file transfer commit

Aug.-Sept./exampleSubmission.md

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+# Sklyer MacDougall
+
+## PHYS-111-03
+
+### Assignment
+
+filename=MacDougall_Skyler_Week#HW.pdf
+
+Notes to use when writing:
+
+turn on page numbers, answers in order. Restate question, like normal. Show work. Units. SciNote. SigFigs. Box final mathematical answers. 

Aug.-Sept./extraActivitesQuestions.md

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+1. What is the general process for an approach of a physics problem?
+    1. Read the question
+        1. Understand (or find) the units. 
+        2. Find the requested answer, and its associated units.
+    2. Use provided equations. (These will be taught in class.)
+    3. Convert units (may be optional)
+    4. Ensure the answer makes sense.
+    5. Reduce to required number of significant figures.
+    6. Change answer to scientific notation (or metric prefixes, if allowed). (May be optional, depending on length of value)

Aug.-Sept./lab1.md

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+Due to multiple technical difficulties, not all questions asked in class were properly received, and as such, have not been answered here.
+
+Why are you in this class?
+
+I am in this class because it is required for my major, and I REALLY didn't want to take calc-based physics because I heard it was extremely difficult.
+
+
+

Aug.-Sept./lecture1.md

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+All notes on computers are completely acceptable. If you know LaTeX, then you'll be perfect
+
+Expect notes to be light for this course. We are already familiar with physics, and have taken algebra based physics in high school
+
+
+
+all answers are to be put in SI (unless otherwise stated)
+
+know your base values for the equations. (meter, kilogram, gram, second, hour, etc, etc)
+
+
+

Aug.-Sept./lecture2.md

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+1. When jumping, a flea reaches a takeoff speed of $1.00m/s$ over a distance of $0.500mm$.
+
+    ![image-20200831144218815](lecture2.assets/image-20200831144218815.png)
+
+    | Variable          | Value                 | Significance                                            |
+    | ----------------- | --------------------- | ------------------------------------------------------- |
+    | $v_0=v_{takeoff}$ | $1.00m/s$ (given)     | velocity achieved when transitioning from no $g$ to $g$ |
+    | $\Delta y_1$      | $0.500mm-0mm$ (given) | change in displacement without being affected by $g$    |
+    | $a_1$             |                       | acceleration due to jump                                |
+    | $a_2$             | $9.81m/s^2$ (given)   | acceleration due to gravity                             |
+    | $v_1$             | $0m/s$ (given)        | velocity at peak of jump                                |
+    | $y_2$             |                       |                                                         |
+    |                   |                       |                                                         |
+    |                   |                       |                                                         |
+    |                   |                       |                                                         |
+
+    
+
+    1. What is the flea's acceleration during the jump phase?
+        $$
+        v_{takeoff}^2=v_i^2+2a_1\Delta y_1\\
+        v_{takeoff}^2=\cancel{v_i^2+}2a_1\Delta y_1\\
+        v_{takeoff}^2=2a_1\Delta y_1\\
+        v_{takeoff}^2=2a_1\Delta y_1\\
+        \frac{v_{takeoff}^2}{2\Delta y_1}=a_1\\
+        \frac{(1.00\cancel{m}/s*\frac{10^3mm}{1\cancel{m}})^2}{2(0.500mm-0mm)}=a_1\\
+        \frac{(1.00*{10^3})^2}{2(0.500-0)}\frac{\frac{mm^2}{s^2}}{mm}=a_1\\
+        \frac{(1.00*{10^3})^2}{2(0.500)}\frac{\frac{mm}{s^2}}{1}=a_1\\
+        \frac{(1.00*{10^3})^2}{1}\frac{mm}{s^2}=a_1\\
+        {(1.00*{10^3})^2}\frac{mm}{s^2}=a_1\\
+        {1.00*{10^6}}\frac{mm}{s^2}=a_1\\
+        {1.00*{10^3}}\frac{m}{s^2}=a_1\\
+        \underline{\overline{|{1.00*{10^3}}m/s^2=a_1|}}
+        $$
+
+    2. How long does the acceleration phase last?
+        $$
+        a=\frac{\Delta v}{\Delta t}\\
+        \Delta t=\frac{\Delta v}{a_1}\\
+        \Delta t=\frac{v_{takeoff}-v_i}{a_1}\\
+        \Delta t=\frac{1m/s-0}{1\times10^3m/s^2}\\
+        \Delta t=\frac{1m/s}{1\times10^3m/s^2}\\
+        \Delta t=\frac{1}{1\times10^3}s\\
+        \underline{\overline{|\Delta t=1.00\times 10^-3s|}}
+        $$
+
+    3. If the flea jumps straight up, how high will it go? Ignore air resistance for ease of math.
+        $$
+        v_{max}^2=v_{takeoff}^2+2a_2\Delta y_2\\
+        v_{max}^2=v_{takeoff}^2+2a_2\Delta y_2\\
+        v_{max}^2-v_{takeoff}^2=2a_2\Delta y_2\\
+        \frac{v_{max}^2-v_{takeoff}^2}{2a_2}=\Delta y_2\\
+        \frac{0-(1.00m/s)^2}{2(-9.81m/s^2)}=\Delta y_2\\
+        \frac{(1.00m/s)^2}{2(9.81m/s^2)}=\Delta y_2\\
+        \frac{(1.00m/s)^2}{2(9.81m/s^2)}\frac{m^2/\cancel{s^2}}{m/\cancel{s^2}}=\Delta y_2\\
+        \frac{(1.00m/s)^2}{2(9.81m/s^2)}m=\Delta y_2\\
+        $$

