OpenLLM-France/wikimedia huggingface dataset:
# University of Florida/Egm6321/f09.team1.gzc/Mtg34

Mtg 34: Wed, 23 Mar 11
    page34-1 1) Supersonic intercaptor: Find angle of attack and thrust to reach a given altitude inmin time Reference: Bryson & Denham 1962 2) Bunt maneuver of aircraft: min altitude to reach target with a bunt maneuver (inverted loop) page34-2 ${\color {blue}V}\ =\ velocity\ of\ P\ (airplane\ =\ pt)$ ${\color {blue}T}\ =\ thrust\ //\ airplace\ axis$ $\alpha \ =\ \angle \ ({\color {blue}V,}\ {\color {red}T})\ =\ angle\ of\ attack$ $\gamma \ =\ \angle \ (x,\ {color{blue}V})\ =\ angle\ between\ honz\ x\ axis\ and\ veloctity$ $D\ =\ axial\ aero\ force\ //\ {\color {red}T}$ $(parall.\ airplane\ axis)$ page34-3 $L\ =\ Tramsv.\ aero.\ force\ \bot \ {\color {red}T}$ $W\ =\ mg\ =\ airplane\ weight.$ $Equation\ of\ motion\ (EOM)$ ${color{blue}Control\ input:}\ {\color {blue}{\underset {time\ dependent}{\underbrace {{\color {red}T}(t)} \ ,\ \alpha (t)}}}$ ${\color {blue}state\ var,:}\ x(t),\ y(t)\ v(t),\ r(t)$ ${\color {blue}{\begin{cases}&\ {\color {black}D={\frac {L}{2}}\ C_{d}\ \rho \ V^{2}\ S_{reft}}\\&\ {\color {black}L={\frac {L}{2}}\ C_{l}\ \rho \ V^{2}\ S_{reft}}\end{cases}}}$ $C_{d}=drg\ coefficient\ {\color {green}|}\ \rho =air\ density$ $C_{l}=lift\ coefficient\ {\color {green}|}\ \S _{ref}=ref.\ area\ of\ airplane$ page34-4 $C_{d}=A_{1}\alpha ^{2}+A_{2}\alpha +A_{3}\ {\color {red}(1)}$ $(A_{1},\ A_{2},\ A_{3})=curve\ fitting\ coefficient$ $C_{l}=B_{1}\alpha +B_{2}\ {\color {red}(2)}$ $(B_{1},\ B_{2})=curve\ fitting\ coefficient$ $\rho =C_{1}h^{2}+C_{2}h+C_{3}\ {\color {red}(3)}$ $(C_{1},\ C_{2},\ C_{3})=curve\ fitting\ coefficient$ Reference: S & Z 2007 ${\underline {\color {blue}{Kinematics:}}}\ {\frac {dx}{dt}}=V_{x}=Vcosr\ {\color {red}(4)}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\frac {dy}{dt}}=V_{y}=Vsinr\ {\color {red}(5)}$ ${\underline {\color {blue}{Kinematics:}}}{\color {blue}Euler\ equations:}\ {\frac {d}{dt}}{\underline {P}}=\sum _{\color {red}i}{\underline {F}}_{\color {red}i}\ {\color {red}(6)}$ page34-5 Note: Parameteriztion of ellipse ${\color {blue}{\underset {cos^{2}t}{\underbrace {\color {black}\left({\frac {x}{a}}\right)^{2}} }}}+{\color {blue}{\underset {sin^{2}t}{\underbrace {\color {black}\left({\frac {y}{b}}\right)^{2}} }}}=1\ {\color {blue}ellipse}$ $<{\color {blue}dl={\color {red}[{\color {blue}dx^{2}+dy^{2}}]^{1/2}}}/math><math>Eccentricity:\ e=\left(1-{\frac {b^{2}}{a^{2}}}\right)^{1/2}$ $C=\int dl=a\int _{t=0}^{2\pi }[1-e^{2}cos^{2}t]^{1/2}dt$ $=41\int _{\alpha =0}^{\frac {\pi }{2}}[1-e^{2}sin^{2}\alpha ]^{\frac {1}{2}}d\alpha =4aE(e)$
# University of Florida/Egm6321/f09.team1.gzc/Mtg6

