OpenLLM-France/wikimedia huggingface dataset:
# AP Biology/DNA

DNA (Deoxyribonucleic acid) is what contains all of the information for life on earth. Every living thing including a bacteria cell to a killer whale has its own unique genetic code, which can be found within its DNA. This information has the ability to replicate itself ensuring that the species can continue by producing offspring exactly identical to its own. This is known as reproduction and is vital for life, as the life span of any being is limited.

## Structure

DNA is a double-helix that is made up of three main components. The monosaccharide Deoxyribose acts as a base, and this is connected to a nitrogenous base, and also to a phosphate group - collectively these are known as 'Nucleotides' and linked together cause a chain which, when linked to another chain of nucleotide becomes DNA. There are four types of nitrogenous bases: adenine, thymine, cytosine, and guanine. Adenine and guanine are purines, meaning that they have a double-ring base. While thymine and cytosine are pyrimidines, meaning that they have a single-ring base. Adenine and thymine always bind together, while cytosine and guanine always bind together. This is called complementarity.

## DNA Replication

DNA replication occurs during interphase in a cell cycle. It is essential for cell division as a way to pass on genetic information to the next "generation" of cells. First, the DNA double-helix is separated into two strands using the enzyme helicase, which unwinds the DNA double-helix to form a y-shaped replication fork. Binding proteins that are single-stranded prevent the newly separated DNA strands from recombining.
# AP Biology/Diagrams

Contents | Preface | Diagrams | Resources | Contributors | Edit Navigation Bar

## Chapter 1

* How Science is Done
* A sucky diagram of a tRNA molecule
* A rendering of a segment of a DNA molecule.
# AP Biology/How Cells Divide

Eukaryotic cells divide through a process called mitosis followed by cytokinesis.

## Mitosis

Mitosis in itself is the division of the nucleus inside the cell. However, once the nucleus divides, the whole cell divides shortly afterwards, known as cytokinesis.
The stages of mitosis are as follows:

### Interphase

Interphase is considered by most to be a resting phase. It consists of G1, S, and G2.

### Prophase

Just prior to mitosis, the pair of centrioles duplicates. During prophase, the two pairs of centrioles migrate to opposite poles. Centrioles form spindle fibers, which become microtubules and eventually attach to the centromeres. Fibers, known as asters, radiate from the centrioles outward.

### Metaphase

The spindle fibers enter the nuclear region, extend from the centrioles to the centromere, and attach at a point known as the kinetechore. Once the spindle fibers are attached, they align the centromeres along the equatorial region of the nucleus known as the metaphase plate, so that the arms of the chromosomes point towards the poles of the cell.

### Anaphase

The centromere divides and the two chromatids separate from each other, forming two identical daughter chromosomes. The spindle fibers attach to the centromere and pull the newly-divided chromosomes towards the poles and away from the metaphase plate. The spindle fibers appear to move, but in fact, the microtubules are continuously formed at one end of the spindle fiber and then disassembled at the other.

### Telophase

After the chromosomes reach the poles, a nuclear membrane forms around each set of daughter nuclei and the chromosomes uncoil and elongate, once again becoming invisible. The spindle fibers break down and disappear. In animal cells, a cleavage furrow, an indentation in the cell membrane, begins to develop between the daughter cells. This marks the end of mitosis.
# AP Biology/LABORATORY 11. Animal Behavior

This is a lab performed by AP Biology students that is used to understand the movement or activity of an organism in response to a stimulus (kinesis) and the direction of the greatest rate of increase of a function under observation and its magnitude is the slope of the graph in a particularly meaningful direction (gradients). The gradiated factor is chosen by the student and is observed with relevance to pill bugs (or meal worms).
# A-level Mathematics/OCR/M1

This is the Mechanics 1 module of the A-level Mathematics book. This module assumes that you are familiar with the C1 and C2 modules. Please post all questions to A-level Mathematics/Discussion.

## Contents

* Force as a Vector
* Equilibrium of a Particle
* Kinematics of Motion in a Straight Line
* Newton's Laws of Motion
* Linear Momentum
* Appendix of Formulae

## Definitions
### Smooth

    a surface without friction

### Rough

    a surface with friction

### Inextensible

    Does not stretch however large the load is

### Elastic

    Stretches when a force is put on it

### String

    string

### Light

    no mass or negligible mass

### Particle

    object with no volume, only mass.

