the system of equations above, we can get\n$$\nv_B=\\sqrt{\\frac{v_A^2+v_C^2}{2}}\n$$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_560",
        "problem": "In general, whenever an electric and a magnetic field are at an angle to each other, energy is transferred; for example, this principle is the reason electromagnetic radiation transfers energy. The power transferred per unit area is given by the Poynting vector:\n\n$$\n\\vec{S}=\\frac{1}{\\mu_{0}} \\vec{E} \\times \\vec{B}\n$$\n\nIn each part of this problem, the last subpart asks you to verify that the rate of energy transfer agrees with the formula for the Poynting vector. Therefore, you should not use the formula for the Poynting vector before the last subpart!\n\nA long, insulating cylindrical rod has radius $R$ and carries a uniform volume charge density $\\rho$. A uniform external electric field $E$ exists in the direction of its axis. The rod moves in the direction of its axis at speed $v$.\n\nCompute the Poynting vector, draw its direction on a diagram, and verify that it agrees with the rate of energy transfer.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nIn general, whenever an electric and a magnetic field are at an angle to each other, energy is transferred; for example, this principle is the reason electromagnetic radiation transfers energy. The power transferred per unit area is given by the Poynting vector:\n\n$$\n\\vec{S}=\\frac{1}{\\mu_{0}} \\vec{E} \\times \\vec{B}\n$$\n\nIn each part of this problem, the last subpart asks you to verify that the rate of energy transfer agrees with the formula for the Poynting vector. Therefore, you should not use the formula for the Poynting vector before the last subpart!\n\nA long, insulating cylindrical rod has radius $R$ and carries a uniform volume charge density $\\rho$. A uniform external electric field $E$ exists in the direction of its axis. The rod moves in the direction of its axis at speed $v$.\n\nCompute the Poynting vector, draw its direction on a diagram, and verify that it agrees with the rate of energy transfer.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": [
            "$\\pi R^{2} \\rho v E$"
        ],
        "solution": "The electric and magnetic fields are perpendicular, so the Poynting vector has magnitude\n\n$$\nS=\\frac{1}{\\mu_{0}} E B=\\frac{1}{2} R \\rho v E\n$$\n\nA quick application of the right hand rule indicates that it points inward along the surface of the cylinder, as it ought. The cylinder has area per unit length $2 \\pi r$, so the rate of energy transfer per unit length is\n\n$$\n\\mathcal{P}=2 \\pi r S=\\pi R^{2} \\rho v E\n$$\n\nin agreement with the previous result.",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_385",
        "problem": "The power radiated in gravitational waves by an orbiting binary system is given by $P\\left(r, m_{1}, m_{2}\\right)=\\frac{32}{5} \\frac{G^{4}}{c^{5}} \\frac{\\left(m_{1} m_{2}\\right)^{2}\\left(m_{1}+m_{2}\\right)}{r^{5}}$ where $r$ is the distance between the centers of the two orbiting masses $m_{1}$ and $m_{2}$. It is known that the most compact object is a black hole. The size of a black hole is defined by its Schwarzschild radius $r_{s}=\\frac{2 G m}{c^{2}}$, where $m$ is the mass of i) (2 points) Estimate the upper limit of the power that can ever be emitted in gravitational waves by an orbiting binary system.\n\nThe gravitational wave detectors on Earth function by measuring the so called gravity wave strain $\\varepsilon(t)$ over time, which characterizes the deformation of spacetime. Data processing then yields the maximum strain $\\varepsilon$ and its corresponding wave frequency $f$. With the help of a theoretical spacetime model, the energy density $u$ associated with the wave can then be determined. We will use the analogy of linear elasticity to examine this model.\n\nDerive the energy density $u=$ $u(\\varepsilon, E)$ in a uniformly stretched elastic band in terms of the strain $\\varepsilon$ and the Elastic (Young's) modulus $E$.