Mtg 6: Sun, 16 Jan 11
page6-1
    Proof of Taylor series continued Since $\displaystyle \int _{x_{0}}^{x}{f}^{1}(t)dt=[{f}^{0}(t)]_{t={x}_{0}}^{t=x}$ (1)
$\displaystyle \Rightarrow \color {red}(e)\color {blue}p.5-3$
Int. by parts (1)
$\displaystyle \color {red}(2){\begin{cases}&\ \color {black}\int _{x_{0}}^{x}\color {blue}{\underset {u^{'}}{\underbrace {1} }}{\underset {v}{\underbrace {\color {black}{{f}^{(1)}(t)}} }}\color {black}dt={[uv]}_{x_{0}}^{x}-\int _^{uv}^{'}\\&\ \color {black}={[tf^{(1)}{t}]}_{t=x_{0}}^{t=x}-\color {blue}{\underset {\alpha }{\underbrace {\color {black}{\int _{x_{0}}^{x}tf^{(2)}dt}} }}\\&\ \color {black}=xf^{1}(x)-x_{0}f^{(1)}(x)0)-\color {blue}\alpha \\&\ \color {red}+\color {blue}xf^{(1)}(x_{0})\color {red}-\color {blue}xf^{(1)}(x_{0})\\&\ \color {black}=\color {blue}{\underset {\int _{x_{0}}^{x}xf^{(2)}(t)dt=:\beta }{\underbrace {\color {black}[xf^{1)(x)-xf^{1)(x_{0})]} }}\color {black}+(x-x_{0}){f}^{(1)}(x_{0})-\color {blue}\alpha \end{cases}}$
page6-2
Combine [ + β - α] into a single int. use (2) p.6-1 in (2) p.5-3:
    $\displaystyle f(x)=f(x_{0})+{\frac {{(x-x_{0})}^{\color {blue}1}}{\color {blue}1!}}{f}^{(1)}(x_{0})+\color {blue}{\underset {\beta -\alpha }{\underbrace {\color {black}\int _{x_{0}}^{x}(x-t)f^{(2)}(t)dt} }}$ (1)
    HW*2.1: 1) Do integration by parts on last term (integration) of (1) to reveal 3 more terms in Taylor series, i.e. , $\displaystyle {\frac {{x-{x}_{0}}^{2}}{2!}}{f}^{(2)}({x}_{0})+{\frac {{x-{x}_{0}}^{3}}{3!}}{f}^{(3)}({x}_{0})+{\frac {{x-{x}_{0}}^{4}}{4!}}{f}^{(4)}({x}_{0})$
plus remainder
2)Use IMVT to expression remainder in terms of f$^{(s)}$(ξ) for s belong[x$_0$,x]
3) Assume(3)&(4)p.3-3 correct , do intergration by parts once more
page6-3
to verify(3)&(4) p.3-3 for (n+1) expansionwith R$_{(n+2)}$(x)
4) UseIMVT on(4) p.3-3to show(5)p.3-3
* IMVT:$\int _{a}^{b}w(x)g(x)dx{\overset {\color {red}(1)}{=}}g(\xi )\int _{a}^{b}w(x)dx$
* $\xi \in [a,b]$
* Use "g(x)" instead of "f(x)" to avoid confusion with "f(x)" is a Taylor Series.
HW$^*$2.2:$f(x)=sinx,x\in [0,\pi ]$ Constrast Taylor Series of f(.) around
$\displaystyle {\begin{cases}&\ {x}_{0}={\frac {\pi }{4}}\ \color {blue}S10\\&\ {x}_{0}={\frac {3\pi }{4}}\ \color {blue}S11\end{cases}}$ for n = 0,1,2,...,10
Plot these series (for each n) Find (estimate) max $\left|{R}_{n+1}(x={\frac {3\pi }{4}})\right|$
${\color {red}(5)}{\color {blue}p.3-3}\Rightarrow \left|{R}_{n+1}(x={\frac {3\pi }{4}})\right|\leqq {\frac {{x-{x}_{0}}^{n+1}}{(n+1)!}}{\color {blue}(\alpha )}$
page6-4
${\color {blue}\alpha }=max\left|{f}^{(n+1)}(t)\right|$
$\color {blue}{\underset {\leqq 1(f(x)=sinx)}{\underbrace {\color {black}t\in [{x}_{0},x]} }}$
Note:Motivation for pf of Taylor series expansion. (similar technique will be used)
* higher order analysis of Trap. rule (not in A.)
* Richardson extrap.
* clenshaw-Cwetis quadrature
* chebyshew poly (orthog.) Recent devel. using chebyshew poly to solveL2_ODE_VC (Linear 2nd order ODE with varying coefficient ) combine of symbolics + numericsReference : Trefethen's chebfun.
page6-5
# University of Florida/Eml4507/s13.team3.ADR6