### Bead

    Particle with a hole drilled through it

### Rigid Body

    a solid body impervious to deformation

### Wire

    rigid thread like body

### Uniform rod

    same width along the rod

### Non-Uniform rod

    different widths along length

### Peg

    pin/ support that can hang things

### Lamina

    flat object

### g

    The force due to gravity and equal to 9.8 N/Kg, or the acceleration due to gravity (9.8ms$^{-2}$).
# A-level Mathematics/OCR/M1/Linear Momentum
## Objectives

By the end of this chapter you should be able to do the following:
Use (and recall) the definition of linear momentum, and show an understanding of its vector nature in 1 dimension.
Understand and use conservation of linear momentum, in 1 dimension, and solve simple problems involving direct collision of two bodies before and after impact.

## Momentum as a vector

Because momentum is defined to be mass times velocity, momentum is the product of a scalar and a vector, which is ultimately a vector.
$p=mv$

## Conservation of Momentum

Within a closed system (i.e. a system not subject to an external influence) momentum is always conserved. Due to this, we can easily derive the following formula that is used in working out momentum:
$(M_{1}*U_{1})+(M_{2}*U_{2})=(M_{1}*V_{1})+(M_{2}*V_{2})$
Where:
M = Mass U = Initial Velocity V = Velocity after collision
Whatever the momentum on the LHS, the total momentum on the RHS will be the same.

## Collisions

The conservation of momentum is best illustrated through an example question:
The system before the collision.
The system after the collision.
As we have all the values before the collision, we can use the formula above to work out the total momentum in the system:
$(2*5)+(5*0)=10{\text{ Ns}}$
Now we know that momentum before = momentum after. So, $10=(2*10)+(5v)$. This may seem to be unworkable at first, but because velocity is a vector and the two balls are rolling in opposite directions, the 5 kg ball has a negative velocity.
Rearranging our formula we get $10-20=5v$. Rearranging this again to ${\frac {10-20}{5}}=v$ we can see that $v=-2{\text{ ms}}^{-1}$

## Impulse

Impulse is the change in momentum, measured in Ns as with momentum. It is a vector quantity, with magnitude and direction. The impulse by B on A is A's change in momentum as a result of collision with B; it can be found with $m_{1}*u_{1}-m_{1}*v_{2}$.
# A-level Mathematics/OCR/M1/Formulae
## Formulae

By the end of this module you will be expected to have learnt the following formulae:

### Forces and Equilibrium

* Weight $=mass\times g$
* Limiting Friction: $F=\mu R\,$
* $F=ma\,$

### Kinematics

* $v=u+at\,$
* $s=ut+{\frac {1}{2}}at^{2}$
* $s={\frac {1}{2}}(u+v)t$
* $v^{2}=u^{2}+2as\,$
* $s=vt-{\frac {1}{2}}at^{2}$
* $v={\frac {ds}{dt}}$ , $a={\frac {dv}{dt}}$

### Linear Momentum

* Momentum of a Particle: $p=mv\,$
Momentum is conserved. Therefore if there is a collision between 2 objects, the momentum before = momentum afterwards.
Impulse is the change in momentum.
# A-level Mathematics/OCR/M1/Equilibrium of a Particle
## Forces

Force is a vector quantity that influences the movement of an object.

## Equilibrium of A Particle

If you look at an object lying on a table it is in equilibrium. An object is in equilibrium if the resultant forces acting on it is zero. Yet this does not mean that the object is not moving. It means that the sum of the forces in the x direction is equal to zero: $\sum {f_{x}}=0$ and the sum of the forces in the y-direction is equal to zero: $\sum {f_{y}}=0$.

### Example One

A 20 kilogram box is suspended from the ceiling by a rope that weighs 1 kilogram. Find the tension at the top of the rope.
* The first step is to draw a diagram.
* The second step is to label all the forces. This is the most crucial step. T is the tension force and F is the weight force.
* Now we are ready to solve the problem. Since the object is in equilibrium, because it is not moving, the components must add up to zero. There are no x forces in this problem.
    $\sum {f_{y}}=0=ma-T$
    $T=ma\,$
    Now be careful we need to add the mass of the rope to the mass of the box.
    $T=(20+1)\times 9.8=205.8N$

## Friction

It is easiest to calculate the forces on an object that lies on a smooth surface or is not moving. A perfectly smooth surface will not have friction. However in the real world objects are not perfectly smooth, they have rough surfaces. The interaction of these surfaces creates friction. Friction resists motion. Friction is a dimensionless quantity represented by the Greek letter mu $\mu$, $\mu \geq 0$. There are two kinds of frictions; there is statical friction that applies while and object is not moving and there is kinetic friction when an object is moving. Static friction is always greater than kinetic friction. Static friction increases with the force until it reaches a maximum value, this value is known as limiting friction. If there is limiting friction but the object is not moving the object is in limiting equilibrium. With statical friction you need to know these two formulae:
       $F\leq \mu R$
       $F=\mu R\,$ when the object is in limiting equilibrium