Aug.-Sept./macdougall_skyler_week2.md

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+# Week 1 Activities Problems
+
+## Skyler MacDougall
+
+## PHYS-111 2pm Lab
+
+### Group Members: Will Caliguri, Nina Chambliss, Caroline Cody
+
+#### Due: 8/24/2020
+
+# 1-D Kinematics with a change in acceleration and/or change in direction
+
+1. You're driving down the highway late one night at $20.0m/s$ when a deer steps out onto the road $35.0m$ in front of you. Your reaction time before stepping on the brakes is $0.500s$, and the maximum deceleration of your car is $10.0m/s^2$. ***Numerical calculations*** are required, show your work.
+
+    1. Carefully sketch the **acceleration time, velocity time, and position time graphs** for your car. Numerical values on all axes are required. Show calculations.
+        $$
+        x=vt\\
+        x=(20.0m/s)(0.500s)\\
+        x=10m\\
+        \rule{100pt}{0.4pt}\\
+        v_f=v_i+a(\Delta t)\\
+        0m/s=20m/s-10m/s^2(t_f-0.5s)\\
+        20m/s=10m/s^2(t_f-0.5s)\\
+        \frac{20m/s}{10m/s^2}=(t_f-0.5s)\\
+        \frac{20}{10}s=t_f-0.5s\\
+        \frac{20}{10}s+0.5s=t_f\\
+        2.5s=t_f\\
+        \rule{100pt}{0.4pt}\\
+        x_f=x_i+v_i(\Delta t)+\frac12a(\Delta t)^2\\
+        x_f=10m+20m/s(2.5s-0.5s)+\frac12(-10m/s^2)(2.5s-0.5s)^2\\
+        x_f=10m+20m/s(2s)+\frac12(-10m/s^2)(2s)^2\\
+        x_f=10m+40m+\frac12(-10m/s^2)(4s^2)\\
+        x_f=10m+40m-20m\\
+        x_f=30m
+        $$
+        ![image-20200831190438503](macdougall_skyler_week2.assets/image-20200831190438503.png)
+
+    2. Do you hit the deer? Why or why not?
+        No. When your speed is zero, your distance from where you were when the deer stepped out into the street is 30m, while the deer is 35m away when it steps out onto the street. This means that there is 5m between you and the deer when you come to a whiplash-inducing halt.
+
+2. A hot-air balloon is **ascending** at a rate of $2.00m/s$ when a passenger drops a camera over the side. The camera is $45.0m$ above the ground when it is dropped.
+
+    1. Choose a positive direction for your positive axis, and make a simple drawing of the camera's motion from the moment it leaves the balloon. Use vector arrows in your drawing to show the directions of the camera's initial and final velocity. You can also use a double-wide arrow to show the direction of the acceleration. 
+
+        ![image-20200901152126535](macdougall_skyler_week2.assets/image-20200901152126535.png)
+
+    2. What is the velocity of the camera just before it hits the ground?
+        $$
+        v_f^2=v_i^2+2a(x_f-x_i)\\
+        v_f^2=(2.00m/s)^2+2(-9.81m/s^2)(0m-45.0m)\\
+        v_f=\sqrt{2.00m/s + 2(-9.81m/s^2)(-45.0m)}\\
+        v_f=\sqrt{2.00m/s+(9.81m/s^2)(90.0m)}\\
+        v_f=\pm29.7m/s\\
+        \overline{\underline{|V_f=-29.7m/s|}}
+        $$
+
+    3. How long does it take for the camera to reach the ground?
+        $$
+        v_f=v_i+a_x(t_f-t_i)\\
+        v_f-v_i=a_x(t_f-t_i)\\
+        \frac{v_f-v_i}{a_x}=t_f-t_i\\
+        \frac{v_f-v_i}{a_x}+t_i=t_f\\
+        t_i+\frac{v_f-v_i}{a_x}=t_f\\
+        0s+\frac{-29.7m/s-2.00m/s}{-9.81m/s}=t_f\\
+        \frac{-31.7m/s}{-9.81m/s^2}=t_f\\
+        \underline{\overline{|t_f=3.23s|}}
+        $$
+    
+4. Carefully sketch the **acceleration time, velocity time, and position time graphs** for the camera. Numerical values on all axes are required. Show calculations.
+        ![image-20200901151825285](macdougall_skyler_week2.assets/image-20200901151825285.png)
+    
+5. Suppose the balloon is **decending** at a rate of $2.00m/s$ when the camera is dropped from a height of $45.0m$ above the ground. What is its speed the instant it hits the ground, and how long does it take to hit the ground?
+        $$
+        v_f^2=v_i^2+2a(x_f-x_i)\\
+        v_f^2=(-2.00m/s)^2+2(-9.81m/s^2)(0m-45.0m)\\
+        v_f=\sqrt{-2.00m/s + 2(-9.81m/s^2)(-45.0m)}\\
+        v_f=\sqrt{-2.00m/s+(9.81m/s^2)(90.0m)}\\
+        v_f=\pm29.7m/s\\
+        \overline{\underline{|V_f=-29.7m/s|}}
+        $$
+    
+    $$
+        v_f=v_i+a_x(t_f-t_i)\\
+        v_f-v_i=a_x(t_f-t_i)\\
+        \frac{v_f-v_i}{a_x}=t_f-t_i\\
+        \frac{v_f-v_i}{a_x}+t_i=t_f\\
+        t_i+\frac{v_f-v_i}{a_x}=t_f\\
+        0s+\frac{-29.7m/s-(-2.00m/s)}{-9.81m/s}=t_f\\
+        \frac{-27.7m/s}{-9.81m/s^2}=t_f\\
+        \underline{\overline{|t_f=2.83s|}}
+    $$
+    
+    
+    
+3. A rocket, initially at rest on the ground, accelerates straight upward from rest with a constant acceleration of $58.8m/s^2$. The acceleration period lasts for $9.00s$. After that, the rocket is in free fall. 
+
+    1. Find the maximum height $y_{max}$ reached by the rocket.
+        $$
+        v_f=v_i+a_x(t_f-t_i)\\
+        v_f=0m/s+(58.8m/s^2)(9.00s-0.00s)\\
+        v_f=(58.8m/s^2)(9.00s)\\
+        v_f=529.2m/s\\
+        \rule{100pt}{0.4pt}\\
+        v_f^2=v_i^2+2a(x_f-x_i)\\
+        v_f^2-v_i^2=2a(x_f-x_i)\\
+        \frac{v_f^2-v_i^2}{2a}=(x_f-x_i)\\
+        \frac{v_f^2-v_i^2}{2a}+x_i=x_f\\
+        \frac{(529.2m/s)^2-(0m/s)^2}{2(58.8m/s^2)}+0m=x_f\\
+        \frac{(529.2m/s)^2}{2(58.8m/s^2)}=x_f\\
+        x_f=2381.4m\\
+        \rule{100pt}{0.4pt}\\
+        v_f=v_i+a_x(t_f-t_i)\\
+        t_i+\frac{v_f-v_i}{a_x}=t_f\\
+        9.00s+\frac{0m/s-529.2m/s}{-9.81m/s^2}=t_f\\
+        9.00s+\frac{529.2m/s}{9.81m/s^2}=t_f\\
+        9.00s+53.94s=t_f\\
+        62.94s=t_f\\
+        \rule{100pt}{0.4pt}\\
+        v_f^2=v_i^2+2a(x_f-x_i)\\
+        v_f^2-v_i^2=2a(x_f-x_i)\\
+        \frac{v_f^2-v_i^2}{2a}=(x_f-x_i)\\
+        \frac{v_f^2-v_i^2}{2a}+x_i=x_f\\
+        \frac{(0m)^2-(529.2m/s)^2}{2(-9.81m/s^2)}+2381.4m=x_f\\
+        \frac{(529.2m/s)^2}{2(9.81m/s^2)}+2381.4m=x_f\\
+        x_f=16655.235m\\
+        \overline{\underline{|y_{max}=16.6km|}}
+        $$
+    
+    2. Carefully sketch the rocket's **velocity time graph** for the time interval between first firing and impact with the ground. Label the velocity and time axes with numerical values. 
+    
+        ![image-20200901164434010](macdougall_skyler_week2.assets/image-20200901164434010.png)
+
+
+
+# Vector Algebra - Component Technique
+
+1. An ant undergoes 3 successive displacements:
+    $$
+    \overrightarrow {d_1}=3.00m,\ 20.0^\circ\ south\ of\ east\\
+    \overrightarrow {d_2}=2.00m,\ north\\
+    \overrightarrow {d_3}=7.00m,\ 35.0^\circ\ west\ of\ south
+    $$
+
+    1. Carefully sketch and label the 3 displacements with their tails at the origin. Use a coordinate system in which the positive x-axis is east and the positive y axis is north.
+    
+        ![image-20200903142547804](macdougall_skyler_week2.assets/image-20200903142547804.png)
+    
+    2. Calculate the x and y components of each of the displacements.
+        $$
+        (d_{1})_x=3.00m(cos(360^\circ-20^\circ));\ east=0^\circ=360^\circ\\
+        (d_{1})_x=3.00m(cos(340^\circ))\\
+        (d_{1})_x=3.00m(0.939)\\
+        (d_{1})_x=2.82m\\
+        \rule{100pt}{0.4pt}\\
+        (d_{1})_y=3.00m(sin(360^\circ-20^\circ));\ east=0^\circ=360^\circ\\
+        (d_{1})_y=3.00m(sin(340^\circ))\\
+        (d_{1})_y=3.00m(-0.342)\\
+        (d_{1})_y=-1.02m\\
+        \rule{100pt}{0.4pt}\\
+        (d_{2})_x=2.00m(cos(90^\circ));\ east=0^\circ=360^\circ\\
+        (d_{2})_x=2.00m(cos(90^\circ))\\
+        (d_{2})_x=2.00m(0)\\
+        (d_{2})_x=0.00m\\
+        \rule{100pt}{0.4pt}\\
+        (d_{2})_y=2.00m(sin(90^\circ));\ east=0^\circ=360^\circ\\
+        (d_{2})_y=2.00m(sin(90^\circ))\\
+        (d_{2})_y=2.00m(1)\\
+        (d_{2})_y=2.00m\\
+        \rule{100pt}{0.4pt}\\
+        (d_{3})_x=3.00m(cos(270^\circ-35^\circ));\ east=0^\circ=360^\circ\\
+        (d_{3})_x=7.00m(cos(235^\circ))\\
+        (d_{3})_x=7.00m(-0.574)\\
+        (d_{3})_x=-4.02m\\