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe power radiated in gravitational waves by an orbiting binary system is given by $P\\left(r, m_{1}, m_{2}\\right)=\\frac{32}{5} \\frac{G^{4}}{c^{5}} \\frac{\\left(m_{1} m_{2}\\right)^{2}\\left(m_{1}+m_{2}\\right)}{r^{5}}$ where $r$ is the distance between the centers of the two orbiting masses $m_{1}$ and $m_{2}$. It is known that the most compact object is a black hole. The size of a black hole is defined by its Schwarzschild radius $r_{s}=\\frac{2 G m}{c^{2}}$, where $m$ is the mass of i) (2 points) Estimate the upper limit of the power that can ever be emitted in gravitational waves by an orbiting binary system.\n\nThe gravitational wave detectors on Earth function by measuring the so called gravity wave strain $\\varepsilon(t)$ over time, which characterizes the deformation of spacetime. Data processing then yields the maximum strain $\\varepsilon$ and its corresponding wave frequency $f$. With the help of a theoretical spacetime model, the energy density $u$ associated with the wave can then be determined. We will use the analogy of linear elasticity to examine this model.\n\nDerive the energy density $u=$ $u(\\varepsilon, E)$ in a uniformly stretched elastic band in terms of the strain $\\varepsilon$ and the Elastic (Young's) modulus $E$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of \\mathrm{~W}, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": [
            "4.5 \\times 10^{50}"
        ],
        "solution": "Black holes will merge at this distance.\n\n$r_{\\text {min }}=r_{s 1}+r_{s 2}=\\frac{2 G\\left(m_{1}+m_{2}\\right)}{c^{2}}$\n\nSubstituting for $r$ in the power equation, we get the maximum power for a system with masses $m_{1}$ and $m_{2}$\n\n$P_{\\text {max }}\\left(m_{1}, m_{2}\\right)=\\frac{1}{5} \\frac{c^{5}}{G} \\frac{\\left(m_{1} m_{2}\\right)^{2}}{\\left(m_{1}+m_{2}\\right)^{4}}$\n\nNow let us consider the effect of masses. Let us define $a:=\\frac{\\left(m_{1} m_{2}\\right)^{2}}{\\left(m_{1}+m_{2}\\right)^{4}}$. The intuition tells us, that more power should be radiated off a bigger system, thus the masses should tend to infinity for maximum power\n\nHowever the limit $\\lim _{m_{1}=m_{2} \\rightarrow \\infty} a=1$ suggests otherwise. So does the dimensional analysis: the dimension of $[a]=\\frac{(M \\cdot M)^{2}}{M^{4}}=1$ means that the absolute values of the masses don't contribute to the equation. This is intuitively explained by the proportionally increasing Schwarzschild radii.\n\nHowever the mass distribution does matter, since $a$ isn't the same for all $m_{1}, m_{2}$ either. Let us determine the maximum value it can take. Let us redefine it in terms of the total mass $M:=m_{1}+m_{2}$ and the mass distribution $k:=\\frac{m_{1}}{M}$. Then\n\n$m_{1}=k M$\n\n$m_{2}=(1-k) M$\n\n$\\Rightarrow a=\\frac{\\left(m_{1} m_{2}\\right)^{2}}{\\left(m_{1}+m_{2}\\right)^{4}}=\\frac{(k M(k-1) M)^{2}}{M^{4}}=[k(k-1)]^{2}$\n\n$\\frac{d}{d k}[k(k-1)]^{2}=0 \\Leftrightarrow k=\\frac{1}{2} \\Leftrightarrow m_{1}=m_{2}$\n\n$\\Rightarrow a_{\\text {max }}=\\left[\\frac{1}{2}\\left(1-\\frac{1}{2}\\right)\\right]^{2}=\\frac{1}{16}$\n\nSo that finally\n\n$P_{\\max }=\\frac{1}{5} \\frac{c^{5}}{G} \\frac{1}{16}=\\frac{1}{80} \\frac{c^{5}}{G}=4.5 \\times 10^{50} \\mathrm{~W}$\n\nInterestingly enough, the peak power emitted in gravitational radiation by any merging binary system will be the same. $\\frac{c^{5}}{G}$ is also called luminosity of the universe.",
        "answer_type": "NV",
        "unit": [
            "\\mathrm{~W}"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_1399",
        "problem": "木星是太阳系内质量最大的行星（其质量约为地球的 318 倍）。假设地球与木星均沿圆轨道绕太阳转动, 两条轨道在同一平面内。将太阳、地球和木星都视为质点, 忽略太阳系内其它星体的引力; 且地球和木星之间的引力在有太阳时可忽略。已知太阳和木星质量分别为 $m_{\\mathrm{s}}$ 和 $m_{\\mathrm{j}}$, 引力常量为 $G$ 。地球和木星绕太阳运行的轨道半径分别是 $r_{\\mathrm{e}}$ 和 $r_{\\mathrm{j}}$ 。假设在某个时刻, 地球与太阳的连线和木星与太阳的连线之间的夹角为 $\\theta$ 。这时若太阳质量突然变为零, 求此时地球相对木星的速度大小 $v_{\\mathrm{ej}}$ 和地球不被木星引力俘获所需要的最小速率 $v_{0}$",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n这里是一些可能会帮助你解决问题的先验信息提示：\n木星是太阳系内质量最大的行星（其质量约为地球的 318 倍）。假设地球与木星均沿圆轨道绕太阳转动, 两条轨道在同一平面内。将太阳、地球和木星都视为质点, 忽略太阳系内其它星体的引力; 且地球和木星之间的引力在有太阳时可忽略。已知太阳和木星质量分别为 $m_{\\mathrm{s}}$ 和 $m_{\\mathrm{j}}$, 引力常量为 $G$ 。地球和木星绕太阳运行的轨道半径分别是 $r_{\\mathrm{e}}$ 和 $r_{\\mathrm{j}}$ 。假设在某个时刻, 地球与太阳的连线和木星与太阳的连线之间的夹角为 $\\theta$ 。这时若太阳质量突然变为零, 求\n\n问题:\n此时地球相对木星的速度大小 $v_{\\mathrm{ej}}$ 和地球不被木星引力俘获所需要的最小速率 $v_{0}$\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": [