### Example Two

A 17 kilogram block is resting on a ramp that makes and angle of 27° with the ground. Calculate how fast the block is accelerating if it is accelerating when a) The surface is smooth b) The surface is rough and $\mu =0.4$.
* The first step is to draw a diagram.
* Next we need to label all the forces. In this case there will be y and x forces.
a) In the case of a smooth surface, friction has the value of zero.
The weight equals 17 X 9.8 = 166.6 N.
Using F=ma,
17 X 9.8 X cos(90-27)°= 17a
17 X 9.8 X cos 63° = 17 a, so
a=9.8cos63°=4.45 ms$^{−2}$ (3sf)
b) The weight equals 17 X 9.8 = 166.6 N
Use F = $\mu$ X R to get F = 0.4 X R
# A-level Mathematics/OCR/M1/Force as a mk Vector
## Vectors

A vector is a quantity that has both a magnitude (or a size) and a direction. The opposite of vectors are scalars. Scalars only have a magnitude. There is no direction. For example, speed is a scalar as speed is the same regardless of direction. This is best illustrated as a triangle:
Our point, P, is a plane travelling along the hypotenuse of this triangle at a speed of $5ms^{-1}$. Its velocity, however, is not 5. As velocity is a vector and has both magnitude and direction, the speed of P is equal to moving at a velocity of $4ms^{-1}$ along the horizontal and $3ms^{-1}$ along the vertical.
There are several different ways of writing this as a vector. One of the most common is the i and j notation. Where i is the horizontal component of the velocity and j is the vertical component of the velocity. Using this notation, our plane would have a velocity of (4i + 3j)
Another common way of writing vectors is in the form of ${x \choose y}$ where x is the horizontal component and y is the vertical component. Using our plane as the example, is this vector form its velocity would be ${4 \choose 3}ms^{-1}$.
To change a Vector into its horizontal and vertical components we:
1. Draw a triangle representing the vector.
2. Label all known values on triangle.
3. Use trigonometry to solve.
E.g. A force, P, with magnitude 25N has a direction of $arcsin$ ${\frac {7}{25}}$ (arcsin is the opposite of $sin$.), find the horizontal and vertical components of P
Triangle
Label triangle:
Use trigonometry: is $\arcsin \theta$ is ${\frac {7}{25}}$ then $\theta$ is $sin$ ${\frac {7}{25}}$. Sin is O/H. Therefore the vertical component of P is 7. The Horizontal component can be found by using Pythagoras' theorem or recognising 7, 24, 25 as a Pythagorean triple. Pythagoras' theorem says that $a^{2}$+$b^{2}$=$c^{2}$ where c is the hypotenuse and a and b are the adjacent and opposite (order does not matter). Therefore $\ 25^{2}-\ 7^{2}\$ = $horizontalcomponent^{2}$ = 576. $576^{\frac {1}{2}}$ = 24. In i and js this is (24i+7j).
# A-level Mathematics/OCR/M1/Newton's Laws of Motion

Newton's first law of motion states that a body acted on by a balanced set of forces will stay at rest if it is at rest, and it will continue at a constant velocity if it is in motion. A common phrasing of the law is if the body is acted on by external forces however this is incorrect, for example a car can be acted on by no external forces and it can accelerate. This law is sometimes referred to as the law of inertia.
Newton's second law of motion states that if a body is acted on by an unbalanced set of forces then the resultant force equals the mass of the object times the acceleration of the object:
$F=ma$
Newton's third law of motion states that, in equilibria, every force has an equal and opposite reaction force.
These laws can be used (by themselves or in conjunction with other methods) to solve problems in M1. If a particle of mass m has a force acting on it of F, then its acceleration would be a. A numerical example of this would be a particle of mass 3 kg, being accelerated at a rate of 2 m/s^2 would require a force of 6N (kgm/s^2).
From Newton's second law of motion we gain the definition of the unit of force: the Newton, as the force required to accelerate a particle of mass one kg at a rate of 1 m/s^2. It has the units N, or in SI base units kg*m/s^2.
# A-level Mathematics/OCR/FP3

This is the Further Pure Mathematics 3 module of the A-level Mathematics book. This module assumes that you are familiar with the C1, C2, C3, C4 and FP1 modules. Please post all questions to A-level Mathematics/Discussion.