+        \rule{100pt}{0.4pt}\\
+        (d_{3})_y=3.00m(sin(270^\circ-35^\circ));\ east=0^\circ=360^\circ\\
+        (d_{3})_y=7.00m(sin(235^\circ))\\
+        (d_{3})_y=7.00m(-0.819)\\
+        (d_{3})_y=-5.73m
+        $$
+    
+        | Variable | X component | Y Component |
+        | -------- | ----------- | ----------- |
+        | $d_1$    | $2.82m$     | $-1.02m$    |
+        | $d_2$    | $0.00m$     | $2.00m$     |
+        | $d_3$    | $-4.02m$    | $-5.73m$    |
+    
+    3. Use the results in part 2 to calculate the distance between where it begins and ends.
+        $$
+        d_x=\sum (d_n)_x\\
+        d_x=(2.82m+0.00m-4.02m)\\
+        d_x=-1.20m\\
+        \rule{100pt}{0.4pt}\\
+        d_y=\sum (d_n)_y\\
+        d_y=(-1.02m+2.00m-5.73m)\\
+        d_y=-4.76m\\
+        \rule{100pt}{0.4pt}\\
+        \theta_x=tan^{-1}(\frac{d_x}{d_y})\\
+        \theta_x=tan^{-1}(\frac{-1.20m}{-4.76m})\\
+        \theta_x=-104^\circ\\
+        \rule{100pt}{0.4pt}\\
+        |d_{total}|=\sqrt{d_x^2+d_y^2}\\
+        |d_{total}|=\sqrt{(-1.20m)^2+(-4.76m)^2}\\
+        |d_{total}|=4.91m\\
+        \overline{\underline{|d_{total}=4.91m\ang-104.^\circ|}}\\
+        \overline{\underline{|d_{total}=4.91m\ang14.1^\circ\ west\ of\ south|}}\\
+        \overline{\underline{|d_{total}=4.91m\ang256.^\circ|}}\\
+        $$
+    
+    4. What displacement is necessary to return the ant to its initial position?
+        $$
+        -4.91m\ang-104.^\circ\\
+        4.91m\ang75.9^\circ
+        $$
+        
+    
+2. 
+
+
+
+# 2-D Motion/Projectile Motion - Part 1
+
+1. In 1780, in what is now referred to as "Brady's Leap", Captain Sam Brady of the US Continental Army escaped certain death from his enemies by running the edge of the cliff above Ohio's Cuyahoga River, which is confined at that spot to a gorge. He landed safely on the far side of the river. 
+
+    ![image-20200903110954934](macdougall_skyler_week2.assets/image-20200903110954934.png)
+
+    1. Representing the horizontal distance jumped as L and the vertical drop as h, as shown in the diagram, derive an expression for the minimum horizontal speed ($v_i$) he would need if he made his leap straight off the cliff.
+        $$
+        L=\overrightarrow {v_t}*t\\
+        h=\frac12gt^2\\
+        \frac{L}{\overrightarrow{v_t}}=t;\ t=\sqrt{\frac{2h}{g}}\\
+        \frac{L}{\overrightarrow{v_t}}=\sqrt{\frac{2h}{g}}\\
+        \frac{L}{1}=\overrightarrow{v_t}\sqrt{\frac{2h}{g}}\\
+        \frac{L}{\sqrt{\frac{2h}{g}}}=\overrightarrow{v_t}\\
+        $$
+        
+    2. Evaluate your expression, given the following. Calculate in $m/s$.
+        $$
+        L=22ft\\
+        h=20ft\\
+        1m=3.28ft
+        $$
+        
+        $$
+        \frac{22\cancel{ft}}{1}\times\frac{1m}{3.28\cancel{ft}}\approx6.707m\\
+        \frac{20\cancel{ft}}{1}\times\frac{1m}{3.28\cancel{ft}}\approx6.098m\\
+        \frac{L}{\sqrt{\frac{2h}{g}}}=\overrightarrow{v_t}\\
+        \frac{6.707m}{\sqrt{\frac{2(6.098m)}{9.81m/s^2}}}=\overrightarrow{v_t}\\
+        \frac{6.707m}{\sqrt{\frac{2(6.098)}{9.81}\frac m{m/s^2}}}=\overrightarrow{v_t}\\
+        \frac{6.707m}{\sqrt{\frac{12.195}{9.81}\frac 1{1/s^2}}}=\overrightarrow{v_t}\\
+        \frac{6.707m}{\sqrt{\frac{12.195}{9.81}s^2}}=\overrightarrow{v_t}\\
+        \frac{6.707m}{\sqrt{\frac{12.195}{9.81}}s}=\overrightarrow{v_t}\\
+        \frac{6.707}{\sqrt{1.243}}m/s=\overrightarrow{v_t}\\
+        \frac{6.707}{1.114}m/s=\overrightarrow{v_t}\\
+        6.0158m/s=\overrightarrow{v_t}\\
+        \overline{\underline{|6.0m/s=\overrightarrow{v_t}|}}
+        $$
+    
+2. Is it reasonable that a person could make this leap? Use the fact that the world record for the 100m dash is approximately $10s$ to estimate the maximum speed such a runner would have.
+    $$
+    v_{max}=\frac dt\\
+    v_{max}=\frac {100m}{10s}\\
+    v_{max}=\frac {100}{10}m/s\\
+    v_{max}=10m/s\\
+    \overrightarrow{v_t}<v_{max}\\
+    \therefore\\
+    It\ is\ reasonable\ for\\ a\ person\ to\ make\\ this\ leap.
+    $$
+
+3. In a shot-put event, an athlete throws the shot with an initial velocity of $\overrightarrow {v_i}=12.0m/s \ang40.0^\circ$. The shot leaves her hand at a height of $1.80m$ above the ground.
+    $$
+    \overrightarrow{v_i}=12.0m/s\ang40^\circ\\
+    v_{i_x}=12m/s(cos(40^\circ))\approx9.192m/s\\
+    v_{i_y}=12m/s(sin(40^\circ))\approx7.713m/s
+    $$
+
+    1. How long is the shot in the air?
+        $$
+        v_{f_y}^2=v_{i_y}^2+2a_y(x_{f_y}-x_{i_y})\\
+        v_{f_y}=-\sqrt{v_{i_y}^2+2a_y(x_{f_y}-x_{i_y})}\\
+        v_{f_y}=v_{i_y}+a_y(t_f-t_i)\\
+        v_{f_y}-v_{i_y}=a_y(t_f-t_i)\\
+        \frac{v_{f_y}-v_{i_y}}{a_y}=(t_f-t_i)\\
+        \frac{v_{f_y}-v_{i_y}}{a_y}+t_i=t_f\\
+        \frac{-\sqrt{v_{i_y}^2+2a_y(x_{f_y}-x_{i_y})}-v_{i_y}}{a_y}+t_i=t_f\\
+        \frac{-\sqrt{(7.713m/s)^2+2(-9.81m/s^2)(0m-1.8m)}-7.713m/s}{-9.81m/s^2}+0s=t_f\\
+        \frac{-\sqrt{((7.713)^2+2(-9.81)(-1.8))m^2/s^2}-7.713m/s}{-9.81m/s^2}=t_f\\
+        \frac{(-\sqrt{(7.713)^2+2(-9.81)(-1.8)})m/s-7.713m/s}{-9.81m/s^2}=t_f\\
+        \frac{(-\sqrt{(7.713)^2+2(-9.81)(-1.8)})m/s-7.713m/s}{-9.81m/s^2}=t_f\\
+        \frac{-\sqrt{(7.713)^2+2(-9.81)(-1.8)}-7.713}{-9.81}\frac{m/s}{m/s^2}=t_f\\
+        \frac{-\sqrt{(7.713)^2+2(-9.81)(-1.8)}-7.713}{-9.81}\frac{1}{1/s}=t_f\\
+        \frac{-\sqrt{(7.713)^2+2(-9.81)(-1.8)}-7.713}{-9.81}s=t_f\\
+        \frac{-\sqrt{(7.713)^2+35.316}-7.713}{-9.81}s=t_f\\
+        \frac{-9.737-7.713}{-9.81}s=t_f\\
+        \frac{-17.451}{-9.81}s=t_f\\
+        1.7789s=t_f\\
+        \overline{\underline{|1.78s=t_f|}}
+        $$
+
+    2. How far horizontally does the shot travel?
+        $$
+        d_x=\frac{v_x}{\Delta t}\\
+        d_x=\frac{9.912m/s}{1.78s-0s}\\
+        d_x=\frac{9.912m/s}{1.78s}\\
+        d_x=\frac{9.912}{1.78}\frac{m/s}{s}\\
+        d_x=\frac{9.912}{1.78}m\\
+        d_x=5.1676m\\
+        \underline{\overline{|d_x=5.17m|}}
+        $$
+
+    3. What is the maximum height of the shot above the ground?
+        $$
+        d_{y_{max}}\ when\ v_y=0m/s\\
+        v_{f_y}^2=v_{i_y}^2+2a_y(x_{f_y}-x_{i_y})\\
+        v_{f_y}^2-v_{i_y}^2=2a_y(x_{f_y}-x_{i_y})\\
+        \frac{v_{f_y}^2-v_{i_y}^2}{2a_y}=x_{f_y}-x_{i_y}\\
+        \frac{v_{f_y}^2-v_{i_y}^2}{2a_y}+x_{i_y}=x_{f_y}\\
+        \frac{0m/s^2-(7.713m/s)^2}{2(-9.81m/s^2)}+1.80m=x_{f_y}\\
+        \frac{-(7.713m/s)^2}{2(-9.81m/s^2)}+1.80m=x_{f_y}\\
+        \frac{-(7.713)^2m^2/s^2}{2(-9.81)m/s^2}+1.80m=x_{f_y}\\
+        \frac{-(7.713)^2}{2(-9.81)}\frac{m^2/s^2}{m/s^2}+1.80m=x_{f_y}\\
+        \frac{-(7.713)^2}{-19.62}\frac{m}{1}+1.80m=x_{f_y}\\
+        \frac{-59.497}{-19.62}m+1.80m=x_{f_y}\\
+        (\frac{59.497}{19.62}+1.80)m=x_{f_y}\\
+        (3.03248+1.80)m=x_{f_y}\\
+        4.832m=x_{f_y}\\
+        \overline{\underline{|4.83m=x_{f_y}|}}
+        $$
+
+    4. How long does it take for the shot to reach its maximum height?
+        $$
+        v_{f_y}=v_{i_y}+a_y(t_f-t_i)\\
+        v_{f_y}-v_{i_y}=a_y(t_f-t_i)\\
+        \frac{v_{f_y}-v_{i_y}}{a_y}=(t_f-t_i)\\
+        \frac{v_{f_y}-v_{i_y}}{a_y}+t_i=t_f\\
+        \frac{0m/s-7.713m/s}{-9.81m/s^2}+0s=t_f\\
+        \frac{-7.713m/s}{-9.81m/s^2}=t_f\\
+        0.786s=t_f\\
+        \overline{\underline{|0.786s=t_f|}}
+        $$
+
+    5. Carefully sketch the horizontal and vertical components of the shot's velocity as a function of time. Label all axes with numerical values.
+
+        ![image-20200903122958061](macdougall_skyler_week2.assets/image-20200903122958061.png)
+
+    6. Carefully sketch the horizontal and vertical components of the shot's position as a function of time. Label all axes with numerical values.
+
+        ![image-20200903121958162](macdougall_skyler_week2.assets/image-20200903121958162.png)