            "$\\frac{\\left(2 G m_{\\mathrm{j}}\\right)^{1 / 2}}{\\left(r_{\\mathrm{e}}^{2}+r_{\\mathrm{j}}^{2}-2 r_{\\mathrm{e}} r_{\\mathrm{j}} \\cos \\theta\\right)^{1 / 4}}$"
        ],
        "solution": "若太阳质量突然变为零, 地球和木星围绕太阳转动速度不会突然改变, 因而应当等于在太阳质量变为零之前的瞬间, 地球和木星围绕太阳转动的速度。设在太阳质量变为零之前, 地球和木星绕太阳转动速度分别是 $v_{\\mathrm{e}}$ 和 $v_{\\mathrm{j}}$ 。以太阳为原点、地球和木星公转轨道平面为 $x-y$ 平面建立坐标系。由万有引力定律和牛顿第二定律有\n\n$$\nG \\frac{m_{\\mathrm{s}} m_{\\mathrm{e}}}{r_{\\mathrm{e}}^{2}}=m_{\\mathrm{e}} \\frac{v_{\\mathrm{e}}^{2}}{r_{\\mathrm{e}}}\n$$\n\n由(1)式得\n\n$$\nv_{\\mathrm{e}}=\\sqrt{\\frac{G m_{\\mathrm{s}}}{r_{\\mathrm{e}}}}\n$$\n\n同理有\n\n$$\nv_{\\mathrm{j}}=\\sqrt{\\frac{G m_{\\mathrm{s}}}{r_{\\mathrm{j}}}}\n$$\n\n现计算地球不被木星引力俘获所需要的最小速率 $v_{0}$ （不考虑太阳引力）。若地球相对木星刚好以速度 $v_{0}$ 运动, 也就是说, 当地球在木星的引力场里运动到无限远时, 速度刚好为零, 此时木\n星-地球系统引力势能为零, 动能也为零, 即总机械能为零。按机械能守恒定律, 在地球离木星距离为 $r_{\\mathrm{ej}}$ 时, 速度 $v_{0}$ 满足\n\n$$\n\\frac{1}{2} m_{\\mathrm{e}} v_{0}^{2}-G \\frac{m_{\\mathrm{e}} m_{\\mathrm{j}}}{r_{\\mathrm{ej}}}=0\n$$\n\n即\n\n$$\nv_{0}=\\sqrt{\\frac{2 G m_{\\mathrm{j}}}{r_{\\mathrm{ej}}}}\n$$\n\n可见, 地球不被木星引力俘获所需要的最小速率 $v_{0}$ 的大小与木星质量和地球离木星的距离有关。\n\n设在太阳质量变为零的瞬间, 木星的位矢为\n\n$$\nr_{j}=\\left(r_{j}, 0\\right)\n$$\n\n地球的位矢为\n\n$$\n\\boldsymbol{r}_{\\mathrm{e}}=\\left(r_{\\mathrm{e}} \\cos \\theta, r_{\\mathrm{e}} \\sin \\theta\\right),\n$$\n\n式中 $\\theta$ 为地球此时的位矢与 $x$-轴的夹角。此时地球和木星的距离为\n\n$$\nr_{\\mathrm{ej}}=\\sqrt{r_{\\mathrm{e}}^{2}+r_{\\mathrm{j}}^{2}-2 r_{\\mathrm{e}} r_{\\mathrm{j}} \\cos \\theta}\n$$\n\n此时地球相对于木星的速度大小为\n\n$$\nv_{\\mathrm{ej}}=\\left|v_{\\mathrm{e}}-v_{\\mathrm{j}}\\right|=\\sqrt{v_{\\mathrm{e}}^{2}+v_{\\mathrm{j}}^{2}-2 v_{\\mathrm{e}} v_{\\mathrm{j}} \\cos \\theta}=\\sqrt{G m_{\\mathrm{s}}} \\sqrt{\\frac{1}{r_{\\mathrm{e}}}+\\frac{1}{r_{\\mathrm{j}}}-\\frac{2 \\cos \\theta}{\\sqrt{r_{\\mathrm{e}} r_{\\mathrm{j}}}}}\n$$\n\n式中 $\\cos \\theta$ 项前面取减号是因为考虑到木星和地球同方向绕太阳旋转的缘故。由(5)(8)式得\n\n$$\nv_{0}=\\sqrt{\\frac{2 G m_{\\mathrm{j}}}{r_{\\mathrm{ej}}}}=\\frac{\\left(2 G m_{\\mathrm{j}}\\right)^{1 / 2}}{\\left(r_{\\mathrm{e}}^{2}+r_{\\mathrm{j}}^{2}-2 r_{\\mathrm{e}} r_{\\mathrm{j}} \\cos \\theta\\right)^{1 / 4}},\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Physics_1672",
        "problem": "如图, 在磁感应强度大小为 $B$ 、方向坚直向上的匀强磁场中, 有一均质刚性导电的正方形线框 abcd, 线框质量为 $m$ ，边长为 $l$, 总电阻为 $R$ 。线框可绕通过 ad 边和 bc 边中点的光滑轴 $\\mathrm{OO}^{\\prime}$ 转动。 $\\mathrm{P} 、 \\mathrm{Q}$ 点是线框引线的两端, $\\mathrm{OO}^{\\prime}$ 轴和 X 轴位于同一水平面内, 且相互垂直。不考虑线框自感。\n\n[图1]$t=0$ 时, 线框静止, 其所在平面与 $\\mathrm{X}$ 轴有一很小的夹角 $\\theta_{0}$, 此时给线框通以大小为 $I$的恒定直流电流, 方向沿 $\\mathrm{P} \\rightarrow \\mathrm{a} \\rightarrow \\mathrm{b} \\rightarrow \\mathrm{c} \\rightarrow \\mathrm{d} \\rightarrow \\mathrm{Q}$, 求此后线框所在平面与 $\\mathrm{X}$ 轴的夹角 $\\theta$ 、线框转动的角速度 $\\dot{\\theta}$ 和角加速度 $\\ddot{\\theta}$ 随时间变化的关系式;",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个方程。\n这里是一些可能会帮助你解决问题的先验信息提示：\n如图, 在磁感应强度大小为 $B$ 、方向坚直向上的匀强磁场中, 有一均质刚性导电的正方形线框 abcd, 线框质量为 $m$ ，边长为 $l$, 总电阻为 $R$ 。线框可绕通过 ad 边和 bc 边中点的光滑轴 $\\mathrm{OO}^{\\prime}$ 转动。 $\\mathrm{P} 、 \\mathrm{Q}$ 点是线框引线的两端, $\\mathrm{OO}^{\\prime}$ 轴和 X 轴位于同一水平面内, 且相互垂直。不考虑线框自感。\n\n[图1]\n\n问题:\n$t=0$ 时, 线框静止, 其所在平面与 $\\mathrm{X}$ 轴有一很小的夹角 $\\theta_{0}$, 此时给线框通以大小为 $I$的恒定直流电流, 方向沿 $\\mathrm{P} \\rightarrow \\mathrm{a} \\rightarrow \\mathrm{b} \\rightarrow \\mathrm{c} \\rightarrow \\mathrm{d} \\rightarrow \\mathrm{Q}$, 求此后线框所在平面与 $\\mathrm{X}$ 轴的夹角 $\\theta$ 、线框转动的角速度 $\\dot{\\theta}$ 和角加速度 $\\ddot{\\theta}$ 随时间变化的关系式;\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个方程，例如ANSWER=\\frac{x^2}{4}+\\frac{y^2}{2}=1",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a47de6806e8da0a0f86dg-02.jpg?height=440&width=674&top_left_y=1322&top_left_x=1222"
        ],