## Contents

* More About Differential Equations
* More About Vectors
* Hyperbolic Functions
* Complex Numbers
* Groups

## Formulae

By the end of this module you will be expected to have learnt the following formulae:

### Differential Equations

* An integrating factor of ${\frac {dy}{dx}}+P(x)y=Q(x)$ is $e^{\int P(x)\,dx}$

### Vectors

* The vector equation of a plane through point $\mathbf {a}$ with normal vector $\mathbf {n}$ is $(\mathbf {r} -\mathbf {a} )\cdot \mathbf {n} =0$

### Complex Numbers

If $z=e^{i\theta }$ then:
   * $\cos \theta ={\frac {1}{2}}\left(z+{\frac {1}{z}}\right)$
   * $\sin \theta ={\frac {1}{2i}}\left(z-{\frac {1}{z}}\right)$
# A-level Mathematics/OCR/FP4
## Contents

* Matrix Algebra
* Vector Product
* Determinants
* Applications of Vector Products
* Inverse Matrices
* Solving Linear Equations
* Eigenvectors

## Formulae

By the end of this module you will have learnt the following formulae:

### Eigenvectors

    $Mv=\lambda v$
Where λ is the Eigenvalue and v is the Eigenvector.
# A-level Physics (Advancing Physics)/Kinetic Theory/Worked Solutions

1. Five molecules are moving at speeds of 1,5,6,8, and 36ms$^{−1}$. What is their mean square speed?
${\bar {c^{2}}}={\frac {1^{2}+5^{2}+6^{2}+8^{2}+36^{2}}{5}}=284.4{\mbox{ m}}^{2}{\mbox{s}}^{-2}$
2. What is the mass of one molecule of N$_2$ (atomic mass 14, 1u = 1.66 x 10$^{−27}$kg)?
$2\times 14\times 1.66\times 10^{-27}=4.648\times 10^{-26}{\mbox{ kg}}$
3. Atmospheric pressure is 101,325 Pa. If one mole of Nitrogen takes up 2.3 m$^3$ at about 10 °C, what is the mean square speed of the molecules in the air outside, assuming that the atmosphere is 100% nitrogen (in reality, it is only 78%)?
$pV={\frac {1}{3}}Nm{\bar {c^{2}}}$
$101235\times 2.3={\frac {1}{3}}\times 6.02\times 10^{23}\times 6.648\times 10^{-22}{\bar {c^{2}}}$
${\bar {c^{2}}}={\frac {3\times 101235\times 2.3}{6.02\times 10^{23}\times 6.648\times 10^{-22}}}=1745{\mbox{ m}}^{2}{\mbox{s}}^{-2}$
4. What is the average speed of a nitrogen molecule under the above conditions?
${\bar {c}}={\sqrt {\bar {c^{2}}}}={\sqrt {1745}}=41.8{\mbox{ ms}}^{-1}$
5. The particles in question 1 are duplicated 3000 times. If they have a completely unrealistic mass of 1g, what is their pressure when they are crammed into a cube with side length 0.5m?
$p={\frac {Nm{\bar {c^{2}}}}{3V}}={\frac {5\times 3000\times 10^{-3}\times 284.4}{3\times 0.5^{3}}}=11376{\mbox{ Pa}}$
# A-level Physics (Advancing Physics)/Ideal Gases

Real-world gases can be modelled as ideal gases. An ideal gas consists of lots of point particles moving at random, colliding with each other elastically. There are four simple laws which apply to an ideal gas which you need to know about:

## Boyle's Law

Boyle's Law states that the pressure of an ideal gas is inversely proportional to its volume, assuming that the mass and temperature of the gas remain constant. If I compress an ideal gas into half the space, the pressure on the outsides of the container will double. So:
$p\propto {\frac {1}{V}}$

## Charles' Law

Charles' Law states that the volume of an ideal gas is proportional to its temperature:
$V\propto T$
T must be measured in kelvin, where a rise of 1°K is equal to a rise 1 °C, and 0 °C = 273°K. If we double the temperature of a gas, the particles move around twice as much, and so the volume also doubles.