Problem 5.1

## Given

$m_{1}=3$
$m_{2}=2$
$k_{1}=10$
$k_{2}=20$
$k_{3}=15$

## Find

Solve by hand the gen. eigenvalue problem for the spring-mass-damper system on p.53-13, using the data for the masses in (2) p.53-13b, and the data for the stiffness coefficients in (4) p.53-13b.

## Solution

First plug given values into the stiffness matrix:
$K={\begin{bmatrix}a_{1}&0&0&-a_{1}&0&0\\0&12a_{2}&6La_{2}&0&-12a_{2}&6La_{2}\\0&6La_{2}&4L_{2}^{2}&0&-6a_{2}&2La_{2}^{2}\\-a_{1}&0&0&a_{1}&0&0\\0&-12a_{2}&-6La_{2}&0&12a_{2}&-6a_{2}\\0&6La_{2}&4L_{2}^{2}&0&-6a_{2}&4La_{2}^{2}\\\end{bmatrix}}$
$K={\begin{bmatrix}(10+20)&-20\\-20&(20+15)\\\end{bmatrix}}$
# University of Florida/Eml4507/s13.team3.GuzyR5

Problem 5.1

## Given

$m_{1}=3$
$m_{2}=2$
$k_{1}=10$
$k_{2}=20$
$k_{3}=15$

## Find

Solve by hand the gen. eigenvalue problem for the spring-mass-damper system on p.53-13, using the data for the masses in (2) p.53-13b, and the data for the stiffness coefficients in (4) p.53-13b.

## Solution

First plug given values into the stiffness matrix:
$K={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}$
$K={\begin{bmatrix}(10+20)&-20\\-20&(20+15)\\\end{bmatrix}}$
$K={\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}$
Next plug stiffness matrix in equation below:
$[K-\gamma _{I}]x=0$
Take the determinant:
$det{\begin{Bmatrix}30-\gamma &-20\\-20&35-\gamma \end{Bmatrix}}=[(30-\gamma )(35-\gamma )]-(-20)(-20)=0$
Simplifying:
$1050-30\gamma -35\gamma +\gamma ^{2}-400$
$\gamma ^{2}-65\gamma +650=0$
Using the Quadratic formula to solve:
$\gamma _{1}={\frac {65+{\sqrt {65^{2}-(4)(650)}}}{2}}$
$\gamma _{2}={\frac {65-{\sqrt {65^{2}-(4)(650)}}}{2}}$
Final gamma values:
$\gamma _{1}=52.66$
$\gamma _{2}=12.34$
To find the eigenvectors, first we:
$[K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}-\gamma *{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}{\begin{bmatrix}]x_{1}\\x_{2}\\\end{bmatrix}}=0$
To solve, lets set $x_{1}$ equal to 1:
$[K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}-52.66*{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}]{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0$
$[K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}+{\begin{bmatrix}-52.66&0\\0&-52.66\\\end{bmatrix}}]{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0$
$[K-\gamma _{I}]x=[{\begin{bmatrix}30-52.66&-20\\-20&35-52.66\\\end{bmatrix}}{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0$
$[K-\gamma _{I}]x=[{\begin{bmatrix}-22.66&-20\\-20&-17.66\\\end{bmatrix}}{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0$
Multiplying out the matrices, we obtain:
$-20-17.66x_{2}=0$
$x_{2}=1.1325$
Using the same process we find $x_{1}$:
$[K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}-52.66*{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}]{\begin{bmatrix}x_{1}\\1\\\end{bmatrix}}=0$
$[K-\gamma _{I}]x=[{\begin{bmatrix}-22.66&-20\\-20&-17.66\\\end{bmatrix}}{\begin{bmatrix}x_{1}\\1\\\end{bmatrix}}=0$
Multiplying out the matrices, we obtain:
$-22.66x_{1}-20=0$
$x_{1}=0.8826$
# University of Florida/Eml4507/s13.team3.DavidPatrickR4
## Problem 4.1
### Given