        "answer": [
            "$$ \\ddot{\\theta}=-\\theta_{0} \\frac{6 B I}{m} \\cos \\sqrt{\\frac{6 B I}{m}} t $$"
        ],
        "solution": "当线框中通过电流 $I$ 时, $\\mathrm{ab}$ 和 $\\mathrm{cd}$ 两边受到大小相等、方向分别指向 $\\mathrm{X}$ 轴正向和反向的安培力。设线框的 $\\mathrm{ab}$ 边转到与 $\\mathrm{X}$ 轴的夹角为 $\\theta$ 时其角速度为 $\\dot{\\theta}$, 角加速度为 $\\ddot{\\theta}, \\mathrm{ab}$ 边和 $\\mathrm{cd}$ 边的线速度则为 $v=\\frac{l}{2} \\dot{\\theta}$, 所受安培力大小为 $F=B I I$, 线框所受力偶矩为\n\n$$\nM=-2 \\cdot \\frac{l}{2} B l I \\sin \\theta \\approx-B l^{2} I \\theta\n$$\n\n式中已利用了小角近似 $\\sin \\theta \\approx \\theta$, 而负号表示力偶矩与线框角位移 $\\theta$ 方向相反。根据刚体转动定理, 有\n\n$$\nJ \\ddot{\\theta}=-B l^{2} I \\theta\n$$\n\n此方程与单摆的动力学方程在形式上完全一致, 所以线框将做简谐运动, 可设\n\n$$\n\\theta=\\theta_{0} \\cos \\omega t\n$$\n\n将其代入(3)式得\n\n$$\n\\omega=l \\sqrt{\\frac{B I}{J}}=\\sqrt{\\frac{6 B I}{m}}\n$$\n\n于是, 角位移 $\\theta$ 随时间 $t$ 而变化的关系式为\n\n$$\n\\theta=\\theta_{0} \\cos \\sqrt{\\frac{6 B I}{m}} t\n$$\n\n角速度 $\\dot{\\theta}$ 随时间 $t$ 而变化的关系式为\n\n$$\n\\dot{\\theta}=-\\theta_{0} \\sqrt{\\frac{6 B I}{m}} \\sin \\sqrt{\\frac{6 B I}{m}} t\n$$\n\n角加速度 $\\ddot{\\theta}$ 随时间 $t$ 而变化的关系式为\n\n$$\n\\ddot{\\theta}=-\\theta_{0} \\frac{6 B I}{m} \\cos \\sqrt{\\frac{6 B I}{m}} t\n$$\n\n由(6)(7)(8)式知, 当 $t=\\bar{t}=\\frac{\\pi}{2} \\sqrt{\\frac{m}{6 B I}}$ 时, $\\theta(\\bar{t})=0, \\dot{\\theta}(\\bar{t})=-\\theta_{0} \\sqrt{\\frac{6 B I}{m}}$ 达到反向最大值, $\\ddot{\\theta}(\\bar{t})=0$ 。可见, 线框的运动是以 $\\theta=0$ 为平衡位置、圆频率 $\\omega=\\sqrt{\\frac{6 B I}{m}}$ 、周期 $T=2 \\pi \\sqrt{\\frac{m}{6 B I}}$ 、振幅为 $\\theta_{0}$的简谐运动。",
        "answer_type": "EQ",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_188",
        "problem": "A uniform stick of mass $m$ is originally on a horizontal surface. One end is attached to a vertical rope, which pulls up with a constant tension force $F$ so that the center of the mass of the stick moves upward with acceleration $a<g$. The normal force $N$ of the ground on the other end of the stick shortly after the right end of the stick leaves the surface satisfies\n\n[figure1]\nA: $N=m g$\nB: $m g>N>m g / 2 $ \nC: $N=m g / 2$\nD: $m g / 2>N>0$\nE: $N=0$\n",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA uniform stick of mass $m$ is originally on a horizontal surface. One end is attached to a vertical rope, which pulls up with a constant tension force $F$ so that the center of the mass of the stick moves upward with acceleration $a<g$. The normal force $N$ of the ground on the other end of the stick shortly after the right end of the stick leaves the surface satisfies\n\n[figure1]\n\nA: $N=m g$\nB: $m g>N>m g / 2 $ \nC: $N=m g / 2$\nD: $m g / 2>N>0$\nE: $N=0$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_6e7e781e6599f99f638bg-07.jpg?height=358&width=448&top_left_y=461&top_left_x=833"
        ],
        "answer": [
            "B"
        ],
        "solution": "Solution\n\nNewton's second law for the vertical motion gives\n\n$$\nF+N-m g=m a\n$$\n\nNow take torques about the left end of the stick. We have\n\n$$\n\\tau=F L-\\frac{m g L}{2}, \\quad I=\\frac{1}{3} m L^{2}, \\quad \\alpha=\\frac{a}{L / 2}\n$$\n\nSolving these two equations for $N$ gives\n\n$$\nN=\\frac{1}{3} m a+\\frac{1}{2} m g\n$$\n\nwhich for the range of $a$ given satisfies choice B.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_1665",
        "problem": "如图, 一质量分布均匀、半径为 $r$ 的刚性薄圆环落到粗糙的水平地面前的瞬间,圆环质心速度 $v_{0}$ 与坚直方向成 $\\theta\\left(\\frac{\\pi}{2}<\\theta<\\frac{3 \\pi}{2}\\right)$ 角, 并同时以\n\n[图1]\n$\\omega_{0}$ ( $\\omega_{0}$ 的正方向如图中箭头所示）绕通过其质心 $\\mathrm{O}$ 、且垂直环面的轴转动。已知圆环仅在其所在的坚直平面内运动, 在弹起前刚好与地面无相对滑动, 圆环与地面碰撞的恢复系数为 $k$,重力加速度大小为 $g$ 。忽略空气阻力。求使圆环在与地面碰后能坚直弹起的条件和在此条件下圆环能上升的最大高度",
        "prompt": "你正在参加一个国际物理竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n这里是一些可能会帮助你解决问题的先验信息提示：\n如图, 一质量分布均匀、半径为 $r$ 的刚性薄圆环落到粗糙的水平地面前的瞬间,圆环质心速度 $v_{0}$ 与坚直方向成 $\\theta\\left(\\frac{\\pi}{2}<\\theta<\\frac{3 \\pi}{2}\\right)$ 角, 并同时以\n\n[图1]\n$\\omega_{0}$ ( $\\omega_{0}$ 的正方向如图中箭头所示）绕通过其质心 $\\mathrm{O}$ 、且垂直环面的轴转动。已知圆环仅在其所在的坚直平面内运动, 在弹起前刚好与地面无相对滑动, 圆环与地面碰撞的恢复系数为 $k$,重力加速度大小为 $g$ 。忽略空气阻力。\n\n问题:\n求使圆环在与地面碰后能坚直弹起的条件和在此条件下圆环能上升的最大高度\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_e3a9fdbbef225ad3aefbg-01.jpg?height=314&width=363&top_left_y=2576&top_left_x=1326"