## Amount Law

This law states that the pressure of an ideal gas is proportional to the amount of gas. If we have twice the number of gas particles N, then twice the pressure is exerted on the container they are in:
$p\propto N$
A mole is a number of particles. 1 mole = 6.02 x 10$^{23}$ particles. So, the pressure of a gas is also proportional to the number of moles of gas present n:
$p\propto n$

## Pressure Law

The pressure law states that the pressure of an ideal gas is proportional to its temperature. A gas at twice the temperature (in °K) exerts twice the pressure on the sides of a container which it is in:
$p\propto T$
These laws can be put together into larger formulae linking p, V, T and N. To do this we require a constant of proportionality, (R) the universal molar gas constant, with an experimental value of 8.31Jmol$^{−1}$K$^{−1}$

## Questions

1. I heat some argon from 250K to 300K. If the pressure of the gas at 250K is 0.1 MPa, what is its pressure after heating?
2. The argon is in a 0.5m long cylindrical tank with radius 10 cm. What volume does it occupy?
3. The argon is then squeezed with a piston so that in only occupies 0.4m of the tank's length. What is its new pressure?
4. What is its new temperature?
5. 25% of the argon is sucked out. What is its pressure now?
Worked Solutions
# A-level Physics (Advancing Physics)/Induction

A magnetic field creates a current in a wire moving through it. This process is known as induction.

## Flux Linkage

A magnetic field going through a coil of wire has a property known as flux linkage. This is the product of the flux Φ and the number of coils in the wire N.

## Faraday's Law

Electric current is only induced in a coil of wire if the magnetic field is moving relative to the coil. Faraday's Law gives the electromotive force (emf) ε produced in a coil by a magnetic field:
$\epsilon =-N{\frac {d\Phi }{dt}}$
In other words, the emf (electric potential) induced in the coil is proportional to the rate of change of flux linkage. In practice, this means that if the coil is stationary relative to the magnetic field, no emf is induced. In order to induce emf, either the coil or the magnetic field must move. Alternatively, we may change the number of coils, for example, by crushing the coil, or pressing a switch which added more coils into the circuit, or moving more of the coils into the magnetic field.
Faraday's Law also works the other way. If we were to integrate both sides and rearrange the formula in terms of Φ, we would find that the flux depends on the integral of the voltage - not on its rate of change. If we put an emf across a coil, it induces a magnetic field. The flux does not depend on the rate of change of emf, but the emf does depend on the rate of change of the flux linkage.

## Lenz's Law

Lenz's Law describes the direction of the current / emf induced by a change in magnetic flux. It states that current induced opposes the magnetic field. It does this by creating its own magnetic field. This explains the minus sign in Faraday's Law. This also means that the flux induced by a current (not a change in current) is proportional to the current, since the flux is produced in response to the current.
So, a change in flux induces a current and a voltage which is proportional to the rate of change of flux. This fits with Ohm's Law (V = IR). A current and a voltage in a coil induce a flux which is proportional to the current and the voltage.

## Questions

1. What is the flux linkage of a 30 cm coil of 0.5mm thick wire with a flux perpendicular to it of 10Wb?
2. If the above coil is crushed steadily over a period of 2s, what emf is maintained?
3. The flux in a flux circuit varies according to the equation Φ = sin ωt. What is the equation for the emf induced?
4. Using a constant k, what is the equation for a current which could induce the flux in the flux circuit above?
5. Draw a graph of the flux, flux linkage, emf and current as deduced in the previous two questions.
Worked Solutions
# A-level Physics (Advancing Physics)/Induction/Worked Solutions

1. What is the flux linkage of a 30 cm coil of 0.5mm thick wire with a flux perpendicular to it of 10Wb?
$N={\frac {30\times 10^{-2}}{0.5\times 10^{-3}}}=600{\mbox{ coils}}$
$N\Phi =600\times 10=6000{\mbox{ Wb}}$
2. If the above coil is crushed steadily over a period of 2s, what emf is maintained?
$\mid \epsilon \mid ={\frac {dN\Phi }{dt}}={\frac {6000}{2}}=3000{\mbox{ V}}$
This principle of removing the flux linkage rapidly (although not usually by destroying one's equipment) is used to create very high voltages for short periods of time.
3. The flux in a flux circuit varies according to the equation Φ = sin ωt. What is the equation for the emf induced?
$\epsilon =-{\frac {dN\Phi }{dt}}=-N{\frac {d}{dt}}\sin {\omega t}=-N\omega \cos {\omega t}$
4. Using a constant k, what is the equation for a current which could induce the flux in the flux circuit above?
$I=-k\sin {\omega t}$
5. Draw a graph of the flux, flux linkage, emf and current as deduced in the previous two questions.
In order to keep the scales sane, I used a 2-coil (N=2) coil to plot the following graph:
Remember that the current here is inducing the flux (and flux linkage), which is in turn inducing the emf.