Aug.-Sept./macdougall_skyler_week3.md

@@ -0,0 +1,189 @@
+# Week 3 Activities Problems
+
+## Skyler MacDougall
+
+## PHYS-111 2pm Lab
+
+### Group Members: Will Caliguri, Nina Chambliss, Caroline Cody
+
+#### Due: 9/11/2020
+
+# 2-D Kinematics
+
+1. A baseball is launched by a pitching machine on the ground at an angle of $30.0^\circ$ from the horizontal with an initial speed of $20.0m/s$.
+
+    ![image-20200907151113320](macdougall_skyler_week3.assets/image-20200907151113320.png)
+
+    1. How long does it take for the ball to hit the ground?
+        $$
+        v_i=20m/s\ang30^\circ\\
+        v_{ix}\approx17.3m/s;\ v_{iy}=10m/s\\
+        v_f=v_i+at\\
+        0=10m/s+-9.81m/s*t\\
+        t\approx1.02s\\
+        2t\approx2.04s\\
+        \overline{\underline{|t=2.04s|}}
+        $$
+
+    2. What is the horizontal range of the ball?
+        $$
+        d=vt\\
+        d=17.3m/s*2.04s\\
+        d=16.135m\\
+        \overline{\underline{|d=16.1m|}}
+        $$
+        
+    3. What is the maximum height?
+        $$
+        x=vt+\frac12at^2\\
+        x=(10m/s)*1.02s+\frac12(-9.81m/s^2)(1.02s)^2\\
+        \overline{\underline{|x=15.3m|}}
+        $$
+    
+4. Sketch the horizontal and vertical components of the balls velocity.
+   
+    ![image-20200908114700619](macdougall_skyler_week3.assets/image-20200908114700619.png)
+    
+5. Sketch the horizontal and vertical components of the balls position.
+   
+    ![image-20200908114727947](macdougall_skyler_week3.assets/image-20200908114727947.png)
+    
+2. You want to throw a baseball over a wall that has a height of $10.0m$ and is $6.00m$ away from you. Assume that the wall is negligibly thick, and the baseball is small enough that you can consider it a point particle. Ignore air resistance.
+
+    1. What is the minimum initial speed that the ball can have and still clear the wall? Assume that the minimum trajectory is the one in which the ball reaches the maximum height the instant it gets to the top of the wall.
+        $$
+        v_f^2=v_i^2+2ax\\
+        (0m/s)^2=v_i^2+2(-9.81m/s^2)(10.0m)\\
+        2(9.81m/s^2)(10.0m)=v_i^2\\
+        v_i=14.01m/s\\
+        v_f=v_i+at\\
+        0m/s=14.01m/s+(-9.81m/s^2)t\\
+        \frac{14.01m/s}{9.81m/s^2}=t\\
+        t=1.43s\\
+        d=vt\\
+        6.00m=v(1.43s)\\
+        \frac{6.00m}{1.43s}=v\\
+        v=4.20m/s\\
+        v_{total}=14.01m/s+i(4.20m/s)\\
+        |v_{total}|=\sqrt{(14.01m/s)^2+(4.20m/s)^2}\\
+        |v_{total}|=14.6m/s\\
+        \ang V=tan^{-1}(\frac{4.20m/s}{14.01m/s})\\
+        \ang V=16.7^\circ\\
+        \overline{\underline{|v_{total}=14.6m/s\ang16.7^\circ|}}
+        $$
+    
+2. If you throw the ball at this speed, what is the angle relative to the horizontal should you throw it?
+        $$
+        \theta=16.7^\circ
+        $$
+        
+    
+3. A snowball rolls off a barn roof that slopes downward at an angle of $40.0^\circ$ as shown in the figure. The edge of the roof is $14.0m$ above the ground, and the snowball has a speed of $7.00 m/s$ as it rolls off the roof. Ignore air resistance.
+
+    ![image-20200907151147523](macdougall_skyler_week3.assets/image-20200907151147523.png)
+
+    1. How far from the edge of the barn does the snowball strike the ground, if it doesn't strike anything else while falling?
+        $$
+        angle\ given=180^\circ-40^\circ\\
+        \therefore\\
+        v_0=-7.00m/s\ang140^\circ\\
+        v_{0x}\approx5.36m/s;\ v_{0y}\approx-4.50m/s\\
+        v_f^2=v_i^2+2ax\\
+        v_f=\sqrt{(-4.5m/s)^2+2(9.81m/s^2)(14.00m)}\\
+        v_f\approx-17.2m/s\\
+        v_f=v_i+at\\
+        -17.2m/s=-4.5m/s+(9.81m/s^2)(t)\\
+        t=\frac{12.7m/s}{9.81m/s^2}\\
+        t\approx1.29s\\
+        d=vt\\
+        d=5.36m/s*1.29s\\
+        \overline{\underline{|d=6.93m|}}
+        $$
+    
+2. A man $1.90m$ tall is standing $4.00m$ at the edge of the barn. Will he be hit by the snowball?
+    $$
+    d=vt\\
+        4.00m=5.36m/s*t\\
+        t=0.745s\\
+        d=d_i+vt+\frac12at^2\\
+        d=14.0m+(-4.5m/s)(0.745s)+\frac12(-9.81m/s^2)(0.745s)^2\\
+        d=14.0m-((4.5m/s)(0.745s)+\frac12(9.81m/s^2)(0.745s)^2)\\
+        d=7.91m\\
+        7.91m>1.90m\\
+        \therefore
+    $$
+    ​    The man will not get hit.
+    
+8. Sketch the graphs for position and velocity vs time.
+    ![image-20200908121336160](macdougall_skyler_week3.assets/image-20200908121336160.png)
+    ![image-20200908120504051](macdougall_skyler_week3.assets/image-20200908120504051.png)
+
+
+
+# Forces and Free Body Diagrams
+
+1. For each situation described, draw a free body diagram. Be sure to use standard notation for the forces and define your coordinate system.
+
+    1. Your textbook is at rest on a horizontal table. 
+
+        ![image-20200907160040681](macdougall_skyler_week3.assets/image-20200907160040681.png)
+
+    2. You push vertically down on the book with a force $\overrightarrow {F_{push}}$. The book does not move.
+
+        ![image-20200907160222877](macdougall_skyler_week3.assets/image-20200907160222877.png)
+
+    3. A crate is held at rest above the floor by a single rope. 
+
+        ![image-20200907160333504](macdougall_skyler_week3.assets/image-20200907160333504.png)
+
+2. A block of weight W is at rest on a rough surface inclined at an angle $\theta$ with respect to the normal. Draw this free body diagram twice, defining axes that are parallel then perpendicular to the incline.
+
+    ![image-20200907163138137](macdougall_skyler_week3.assets/image-20200907163138137.png)
+
+
+
+# Ball Drop Pre-Lab
+
+1. As part of the Ball Drop experiment, you will use LoggerPro to fit data for the position of the ball as a function of time during free fall to the quadratic equation.
+
+    1. Write the SI units for each of the constants.
+        $$
+        A:m/s^2;\ B:m/s;\ C:m
+        $$
+
+    2. Name the physical quantity that has the same units as you wrote above for each of the constants.
+        $$
+        A: Acceleration;\ B:initial\ velocity;\ C:initial\ position
+        $$
+
+2. The vertical component of the velocity of a bouncing ball as a function of time is shown in the graph below. The positive y direction is vertically up. The ball deforms slightly while it is in contact with the ground.
+
+    ![image-20200908121633283](macdougall_skyler_week3.assets/image-20200908121633283.png)
+    
+    1. Identify at least 2 non-zero instances of time at which the ball is at its maximum height. 
+        $$
+    t=2.25s;\ t=4.5s
+        $$
+
+    2. Calculate the acceleration of the ball while it is in the air. 
+    $$
+    a=\frac{\Delta v}{\Delta t}\\
+        a=\frac{-9.0m/s-9.0m/s}{3.0s-1.0s}\\
+        \overline{\underline{|a=-9.0m/s^2|}}
+    $$
+    
+    3. Calculate the acceleration of the ball while it is in contact with the floor. 
+        $$
+        a=\frac{\Delta v}{\Delta t}\\
+        a=\frac{9.0m/s-(-9.0m/s)}{1.25s-1.0s}\\
+        \overline{\underline{|a=72.m/s^2|}}
+        $$
+        
+    4. Using the area under the velocity time graph, calculate the maximum height above the floor that the ball reaches. Be sure that you only utilize the portion of the velocity time graph that corresponds to the ball in free fall.
+        $$
+        x_f=vt+\frac12at^2\\
+        x_f=(9.0m/s)(1s)+\frac12(-9.0m/s^2)(1s)^2\\
+        x_f=9.0m-4.5m\\
+        \overline{\underline{|x_f=4.5m|}}
+        $$
+        