Spring-damper-body arrangement as shown. Two separate forces applied to masses.
$M={\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}$
$d={\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}$
$C={\begin{bmatrix}C_{1}+C_{2}&-C_{2}\\-C_{2}&C_{2}+C_{3}\\\end{bmatrix}}$
$K={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}$
$k_{1}=1,k_{2}=3,k_{3}=3$
$K={\begin{bmatrix}3&-2\\-2&5\end{bmatrix}}$
$[K-$γ$I]x=$ ${\begin{bmatrix}3&-2\\-2&5\end{bmatrix}}$ $-$γ ${\begin{bmatrix}1&0\\0&1\end{bmatrix}}$ $)$ ${\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}$ $=$ ${\begin{Bmatrix}0\\0\end{Bmatrix}}$
$det{\begin{Bmatrix}3-\gamma &-2\\-2&5-\gamma \end{Bmatrix}}=\gamma ^{2}-8\gamma +11=0$

### Find

Find the eigenvector $x_{2}$ corresponding to the eigenvalue $\gamma _{2}$ for the spring-mass-damper system on p.53-113. Plot and comment on this mode shape. Verify that the eigenvectors are orthogonal to each other

### Solution

Eigenvalues are found $\gamma _{1}=4+{\sqrt {5}}>0$ $\gamma _{2}=4-{\sqrt {5}}>0$
We find the eigenvectors from $\gamma _{2}$
$\gamma _{2}=4+{\sqrt {5}}$
$[K-\gamma _{2}I]x=$ ${\begin{bmatrix}-1-{\sqrt {5}}&-2\\-2&1-{\sqrt {5}}\end{bmatrix}}$ ${\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}$ $=$ ${\begin{Bmatrix}0\\0\end{Bmatrix}}$
Set $x_{2}=1$
$(-1-{\sqrt {5}})x_{1}-(2)x_{2}=0$
$x_{1}={\frac {2}{-1-{\sqrt {5}}}}$
Eigenvectors are orthogonal to each other:

## Problem 4.2
### Given

Use same given values as in problem 4.1

### Find

Find the eigenvectors for $\gamma _{1}$ and $\gamma _{2}$ when setting $x_{1}=1$

### Solution

We find the eigenvectors from $\gamma _{1}$
$\gamma _{1}=4-{\sqrt {5}}$
$[K-\gamma _{1}I]x=$ ${\begin{bmatrix}-1+{\sqrt {5}}&-2\\-2&1+{\sqrt {5}}\end{bmatrix}}$ ${\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}$ $=$ ${\begin{Bmatrix}0\\0\end{Bmatrix}}$
Set $x_{1}=1$
$(-1-{\sqrt {5}})x_{1}-(2)x_{2}=0$
$x_{2}={\frac {-1+{\sqrt {5}}}{2}}$
We find the eigenvectors from $\gamma _{2}$
$\gamma _{2}=4+{\sqrt {5}}$
$[K-\gamma _{2}I]x=$ ${\begin{bmatrix}-1-{\sqrt {5}}&-2\\-2&1-{\sqrt {5}}\end{bmatrix}}$ ${\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}$ $=$ ${\begin{Bmatrix}0\\0\end{Bmatrix}}$
Set $x_{1}=1$
$(-1-{\sqrt {5}})x_{1}-(2)x_{2}=0$
$x_{2}={\frac {-1-{\sqrt {5}}}{2}}$
# University of Florida/Eml4507/s13.team3.GuzyR3

Problem 2: Dampened and Undamped System
On my honor, I have neither given nor received unauthorized aid in doing this assignment.