        ],
        "answer": [
            "\\frac{k^{2}\\left(v_{0}^{2}-r^{2} \\omega_{0}^{2}\\right)}{2 g}"
        ],
        "solution": "若圆环与地而碰后能坚直弹起, 则其速度与坚直方向的夹角\n\n$$\n\\beta=0\n$$\n\n将上式代入(6)式得, 使圆环在与地面碰后能坚直弹起的条件为\n\n$$\n\\sin \\theta=-\\frac{r \\omega_{0}}{v_{0}}\n$$\n\n在此条件下，在与地面刚刚碰后的瞬间有\n\n$$\n\\omega=0, \\quad v=-v_{0} k \\cos \\theta\n$$\n\n即圆环做坚直上抛运动。圆环上升的最大高度为\n\n$$\nh=\\frac{v^{2}}{2 g}=\\frac{v_{0}^{2} k^{2} \\cos ^{2} \\theta}{2 g}=\\frac{k^{2}\\left(v_{0}^{2}-r^{2} \\omega_{0}^{2}\\right)}{2 g}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_829",
        "problem": "Consider a gas of ultra relativistic electrons $(\\gamma \\gg 1)$, with an isotropic distribution of velocities (does not depend on direction). The proper number density of particles with energies between $\\epsilon$ and $\\epsilon+d \\epsilon$ is given by $f(\\epsilon) d \\epsilon$, where $\\epsilon$ is the energy per particle. Consider also a wall of area $\\Delta A$, which is in contact with the gas.\n\nDerive an equation of state for an ultra relativistic electron gas, relating the $0.6 \\mathrm{pt}$ pressure, volume and total internal energy.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an equation.\n\nproblem:\nConsider a gas of ultra relativistic electrons $(\\gamma \\gg 1)$, with an isotropic distribution of velocities (does not depend on direction). The proper number density of particles with energies between $\\epsilon$ and $\\epsilon+d \\epsilon$ is given by $f(\\epsilon) d \\epsilon$, where $\\epsilon$ is the energy per particle. Consider also a wall of area $\\Delta A$, which is in contact with the gas.\n\nDerive an equation of state for an ultra relativistic electron gas, relating the $0.6 \\mathrm{pt}$ pressure, volume and total internal energy.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an equation, e.g. ANSWER=\\frac{x^2}{4}+\\frac{y^2}{2}=1",
        "figure_urls": null,
        "answer": [
            "$E=3 P V$"
        ],
        "solution": "As the remaining integral in the expression above was identified as the energy per volume in B1, $\\Delta p_{z}=\\Delta t \\Delta A \\frac{1}{3} \\frac{E}{V}$. The pressure is the force per area normal to the wall, so $P=\\frac{\\Delta p_{z}}{\\Delta t} \\frac{1}{\\Delta A}$. Combining these gives $P=\\frac{E}{3 V}$, or $E=3 P V$, which is the equation of state.",
        "answer_type": "EQ",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Physics_844",
        "problem": "## Introduction\n\nActive galactic nuclei (AGN) are supermassive black holes which form the centres of galaxies, and emit large amounts of energy in radiation and particle flows. One feature of many AGN are jetted outflows, which can be observed through radio emission, and sometimes also in other parts of the electromagnetic spectrum, including $x$-rays. These jets are large flows of plasma at relativistic speeds, over lengths of order $10^{20} \\mathrm{~m}$, which is tens of thousands of light years. The $\\mathrm{x}$-ray emission from jets is usually dominated by synchrotron emission from relativistic electrons gyrating in the magnetic field of the jet.\n\n[figure1]\n\nFigure 1: X-ray image of the jet from the Centaurus A AGN. Darker regions represent regions of higher intensity $x$-rays. Brighter regions within the fainter jet are called knots. (Snios et al., 2019)\n\nA simple model of the flow of jets assumes that the flow is steady and directed radially away from the central AGN, so approximately one dimensional, and that the plasma in the jet is in pressure equilibrium with its surroundings. There is assumed to be a constant rate per volume of mass injected into the jet from stars which lose their outer layers as they move through their life cycle.