Aug.-Sept./macdougall_skyler_week4.md

@@ -0,0 +1,410 @@
+# Week 4 Activities Problems
+
+## Skyler MacDougall
+
+## PHYS-111 2pm Lab
+
+### Group Members: Will Caliguri, Nina Chambliss, Caroline Cody
+
+#### Due: 9/20/2020
+
+# Forces and Newton's Second Law - Part 1
+
+For each of the situations described below, before answering any questions:
+
+- Draw a simple schematic of the physical situation
+- identify the object of interest
+- identify the forces acting on the object
+- draw a free body diagram
+- Apply Newton's Second Law in component form
+
+1. A ball of mass $m$ is tied to a string fixed to the ceiling. The ball is pulled to one side and held at rest by an unknown horizontal force $\overrightarrow F$ while the string makes angle $\theta$ with respect to the horizontal (negative from the $0^\circ$ standard)[^2]. 
+
+    ![free body diagram 1](macdougall_skyler_week4.assets/question1FreeBodyDiagram.png)
+
+    1. What is the magnitude of the tension T on the string?[^1] 
+        $$
+        \sum F=0=\overrightarrow W+\overrightarrow F-\overrightarrow T\\
+        \overrightarrow W+\overrightarrow F=\overrightarrow T\\
+        gm\ang-90^\circ+|\overrightarrow F|\ang0^\circ=\overrightarrow T\\
+        |\overrightarrow T|=\sqrt{{(gm)}^2+{\overrightarrow F}^2}
+        $$
+
+    2. What is the magnitude of the Force holding the ball in position?[^1] 
+        $$
+        |\overrightarrow F|=cos(\theta)
+        $$
+        Determine the numerical values for T and F if:
+        $$
+        m=2.00kg\\
+        \theta=30.0^\circ
+        $$
+
+        $$
+        |\overrightarrow F|=cos(\theta);\ |\overrightarrow T|=\sqrt{{(gm)}^2+{\overrightarrow F}^2}\\
+        |\overrightarrow F|=cos(30.0^\circ)*1N={\sqrt3\over2}N\approx0.866N\\
+        |\overrightarrow T|=\sqrt{{(gm)}^2+{\overrightarrow F}^2}\\
+        |\overrightarrow T|=\sqrt{{((9.81m/s^2)(2.00kg))}^2+{(\frac{\sqrt3}2N)}^2}\\
+        |\overrightarrow T|=\sqrt{(19.6133N)^2+{\frac34N^2}}\\
+        |\overrightarrow T|=19.632N\\
+        \underline{\overline{||\overrightarrow F|=0.866N;\ |\overrightarrow T|=19.6N|}}
+        $$
+
+2. A student wants to hang an RIT sign with weight $W$ at rest from 2 cables that make angles $\theta_1$ and $\theta_2$ with respect to the horizontal. (Note that $\theta_1$ is negative from the $0^\circ$ standard, while $\theta_2$ is positive from the $180^\circ$ standard) [^2]Let $T_1$ be the unknown tension in cable 1 and $T_2$ the tension in cable 2. Treat the sign as a standard point mass.
+
+    1. Draw a free body diagram for the sign. 
+
+        ![free body diagram 2](macdougall_skyler_week4.assets/question2FreeBodyDiagram.png)
+
+    2. Determine the expression for the tension $T_1$. [^1]
+        $$
+        \overrightarrow W=\overrightarrow {T_1}+\overrightarrow {T_2}\\
+        \overrightarrow{T_{1x}} =\overrightarrow{T_1}\cos(\theta_1);\ \overrightarrow{T_{1y}} =\overrightarrow{T_1}\sin(\theta_1)\\
+        \overrightarrow{T_{2x}} =\overrightarrow{T_2}\cos(\theta_2);\ \overrightarrow{T_{2y}} =\overrightarrow{T_2}\sin(\theta_2)\\
+        \overrightarrow {W_x}=\overrightarrow {T_{1x}}+\overrightarrow {T_{2x}}\\
+        \overrightarrow {W_y}=\overrightarrow {T_{1y}}+\overrightarrow {T_{2y}}\\
+        0=\overrightarrow {T_{1x}}+\overrightarrow {T_{2x}}\\\rule{100pt}{0.4pt}\\
+        -\overrightarrow {T_{1x}}=\overrightarrow {T_{2x}}\\
+        \overrightarrow {W}=\overrightarrow {T_{1y}}+\overrightarrow {T_{2y}}\\
+        \overrightarrow {W}-\overrightarrow {T_{1y}}=\overrightarrow {T_{2y}}\\
+        \overrightarrow{T_1}=\overrightarrow{T_2}\frac{\cos(\theta_2)}{\cos(\theta_1)}\\
+        \overrightarrow{T_2}=\frac{\overrightarrow{W}-\overrightarrow {T_1}\sin(\theta_1)}{\sin(\theta_2)}\\
+        \overrightarrow{T_1}=\frac{\overrightarrow{W}-\overrightarrow {T_1}\sin(\theta_1)}{\sin(\theta_2)}\frac{\cos(\theta_2)}{\cos(\theta_1)}\\
+        \overrightarrow{T_1}=\frac{\overrightarrow{W}-\overrightarrow {T_1}\tan(\theta_1)}{\tan(\theta_2)}\\
+    \overrightarrow{T_1}\tan(\theta_2)=\overrightarrow W-\overrightarrow {T_1}\tan(\theta_1)\\
+        \overrightarrow{T_1}(\tan(\theta_2)+\tan(\theta_1))=\overrightarrow W\\
+        \overrightarrow{T_1}=\frac{\overrightarrow W}{\tan(\theta_1)+\tan(\theta_2)}\\
+        \overline{\underline{|\overrightarrow{T_1}=\frac{\overrightarrow W}{\tan(\theta_1)+\tan(\theta_2)}|}}
+        $$
+        
+    3. Determine the expression for the tension $T_2$. [^1]
+        $$
+        \overrightarrow{T_2}=\frac{\overrightarrow{W}-\overrightarrow {T_1}\sin(\theta_1)}{\sin(\theta_2)}\\
+        \overrightarrow{T_1}=\overrightarrow{T_2}\frac{\cos(\theta_2)}{\cos(\theta_1)}\\
+    \overrightarrow{T_2}=\frac{\overrightarrow{W}-\overrightarrow{T_2}\frac{\cos(\theta_2)}{\cos(\theta_1)}\sin(\theta_1)}{\sin(\theta_2)}\\
+        \overrightarrow{T_2}\sin(\theta_2)={\overrightarrow{W}-\overrightarrow{T_2}\frac{\cos(\theta_2)}{\cos(\theta_1)}\sin(\theta_1)}\\
+        \overrightarrow{T_2}=\frac{\overrightarrow W}{\sin(\theta_2)+\frac{\cos(\theta_2)}{\cos(\theta_1)}\sin(\theta_1)}\\
+        \overline{\underline{|\overrightarrow{T_2}=\frac{\overrightarrow W}{\sin(\theta_2)+\tan(\theta_1)\cos(\theta_2)}|}}
+        $$
+        
+    4. The weight of the sign is $W=225N$ and the angles are $\theta_1=25.0^\circ;\ \theta_2=55.0^\circ$. If the student chooses cables that are each capable of supporting a maximum tension of $150N$, will the design work?
+        $$
+        \overrightarrow{T_1}=\frac{\overrightarrow W}{\tan(\theta_1)+\tan(\theta_2)};\ \overrightarrow{T_2}=\frac{\overrightarrow W}{\sin(\theta_2)+\tan(\theta_1)\cos(\theta_2)}\\
+        \overrightarrow{T_1}=\frac{225N}{\tan(25.0^\circ)+\tan(55.0^\circ)}\\
+        \overrightarrow{T_1}=\frac{225N}{1.89}\\
+        \overrightarrow {T_1}=118.767N\\
+        \overrightarrow{T_2}=\frac{\overrightarrow W}{\sin(\theta_2)+\tan(\theta_1)\cos(\theta_2)}\\
+        \overrightarrow{T_2}=\frac{225N}{\sin(55.0^\circ)+\tan(25.0^\circ)\cos(55.0^\circ)}\\
+        \overrightarrow {T_2}=207.065N\\
+        \overline{\underline{|\overrightarrow {T_1}=119N;\ \overrightarrow {T_2}=207N|}}
+        $$
+
+    This design will not work.
+
+3. A block of weight $W$ is at rest on a rough surface inclined at an angle $\theta$ with respect to the horizontal (positive from the $0^\circ$ standard)[^2]. 
+
+    1. What is the direction of friction force acting on the block? What direction is the normal force acting on the block? 
+        The friction force is in the positive X. (Up the slope of the sloped surface.) The normal force is acting in the positive Y. (Away from the sloped surface.) 
+
+    2. Draw a free body diagram.
+
+        ![free body diagram 3](macdougall_skyler_week4.assets/question3FreeBodyDiagram.png)
+
+    3. Determine the expression for the magnitude of the friction force acting on the block. [^1]
+        $$
+        \overrightarrow {F_F}=|\overrightarrow{F_F}|\ang0^\circ\\
+        \overrightarrow {F_N}=|\overrightarrow {F_N}|\ang90^\circ\\
+        \overrightarrow W=|\overrightarrow W|\ang(270-\theta)^\circ\\
+        \overrightarrow {F_F}=|\overrightarrow W|\cos(\theta)
+        $$
+        
+    4. Determine the expression for the magnitude of the normal force acting on the block. [^1]
+        $$
+        \overrightarrow {F_N}=|\overrightarrow W|\sin(\theta)
+        $$
+        
+    5. Determine the numerical values for the magnitude of the frictional force and the normal force if
+    
+    $$
+        W=25.0N\\
+        \theta=30.0^\circ