## Given

Spring-damper-body arrangement as shown. Two separate forces applied to masses.
$M={\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}$
$d={\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}$
$C={\begin{bmatrix}C_{1}+C_{2}&-C_{2}\\-C_{2}&C_{2}+C_{3}\\\end{bmatrix}}$
$K={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}$

## Find

1.Draw the FBDs of all components in the spring-mass-damper system on p.53-13, with the known disp dofs being at the left and right supports:
$d_{3}=0$
$d_{4}=0$
2. Derive (1) p.53-12, and (1)-(3) p. 53-13.

## Solution
### Free Body Diagrams (Question 1)

The free body diagrams for a dampened system and undamped system can be seen to the right.

### Solving the System (Question 2)
#### DAMPENED

Analyzing the forces on mass one, we obtain:
    $\displaystyle m_{1}d''_{1}+k_{1}d_{1}+C_{1}d'_{1}-k_{2}d_{2}+k_{2}d_{1}-C_{2}d'_{2}+C_{2}d'_{1}=F_{1}$
    $\displaystyle m_{1}d''_{1}+C_{1}d'_{1}+C_{2}(d'_{1}-d'_{2})+k_{1}d_{1}+k_{2}(d_{1}-d_{2})=F_{1}$
Analyzing the forces on mass two, we obtain:
    $\displaystyle m_{2}d''_{2}+C_{2}(d'_{2}-d'_{1})+k_{2}(d_{2}-d_{1})+C_{3}d'_{2}+k_{3}d_{2}=F_{2}$
    $\displaystyle m_{2}d''_{2}+C_{2}(d'_{2}-d'_{1})+C_{3}d'_{2}+k_{2}(d_{2}-d_{1})+k_{3}d_{2}=F_{2}$
Plugging equations into matrix form we obtain:
${\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}{\begin{bmatrix}d''_{1}\\d''_{2}\\\end{bmatrix}}+{\begin{bmatrix}C_{1}+C_{2}&-C_{2}\\-C_{2}&C_{2}+C_{3}\\\end{bmatrix}}{\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}{\begin{bmatrix}d'_{1}\\d'_{2}\\\end{bmatrix}}+{\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}$\
Notice how this form matches each matrix given in the problem statement. Plugging in the original yields and proves the following equation of motion for an undamped system:
$Md''+Cd'+kd=0$

#### UNDAMPED

Analyzing the forces on mass one, we obtain:
    $\displaystyle m_{1}d''_{1}+k_{1}d_{1}-k_{2}d_{2}+k_{2}d_{1}=F_{1}$
    $\displaystyle m_{1}d''_{1}-k_{2}d_{2}+(k_{1}+k_{2})d_{1}=F_{1}$
Analyzing the forces on mass two, we obtain:
    $\displaystyle m_{2}d''_{2}+k_{2}d_{2}-k_{2}d_{1}+k_{3}d_{2}=F_{2}$
    $\displaystyle m_{2}d''_{2}+(k_{2}+k_{3})d_{2}-k_{2}d_{1}=F_{2}$
Plugging equations into matrix form we obtain:
${\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}{\begin{bmatrix}d''_{1}\\d''_{2}\\\end{bmatrix}}+{\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}{\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}$\
Notice how this form matches each matrix given in the problem statement. Plugging in the original yields and proves the following equation of motion for an undamped system:
$Md''+kd=0$
# University of Florida/Eml4507/s13.team3.GuzyR6
### CALFEM Verification