\n\nThe jet is described in terms of the coordinate representing distance from the AGN, $s$, and the opening radius $r$ of the conical jet. These distances are measured in parsecs, where $1 \\mathrm{pc}=3.086 \\times 10^{16} \\mathrm{~m}$. The speed of the jet flow is assumed to be directed radially away from the central AGN, and be a function of $s$ only. The plasma in the jet is comprised of electrons, protons, and some heavier ionised nuclei. The average energy carried by each particle in the jet, in the reference frame of the bulk flow of the jet (which we will call the jet frame), is $\\epsilon_{\\mathrm{av}}=\\mu_{\\mathrm{pp}} c^{2}+h$, where the term $h$ includes all thermal kinetic energy and potential energies in terms of the pressure $P$ and $n$ is the number density of the plasma.\n\nAs the stars, which the jet flows past, move through their life cycles they can lose part of their atmosphere. This results in a uniform rate of injection of mass per unit volume $\\alpha$ into the jet, and the injected particles are assumed to be at rest relative to the AGN.\n\nThis model can be applied to the Centaurus A jet. Centaurus A is one of the nearest AGN, so it is possible to observe its jet at relatively high spatial resolution. The total power carried by the jet is estimated to be $P_{\\mathrm{j}}=1 \\times 10^{36} \\mathrm{~J} \\cdot \\mathrm{s}^{-1}$. See below for a diagram of a simple geometrical description of the Centaurus A jet, including measurements of some jet parameters. $s_{1}$ is the coordinate of the start of the jet, and $s_{2}$ the coordinate of the end of the jet. In Centuarus A the average mass per particle is $\\mu_{\\mathrm{pp}}=0.59 m_{\\mathrm{p}}$ and $h=\\frac{13}{4} P / n$. The pressure in the plasma surrounding the jet is $P(s)=5.7 \\times 10^{-12}\\left(\\frac{s}{s_{0}}\\right)^{-1.5} \\mathrm{~Pa}$, where $s_{0}=1 \\mathrm{kpc}$.\n\n[figure2]\n\nFigure 2: The Centaurus A jet, showing the geometry compared to the active galactic nucleus (AGN).\n\nThe jet is described by the following parameters, all of which depend on the distance $s$ from the AGN:\n\n- the opening radius of the jet $r(s)$ in the AGN frame\n- the cross sectional area of the jet $A(s)$ in the AGN frame\n- the speed of the jet $v(s)$ in the AGN frame\n- the lorentz gamma factor of the jet $\\gamma(s)$ in the AGN frame\n- the number density $n(s)$ in the frame of the jet\n\nThe power carried by a jet is defined to be the sum of the total bulk kinetic energy flux and the total thermal energy flux, so\n\n$$\nP_{\\mathrm{j}}(s)=F_{\\mathrm{E}}(s)-\\dot{M} c^{2}\n$$\n\nwhere $F_{\\mathrm{E}}(s)$ is the flux of energy through the cross section of the jet at $s$, and $\\dot{M}$ is the mass flux through the jet cross section at the same distance $s$ from the AGN.\n\nWrite the expected relationship between $\\Pi$ and $F_{\\mathrm{Pr}}$. Also, calculate the percent-age difference between the model value of $\\Pi$, which you found in $A 7$, and the expected value.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\n## Introduction\n\nActive galactic nuclei (AGN) are supermassive black holes which form the centres of galaxies, and emit large amounts of energy in radiation and particle flows. One feature of many AGN are jetted outflows, which can be observed through radio emission, and sometimes also in other parts of the electromagnetic spectrum, including $x$-rays. These jets are large flows of plasma at relativistic speeds, over lengths of order $10^{20} \\mathrm{~m}$, which is tens of thousands of light years. The $\\mathrm{x}$-ray emission from jets is usually dominated by synchrotron emission from relativistic electrons gyrating in the magnetic field of the jet.\n\n[figure1]\n\nFigure 1: X-ray image of the jet from the Centaurus A AGN. Darker regions represent regions of higher intensity $x$-rays. Brighter regions within the fainter jet are called knots. (Snios et al., 2019)\n\nA simple model of the flow of jets assumes that the flow is steady and directed radially away from the central AGN, so approximately one dimensional, and that the plasma in the jet is in pressure equilibrium with its surroundings. There is assumed to be a constant rate per volume of mass injected into the jet from stars which lose their outer layers as they move through their life cycle.