+    $$
+    
+    $$
+    \overrightarrow {F_F}=|\overrightarrow W|\cos(\theta);\ \overrightarrow {F_N}=|\overrightarrow W|\sin(\theta)\\
+    \overrightarrow {F_F}=25.0N*\cos(30.0^\circ);\ \overrightarrow {F_N}=25.0N\sin(30.0^\circ)\\
+    \overline{\underline{|\overrightarrow{F_F}=21.7N;\ \overrightarrow{F_N}=12.5N|}}
+    $$
+    
+    
+    
+4. A $25.0N$ book is held at rest against a rough vertical surface by force $\overrightarrow{F_{push}}$ of magnitude $18.0N$, directed at $60.0^\circ$ (positive relative to $0^\circ$ standard)[^2]. 
+
+    1. Draw a free body diagram for the book.
+
+        ![free body diagram 4](macdougall_skyler_week4.assets/question4FreeBodyDiagram.png)
+        
+    2. What is the magnitude and direction of the friction force acting on the book?
+        $$
+        |\overrightarrow{F_F}|=|\overrightarrow{F_{push}}|sin(\theta)-\overrightarrow W\\
+        |\overrightarrow{F_F}|=18.0Nsin(60^\circ)-25.0N\\
+        |\overrightarrow{F_F}|=-9.41N\\
+        \ang\overrightarrow{F_F}=-90^\circ\\
+        \therefore\\
+        \overline{\underline{|\overrightarrow{F_F}=9.41N\ang90^\circ|}}
+        $$
+        Note: $90^\circ$ in this instance refers to directly up on the diagram shown above.
+
+    3. If the magnitude of the push force $\overrightarrow{F_{push}}$ is instead $36.0N$, what is the magnitude and direction of the friction force?
+        $$
+        |\overrightarrow{F_F}|=|\overrightarrow{F_{push}}|sin(\theta)-\overrightarrow W\\
+        |\overrightarrow{F_F}|=36.0Nsin(60^\circ)-25.0N\\
+        |\overrightarrow{F_F}|=6.1769N\\
+        \ang\overrightarrow{F_F}=-90^\circ\\
+        \therefore\\
+        \overline{\underline{|\overrightarrow{F_F}=6.18N\ang-90^\circ|}}
+        $$
+        
+
+5. You are standing on a bathroom scale inside an elevator. Your weight is $140lb$, but the reading of the scale is $120lb$. The scale reads the magnitude of the normal force. ($1lb=4.448N$).
+
+    ![image-20200917144225737](macdougall_skyler_week4.assets/image-20200917144225737.png)
+
+    1. What is the magnitude and direction of the acceleration of the elevator?
+        Your weight decreases, meaning the normal force decreases. Therefore, the direction of the acceleration is up.
+        $$
+        |\overrightarrow {F_a}|=\overrightarrow W-\overrightarrow {F_{normal}}\\
+        |\overrightarrow {F_a}|=(140\cancel{lb}*4.448N/\cancel{lb})-(120\cancel{lb}*4.448N/\cancel{lb})\\
+        |\overrightarrow {F_a}|=88.96N\\
+        \overrightarrow {F_a}=ma\\
+        \overrightarrow {F_a}=(\frac{\overrightarrow W}{g})a\\
+        (88.95\cancel{N})(\frac{-9.81m/s^2}{622.72\cancel{N}})=a\\
+    (88.95)(\frac{-9.81}{622.72})m/s^2=a\\
+        a=-1.401m/s^2\\
+        \underline{\overline{|a=-1.4m/s^2|}}
+        $$
+        
+    2. Can you tell whether the elevator is speeding up or slowing down? Explain.
+        There is no way to know, with the current information available, if the elevator is speeding up or slowing down. It could be speeding up, which would mean its going up. Alternatively, it could be slowing down, which would mean its going down.
+
+6. A $425g$ shuffleboard disk is given an initial speed of $3.20m/s$ across a horizontal court. The disk travels $6.00m$ in a straight line before coming to rest. Assume the acceleration of the disk is constant.
+
+    1. Draw a free body diagram. 
+
+        ![image-20200917000648342](macdougall_skyler_week4.assets/image-20200917000648342.png)
+
+    2. What is the net force exerted on the disk while it is coming to rest?
+        "Assume the acceleration of the disk is constant."
+        Because acceleration is not zero, then the sum of forces is not zero. The normal force and the weight force will cancel out though, leaving only the drag or kinetic friction force.
+        Therefore, the net force is the kinetic friction force, noted for the rest of the problem as $\overrightarrow {F_{drag}}$.
+
+    3. What is the magnitude of the kinetic friction force exerted on the disk by the court?
+        $$
+        v_f^2=v_i^2+2a\Delta x\\
+        v_f^2-v_i^2=2a\Delta x\\
+        \frac{v_f^2-v_i^2}{2\Delta x}=a\\
+        \frac{(0m/s)^2-(3.20m/s)^2}{2(6.00m)}=a\\
+        a=0.85\overline3m/s^2\\
+        \overrightarrow {F_{drag}}=ma\\
+        \overrightarrow {F_{drag}}=(0.425kg)(0.85\overline3m/s^2)\\
+        \overline{\underline{| \overrightarrow {F_{drag}}=363.mN |}}
+        $$
+    
+4. What is the coefficient of kinetic friction between the disk and the court?
+        $$
+        \mu_k=\frac{\overrightarrow {F_k}}{\overrightarrow {F_N}}\\
+        \overrightarrow {F_N}=ma=0.425kg*9.81m/s^2=4.169N\\
+        \overrightarrow {F_k}=\overrightarrow {F_{drag}}=363.mN\\
+        \mu_k=\frac{363.mN}{4.169N}\\
+        \mu_k=0.086986\\
+        \overline{\underline{| \mu_k=8.70*10^{-2}|}}
+        $$
+        
+    
+7. A box of mass $m$ is released from rest on a smooth ramp inclined at an angle $\theta$ above the horizontal (negative relative to $180^\circ$ standard). Orient the positive x-axis down the ramp.
+
+    1. What does the term "smooth ramp" mean?
+        "Smooth ramp" means that friction is negligible.
+        
+    2. Draw a free body diagram.
+    
+        ![image-20200917115108763](macdougall_skyler_week4.assets/image-20200917115108763.png)
+    
+    3. Determine an expression for the acceleration of the box. [^1]Does the acceleration of the box depend on the mass $m$?
+        $$
+        F=ma;\ \overrightarrow {F_{net}}=\overrightarrow{F_{normal}}\cos{\theta}\\
+        \overrightarrow {F_{normal}}=\frac{\sin(\theta)}{\overrightarrow W}\\
+        \overrightarrow{F_{net}}=\frac{\overrightarrow W}{\sin(\theta)}\cos(\theta)\\
+        a_{F_{net}}=\frac{{\overrightarrow W}{\sin(\theta)}\cos(\theta)}{m}\\
+        \overrightarrow W=gm\\
+        a_{F_{net}}=\frac{{gm*}{\sin(\theta)}\cos(\theta)}{m}\\a_{F_{net}}=\frac{{g\cancel{m}*}{\sin(\theta)}\cos(\theta)}{\cancel{m}}\\
+        \overline{\underline{|a_{F_{net}}=g\sin(\theta)\cos(\theta)|}}
+        $$
+        No, the acceleration is not dependent on the mass.
+    
+    4. If the mass is $10.0kg$ and $\theta=55.0^\circ$, what is the magnitude of the acceleration of the box?
+        $$
+        a_{F_{net}}=g\sin(\theta)\cos(\theta)\\
+        a_{F_{net}}=(9.81m/s^2)\sin(55^\circ)\cos(55^\circ)\\
+        a_{F_{net}}=4.60919m/s^2\\
+        \overline{\underline{|a_{F_{net}}=4.61m/s^2|}}
+        $$
+    
+    5. How long does it take for the box to reach a speed of $10.0m/s$?
+        $$
+        t=\frac av\\
+        t=\frac{4.61m/s^2}{10m/s}\\
+        t=0.461s\\
+        x_f=x_i+v_it+\frac12at^2\\
+        x_f=\cancel{0m+(0m/s)(0.461s)+}\frac12(4.61m/s^2)(0.461s)^2\\
+        x_f=\frac12(4.61m/s^2)(0.461s)^2\\
+        x_f=0.489m\\
+        \overline{\underline{|t=0.461s;\ x_f=0.489m|}}
+        $$
+    
+    6. How far down the ramp has the box slid when it reaches a speed of $10.0m/s$?
+        $$
+        \overline{\underline{|x_f=0.489m|}}
+        $$
+        
+    7. Draw a velocity-time graph for values of speed between $0$ and $10.0m/s$.
+    
+        ![image-20200917114126804](macdougall_skyler_week4.assets/image-20200917114126804.png)
+    
+8. A horizontal force $\overrightarrow F$ causes a block of mass $m$ to slide up a rough incline surface with an acceleration of magnitude $a$. The surface is inclined at an angle $\theta$ above the horizontal (positive from the $0^\circ$ standard). Orient the positive x-axis up the ramp. 
+