First begin by constructing the Edof matrix. Column 1 corresponds to the element number; columns 2 and 3 correspond to the global node that corresponds to the first and second local nodes respectively.
Construct an empty stiffness matrix K that has side lengths equal to the number of degrees of freedom times the number of global nodes. For this problem, there are 10 global nodes and motion has 3 degrees of freedom.
Constructing the force matrix and constructing the displacement matrix Q. Global nodes and known displacements as shown.
The function bar2e generates the element stiffness matrices for a bar with 3 degrees of freedom.
The Vector containing the Young's modulus and cross sectional area are found and and the x and y coordinates for each element are shown below:
The commands:
This yields 25 stiffness matrices. The global stiffness matrix can be assembled with the element stiffness matrices. This results in a stiffness matrix that is in accord with that computed in Matlab in the previous section. The function solveq takes the stiffness matrix, known force matrix, and displacement matrix and returns the complete force and displacement matrices.
The displacement and force matrices are in accord with those computed in Matlab in the previous section. The function extract is used to evaluate the displacements of the local nodes for each beam, which is needed to determine the force in each beam.
This results in the displacements. The function bar2s can be used to determine the force in each beam. Its input parameters are the x and y coordinates for the beam (ex1 and ey1), the physical properties of the beam (Young's modulus and cross sectional area - ep) and the displacement (ed).
This results in the following spring forces.
This is in accord with the forces computed in Matlab in the previous section. The axial stresses are given by dividing the axial forces by the area.
# Volcanoes/Mount Rainier/Quiz

Mount Rainier is a lecture from the school of geology about Mount Rainier as a volcano.
You are free to take this quiz based on Mount Rainier at any time.
To improve your score, read and study the lecture, the links contained within, listed under See also, External links, and in the {{geology resources}} template. This should give you adequate background to get 100 %.
As a "learning by doing" resource, this quiz helps you to assess your knowledge and understanding of the information, and it is a quiz you may take over and over as a learning resource to improve your knowledge, understanding, test-taking skills, and your score.
Suggestion: Have the lecture available in a separate window.
To master the information and use only your memory while taking the quiz, try rewriting the information from more familiar points of view, or be creative with association.
Enjoy learning by doing!

## Hypotheses

* There are approximately as many volcanoes per 10° of longitude and 5° of latitude as there are glaciers.
* The same or similar forces producing volcanoes on Io produce volcanoes on Earth including Mount Rainier.
# Volcanoes/Io/Quiz

The volcanoes of Io is a lecture regarding a specific astronomical object (Io) and its volcanoes. It is an offering from the school of geology and the radiation astronomy department of the school of physics and astronomy.
You are free to take this quiz based on the volcanoes of Io at any time.
To improve your score, read and study the lecture, the links contained within, listed under See also, External links, and in the {{radiation astronomy resources}} and the {{geology resources}} templates at the bottom of the page. This should give you adequate background to get 100 %.
As a "learning by doing" resource, this quiz helps you to assess your knowledge and understanding of the information, and it is a quiz you may take over and over as a learning resource to improve your knowledge, understanding, test-taking skills, and your score.
Suggestion: Have the lecture available in a separate window.
To master the information and use only your memory while taking the quiz, try rewriting the information from more familiar points of view, or be creative with association.
Enjoy learning by doing!

## Hypotheses

* The volcanoes of Io and of Earth have the same origin.
# Volcanoes/Mount Redoubt

The mushroom-shaped plume in the image on the right rose from avalanches of hot debris that cascaded down the north flank, when Mount Redoubt erupted on April 21, 1990.
"Redoubt Volcano is a steep-sided cone about 10 km in diameter at its base and with a volume of 30-35 cubic kilometers. The volcano is composed of intercalated pyroclastic deposits and lava flows and rests on Mesozoic granitic rocks of the Alaska-Aleutian Range batholith (Till and others, 1993; 1994). It has been moderately dissected by the action of numerous alpine glaciers. A 1.8-km-wide, ice-filled summit crater is breached on the north side by a northward-flowing glacier, informally known as the Drift Glacier, which spreads into a piedmont lobe in the upper Drift River Valley. The most recently active vent is located on the north side of the crater at the head of the Drift glacier. Holocene lahar deposits in the Crescent River and Drift River valleys extend downstream as far as Cook Inlet."