\n\nThe jet is described in terms of the coordinate representing distance from the AGN, $s$, and the opening radius $r$ of the conical jet. These distances are measured in parsecs, where $1 \\mathrm{pc}=3.086 \\times 10^{16} \\mathrm{~m}$. The speed of the jet flow is assumed to be directed radially away from the central AGN, and be a function of $s$ only. The plasma in the jet is comprised of electrons, protons, and some heavier ionised nuclei. The average energy carried by each particle in the jet, in the reference frame of the bulk flow of the jet (which we will call the jet frame), is $\\epsilon_{\\mathrm{av}}=\\mu_{\\mathrm{pp}} c^{2}+h$, where the term $h$ includes all thermal kinetic energy and potential energies in terms of the pressure $P$ and $n$ is the number density of the plasma.\n\nAs the stars, which the jet flows past, move through their life cycles they can lose part of their atmosphere. This results in a uniform rate of injection of mass per unit volume $\\alpha$ into the jet, and the injected particles are assumed to be at rest relative to the AGN.\n\nThis model can be applied to the Centaurus A jet. Centaurus A is one of the nearest AGN, so it is possible to observe its jet at relatively high spatial resolution. The total power carried by the jet is estimated to be $P_{\\mathrm{j}}=1 \\times 10^{36} \\mathrm{~J} \\cdot \\mathrm{s}^{-1}$. See below for a diagram of a simple geometrical description of the Centaurus A jet, including measurements of some jet parameters. $s_{1}$ is the coordinate of the start of the jet, and $s_{2}$ the coordinate of the end of the jet. In Centuarus A the average mass per particle is $\\mu_{\\mathrm{pp}}=0.59 m_{\\mathrm{p}}$ and $h=\\frac{13}{4} P / n$. The pressure in the plasma surrounding the jet is $P(s)=5.7 \\times 10^{-12}\\left(\\frac{s}{s_{0}}\\right)^{-1.5} \\mathrm{~Pa}$, where $s_{0}=1 \\mathrm{kpc}$.\n\n[figure2]\n\nFigure 2: The Centaurus A jet, showing the geometry compared to the active galactic nucleus (AGN).\n\nThe jet is described by the following parameters, all of which depend on the distance $s$ from the AGN:\n\n- the opening radius of the jet $r(s)$ in the AGN frame\n- the cross sectional area of the jet $A(s)$ in the AGN frame\n- the speed of the jet $v(s)$ in the AGN frame\n- the lorentz gamma factor of the jet $\\gamma(s)$ in the AGN frame\n- the number density $n(s)$ in the frame of the jet\n\nThe power carried by a jet is defined to be the sum of the total bulk kinetic energy flux and the total thermal energy flux, so\n\n$$\nP_{\\mathrm{j}}(s)=F_{\\mathrm{E}}(s)-\\dot{M} c^{2}\n$$\n\nwhere $F_{\\mathrm{E}}(s)$ is the flux of energy through the cross section of the jet at $s$, and $\\dot{M}$ is the mass flux through the jet cross section at the same distance $s$ from the AGN.\n\nWrite the expected relationship between $\\Pi$ and $F_{\\mathrm{Pr}}$. Also, calculate the percent-age difference between the model value of $\\Pi$, which you found in $A 7$, and the expected value.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_2416d49d47cb88c0a72bg-1.jpg?height=651&width=805&top_left_y=1045&top_left_x=631",
            "https://cdn.mathpix.com/cropped/2024_03_14_2416d49d47cb88c0a72bg-2.jpg?height=654&width=1356&top_left_y=821&top_left_x=356"
        ],
        "answer": [
            "40"
        ],
        "solution": "As there are no other forces on the jet, it is expected that $\\Pi=F_{\\mathrm{Pr}}$.\n\nThe $\\%$ deviation is $\\left|\\left(\\Pi-F_{\\operatorname{Pr}}\\right) / F_{\\mathrm{Pr}}\\right| \\approx 40 \\%$",
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Physics_850",