+    1. What does the term "rough surface" mean?
+        "Rough surface" means that friction is not negligible.
+        
+    2. What type of friction (kinetic or static) acts on the block? What is the direction of the frictional force acting on the block?
+        Because the block is moving, its kinetic friction. It is in the opposite direction of the movement.
+        
+    3. Draw a free body diagram.
+
+        ![image-20200917120040907](macdougall_skyler_week4.assets/image-20200917120040907.png)
+
+    4. Determine an equation for the magnitude of the frictional force. [^1]
+        $\overrightarrow {F_{normal}}=\overrightarrow Wsin(\theta+90^\circ)$ because $\theta$ is the angle from zero, and the normal force is $90^\circ$ farther around the circle (counterclockwise) than $\theta$. This is because the normal force is alway perpendicular to the surface of interaction.
+        $$
+        \overrightarrow {F_{net}}=ma\ang\theta;\ a\ne g\\
+        \overrightarrow {F_{normal}}=\overrightarrow W\sin(\theta+90^\circ)\\
+        ma\cos(\theta)=\overrightarrow F-(\overrightarrow {F_{drag}}cos(\theta)+\overrightarrow {F_{normal}}\cos(\theta))\\
+        ma\cos(\theta)+\overrightarrow {F_{drag}}cos(\theta)=\overrightarrow F-\overrightarrow {F_{normal}}\cos(\theta)\\
+        \overrightarrow {F_{drag}}\cos(\theta)=\overrightarrow F-\overrightarrow{F_{normal}}\cos(\theta)-ma\cos(\theta)\\
+        \overrightarrow {F_{drag}}=\frac{\overrightarrow F}{\cos\theta}-\overrightarrow {F_{normal}}-ma\\
+        \overline{\underline{|\overrightarrow {F_{drag}}=\frac{\overrightarrow F}{\cos\theta}-mg\sin(\theta+90^\circ)-ma|}}
+        $$
+
+    5. Determine an equation for the magnitude of the normal force. [^1]
+        $$
+        \overrightarrow {F_{normal}}=\overrightarrow W\sin(\theta)\\
+        \underline{\overline{|\overrightarrow {F_{normal}}=mg\sin(\theta)|}}
+        $$
+
+    6. If $\overrightarrow F=200N;\ m=15.0kg;\ a=25.0m/s^2;\ \theta=20.0^\circ$, calculate the numerical values for the magnitudes of the normal and friction forces.
+        $$
+        \overrightarrow {F_{drag}}=\frac{\overrightarrow F}{\cos\theta}-mg\sin(\theta+90^\circ)-ma;\ \overrightarrow {F_{normal}}=mg\sin(\theta)\\
+        \overrightarrow {F_{drag}}=\frac{200N}{\cos(20^\circ)}-(15kg)(-9.81m/s^2)\sin(20^\circ+90^\circ)-(15kg)(25m/s^2)\\
+        \overrightarrow {F_{drag}}=23.88868N\\
+        \overrightarrow {F_{normal}}=(15.0kg*9.81m/s^2)\sin(20.0^\circ)\\
+        \overrightarrow {F_{normal}}=50.328N\\
+        \overline{\underline{|\overrightarrow {F_{drag}}=23.9N;\ \overrightarrow {F_{normal}}=50.3N|}}
+        $$
+
+    7. What is the coefficient of friction between the block and the incline surface?
+        $$
+        \mu_k=\frac{\overrightarrow {F_k}}{\overrightarrow {F_N}};\ \overrightarrow{F_k}=\overrightarrow{F_{drag}}\\
+        \mu_k=\frac{23.9N}{50.3N}\\
+        \mu_k=0.474657\\
+        \overline{\underline{|\mu_k=0.475|}}
+        $$
+
+9. A box of box is initially at rest. At $t=0$, you start pulling with an applied horizontal force $\overrightarrow {F_a}$. The mass of the box is $80.0kg$. The coefficient of static friction between the bottom of the box and the wooden floor is $0.751$, and the coefficient of kinetic friction is $0.691$. 
+    For each of the following applied forces, determine the magnitude of the friction force on the box and the corresponding horizontal acceleration of the box.
+
+    ![image-20200917124452286](macdougall_skyler_week4.assets/image-20200917124452286.png)
+    
+    Static and kinetic friction values are based off of the normal force, which is based off of the weight force, which does not change in this problem. Therefore, the friction forces can be calculated without the given problems.
+    $$
+    \overrightarrow{F_{static}}=\mu_k\overrightarrow{F_N}\\
+    \overrightarrow{F_{static}}=\mu_k(m)(-g)\\
+    \overrightarrow{F_{static}}=0.751(80.0kg)(9.81m/s^2)\\
+    \overrightarrow{F_{static}}=589.3848N\\\rule{100pt}{0.6pt}\\
+    \overrightarrow{F_{kinetic}}=\mu_k\overrightarrow{F_N}\\
+    \overrightarrow{F_{kinetic}}=\mu_k(m)(-g)\\
+    \overrightarrow{F_{kinetic}}=0.751(80.0kg)(9.81m/s^2)\\
+    \overrightarrow{F_{kinetic}}=542.2968N\\\rule{100pt}{0.6pt}\\
+    \overline{\underline{|\overrightarrow{F_{static}}=589.3848N;\ \overrightarrow{F_{kinetic}}=542.2968N|}}
+    $$
+    
+    1. Applied force is $242N$
+        $$
+        \overrightarrow{F_{static_{max}}}>\overrightarrow{F_{applied}}\\\therefore\\
+        \overrightarrow{F_{static}}=\overrightarrow{F_{applied}}=242N\\
+        \sum\overrightarrow F=0\\\therefore\\
+        a=0\\
+        \overline{\underline{|\overrightarrow{F_{static}}=242N;\ a=0|}}
+        $$
+    
+    2. $407N$
+        $$
+        \overrightarrow{F_{static_{max}}}>\overrightarrow{F_{applied}}\\\therefore\\
+        \overrightarrow{F_{static}}=\overrightarrow{F_{applied}}=407N\\
+        \sum\overrightarrow F=0\\\therefore\\
+        a=0\\
+        \overline{\underline{|\overrightarrow{F_{static}}=407N;\ a=0|}}
+        $$
+    
+    3. $545N$
+        $$
+        \overrightarrow{F_{static_{max}}}>\overrightarrow{F_{applied}}\\\therefore\\
+        \overrightarrow{F_{static}}=\overrightarrow{F_{applied}}=545N\\
+        \sum\overrightarrow F=0\\\therefore\\
+        a=0\\
+        \overline{\underline{|\overrightarrow{F_{static}}=545N;\ a=0|}}
+        $$
+    
+    4. $559N$
+        $$
+        \overrightarrow{F_{static_{max}}}>\overrightarrow{F_{applied}}\\\therefore\\
+        \overrightarrow{F_{static}}=\overrightarrow{F_{applied}}=559N\\
+        \sum\overrightarrow F=0\\\therefore\\
+        a=0\\
+        \overline{\underline{|\overrightarrow{F_{static}}=559N;\ a=0|}}
+        $$
+    
+    5. $619N$
+        $$
+        \overrightarrow {F_{static_{max}}}<\overrightarrow{F_{applied}}\\\therefore\\
+        \overrightarrow{F_{kinetic}}=542.2968N\\
+        \overrightarrow {F_{net}}=\overrightarrow{F_{applied}}-\overrightarrow{F_{kinetic}}\\
+        a=\frac{\overrightarrow{F_{net}}}{m}\\
+        a=\frac{\overrightarrow{F_{applied}}-\overrightarrow{F_{kinetic}}}{m}\\
+        a=\frac{619N-542.2968N}{80kg}\\
+        a=0.65879m/s^2\\
+        \overline{\underline{|\overrightarrow{F_{kinetic}}=542.N;\ a=0.659m/s^2|}}
+        $$
+    
+    6. $719N$
+        $$
+        \overrightarrow {F_{static_{max}}}<\overrightarrow{F_{applied}}\\\therefore\\
+        \overrightarrow{F_{kinetic}}=542.2968N\\
+        \overrightarrow {F_{net}}=\overrightarrow{F_{applied}}-\overrightarrow{F_{kinetic}}\\
+        a=\frac{\overrightarrow{F_{net}}}{m}\\
+        a=\frac{\overrightarrow{F_{applied}}-\overrightarrow{F_{kinetic}}}{m}\\
+        a=\frac{719N-542.2968N}{80kg}\\
+        a=2.20879m/s^2\\
+        \overline{\underline{|\overrightarrow{F_{kinetic}}=542.N;\ a=2.21m/s^2|}}
+        $$
+        
+    7. For $719N$, how far does the box go in $2.32s$ if the box was initially at rest? 
+        $$
+        x_f=x_i+v_it+\frac12at^2\\
+        x_f=\cancel{x_i+v_it+}^0+\frac12(2.21m/s^2)(2.32s)^2\\
+        \overline{\underline{|x_f=5.94m}}|
+        $$
+        
+
+[^1]: Write the result in terms of given quantities and constants.
+
+[^2]:All angles are taken as absolute, with "$0^\circ$ standard" referring to directly right, and "$180^\circ$ standard" referring to directly left in all diagrams. Furthermore, "positive from" refers counterclockwise movement from the noted standard, while "negative from" refers to clockwise movement from the noted standard.