## Eruptions

In the first image on the right, the view is looking "south at the north flank of Redoubt volcano. Ashfall [is] limited to south crater floor, rim, and extending south-southeast. Source of ash is a vent south of the 1990 dome at an elevation of ~8300 ft." This eruption occurred on March 15, 2009.
In the second image down on the right, Mount Redoubt erupts on April 4, 2009. The eruption started at approximately 6:00 a.m. The image was taken at 8:11 a.m. (ADT). The eruption sent ash more than 50,000 feet into the air.
The third image down on the right shows an eruption of Mt Redoubt on June 9, 2009, in Alaska: the lava dome in the summit crater is spilling over from the crater. Size of dome at time of this image is about 1000 m long, 450 m wide, and 200 m high.

## Meltwaters

"On a summit flight, we were able to see the source of the meltwater stream. It appears to starts to the right of the dome where a fall face can be seen and continues down."
The image on the left shows flooding some 7.6 m deep caused by the melting of Drift Glacier.

## Glaciers

"The [Ice Piston] glacier that filled the crater was collapsing because of the increase in ground temperature underneath."
Glaciers and snowfields appear to be numerous in this late spring image on the left of Mount Redoubt.
On the lower right is an image of the Drift Glacier area. The view is downstream to the northeast.
On the lower left is the Redoubt Glacier and below it is Ouzel Lake which the glacier drains into. The image is dated to August so the glacier has probably retreated from its winter advance maximum.

## Recent history

The recent history period dates from around 1,000 b2k to present.
Redoubt Volcano, 3,108 m (10, 197 ft) high, is one of the active volcanoes of the Cook Inlet region. The breached summit crater is drained by Drift glacier, which empties into the Drift River valley forming a piedmont lobe. View is to the south on September 1, 1980.

## Hypotheses

* Mount Redoubt has a VEI < 4.
* Mount Redoubt erupts to allow charge to escape through this conductor.
# Volcanoes/Mount Redoubt/Quiz

Mount Redoubt is an article and a lecture from the school of geology about Mount Redoubt as a volcano.
You are free to take this quiz based on Mount Redoubt at any time.
To improve your score, read and study the lecture, the links contained within, listed under See also, External links, and in the {{geology resources}} template. This should give you adequate background to get 100 %.
As a "learning by doing" resource, this quiz helps you to assess your knowledge and understanding of the information, and it is a quiz you may take over and over as a learning resource to improve your knowledge, understanding, test-taking skills, and your score.
Suggestion: Have the lecture available in a separate window.
To master the information and use only your memory while taking the quiz, try rewriting the information from more familiar points of view, or be creative with association.
Enjoy learning by doing!

## Hypotheses

* There are approximately as many volcanoes per 10° of longitude and 5° of latitude as there are glaciers.
# Volcanoes/Mount Somma/Quiz

Mount Somma is a lecture from the school of geology about the volcano Mount Somma.
You are free to take this quiz based on Mount Somma at any time.
To improve your score, read and study the lecture, the links contained within, listed under See also, External links, and in the {{geology resources}} template. This should give you adequate background to get 100 %.
As a "learning by doing" resource, this quiz helps you to assess your knowledge and understanding of the information, and it is a quiz you may take over and over as a learning resource to improve your knowledge, understanding, test-taking skills, and your score.
Suggestion: Have the lecture available in a separate window.
To master the information and use only your memory while taking the quiz, try rewriting the information from more familiar points of view, or be creative with association.
Enjoy learning by doing!

## Hypotheses

* There are approximately as many volcanoes per 10° of longitude and 5° of latitude as there are glaciers.
# Volcanoes/Mount St. Helens/Quiz

Mount St. Helens is a lecture from the school of geology about the volcano Mount St. Helens.
You are free to take this quiz based on Mount St. Helens at any time.
To improve your score, read and study the lecture, the links contained within, listed under See also, External links, and in the {{geology resources}} template. This should give you adequate background to get 100 %.
As a "learning by doing" resource, this quiz helps you to assess your knowledge and understanding of the information, and it is a quiz you may take over and over as a learning resource to improve your knowledge, understanding, test-taking skills, and your score.
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## Hypotheses

* There are approximately as many volcanoes per 10° of longitude and 5° of latitude as there are glaciers.