        "problem": "A single electron transistor (SET) consists of a quantum dot, which is a small isolated conductor where electrons can be localised, and of several electrodes in its vicinity. The gate electrode couples capacitatively to the quantum dot, while the two other electrodes --- the source and the drain --- are connected via tunnel junctions, through which electrons can tunnel due to quantum mechanics. A simplified circuit diagram for an SET is shown in the figure.\n[figure1]\n\nCircuit diagram representation of an SET. QD is the quantum dot, $\\mathrm{S}$ is the source, $\\mathrm{D}$ is the drain and $\\mathrm{G}$ is the gate.\n\nThe capacitance of the gate is $C_{g}$ and the capacitance of the tunnel junctions is $C_{t} \\ll C_{g}$. Consider $C_{g}$ to be the total capacitance of the quantum dot. In this part of the problem, the source and the drain are held at zero potential, and the voltage on the gate electrode is fixed at $V_{g}$.\n\nConsider a state of the SET in which the quantum dot contains $n$ electrons.\n\nIf $\\Delta E_{n}<0$ then electrons will spontaneously tunnel into the quantum dot until such a number $\\mathcal{N}>n$ is reached that $\\Delta E_{\\mathcal{N}} \\geq 0$. The equilibrium number of electrons $\\mathcal{N}$ and the corresponding addition energy $\\Delta E_{\\mathcal{N}}$ can be controlled by choosing the appropriate voltage $V_{g}$.\n\nFind an expression for the maximal possible value $E_{c}=\\max \\Delta E_{\\mathcal{N}}\\left(V_{g}\\right)$ of the  equilibrium addition energy that can be achieved by tuning the gate voltage of the SET.",
        "prompt": "You are participating in an international Physics competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nA single electron transistor (SET) consists of a quantum dot, which is a small isolated conductor where electrons can be localised, and of several electrodes in its vicinity. The gate electrode couples capacitatively to the quantum dot, while the two other electrodes --- the source and the drain --- are connected via tunnel junctions, through which electrons can tunnel due to quantum mechanics. A simplified circuit diagram for an SET is shown in the figure.\n[figure1]\n\nCircuit diagram representation of an SET. QD is the quantum dot, $\\mathrm{S}$ is the source, $\\mathrm{D}$ is the drain and $\\mathrm{G}$ is the gate.\n\nThe capacitance of the gate is $C_{g}$ and the capacitance of the tunnel junctions is $C_{t} \\ll C_{g}$. Consider $C_{g}$ to be the total capacitance of the quantum dot. In this part of the problem, the source and the drain are held at zero potential, and the voltage on the gate electrode is fixed at $V_{g}$.\n\nConsider a state of the SET in which the quantum dot contains $n$ electrons.\n\nIf $\\Delta E_{n}<0$ then electrons will spontaneously tunnel into the quantum dot until such a number $\\mathcal{N}>n$ is reached that $\\Delta E_{\\mathcal{N}} \\geq 0$. The equilibrium number of electrons $\\mathcal{N}$ and the corresponding addition energy $\\Delta E_{\\mathcal{N}}$ can be controlled by choosing the appropriate voltage $V_{g}$.\n\nFind an expression for the maximal possible value $E_{c}=\\max \\Delta E_{\\mathcal{N}}\\left(V_{g}\\right)$ of the  equilibrium addition energy that can be achieved by tuning the gate voltage of the SET.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_c16db445016d4c0e9a0ag-3.jpg?height=582&width=868&top_left_y=2076&top_left_x=594"
        ],
        "answer": [
            "$$\\frac{e^{2}}{C_{g}} $$"
        ],
        "solution": "$\\mathcal{N}$ is a minimal integer $n$ for which $\\Delta E_{n} \\geq 0$. Consider the marginal case of $\\Delta E_{\\mathcal{N}}=0$ which is achieved at some $V_{g}=V_{0}$,\n\n$$\n\\Delta E_{\\mathcal{N}}\\left(V_{0}\\right)=0=\\frac{e^{2}}{C_{g}}\\left(\\mathcal{N}+\\frac{1}{2}\\right)-e V_{0}\n$$\n\nIf $V_{g}$ would go slightly larger than $V_{0}$, then $\\Delta E_{n}$ would go negative and then minimal $n$ that makes a positive $\\Delta E_{n}$ would jump from $\\mathcal{N}$ to $\\mathcal{N}+1$. Hence $E_{c}=\\Delta E_{\\mathcal{N}+1}\\left(V_{0}\\right)$. This gives\n\n$$\n\\Delta E_{\\mathcal{N}+1}\\left(V_{0}\\right)=E_{c}=\\frac{e^{2}}{C_{g}}\\left(\\mathcal{N}+1+\\frac{1}{2}\\right)-e V_{0}=\\frac{e^{2}}{C_{g}}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Physics",
        "language": "EN",
        "modality": "multi-modal"
    }
]