Aug.-Sept./macdougall_skyler_week5.md

@@ -0,0 +1,257 @@
+# Week 5 Activities Problems
+
+## Skyler MacDougall
+
+## PHYS-111 2pm Lab
+
+### Group Members: Will Caliguri, Nina Chambliss, Caroline Cody
+
+#### Due: 9/25/2020
+
+# Hooke's Law
+
+1. A spring with a force constant $k$ is used to push a block of wood of mass $m$ against a wall as shown in the diagram. The coefficient of static friction between the block and the wall is $\mu_s$
+
+    ![image-20200921145147462](macdougall_skyler_week5.assets/image-20200921145147462.png)
+
+    1. What is the minimum amount of compression of the spring needed to keep the block from falling? [^1]
+        $$
+        \overrightarrow {F_{static}}=\mu_s\overrightarrow {F_N}\\
+        \overrightarrow {F_N}=\overrightarrow {F}\\
+        \overrightarrow F=-kx\\
+        \overrightarrow {F_{static}}=\overrightarrow W\\
+        \overrightarrow {W}=mg\\
+        \therefore\\
+        mg=\mu_s(-kx)\\
+        -\frac{mg}{\mu_sk}=x
+        $$
+    
+2. Calculate the numerical value for the minimum amount of compression if the spring needed to keep the block from falling if:
+        $$
+        k=120N/m\\
+        m=0.270kg\\
+        \mu_s=0.460
+        $$
+    
+    $$
+        -\frac{mg}{\mu_sk}=x\\
+        -\frac{(0.270kg)(-9.81m/s^2)}{0.460(120N/m)}=x\\
+        -\frac{(0.270)(-9.81)N}{0.460(120)N/m}=x\\
+        -\frac{(0.270)(-9.81)}{0.460(120)}m=x\\
+        \frac{(0.270)(9.81)}{0.460(120)}m=x\\
+        x=0.047983696m\\
+        \overline{\underline{|x=48.0mm|}}
+    $$
+    
+    
+    ​    
+    3. Does your answer for part 1.1 change if the mass of the block of wood is doubled? Why?
+        The answer is dependent on the mass proportionally, so if the mass of the block of wood doubles, the spring compression distance must also double.
+    
+2. A block of mass $m$ rests on a rough plank that is inclined at an angle $\theta$ above the horizontal. The upper end of the box is attached to a spring with force constant k, as shown in the diagram. The coefficient of static friction between the box and the plank is $\mu_s$.
+
+    1. Draw the free body diagram for the block.
+
+        ![image-20200921152216504](macdougall_skyler_week5.assets/image-20200921152216504.png)
+
+    2. What is the maximum amount the spring can be stretch and the box remain at rest? [^1]
+        $$
+        \overrightarrow{F_{normal}}=\overrightarrow W\sin(270-\theta)\\
+        \overrightarrow {F_{spring}}=-kx\\
+        \overrightarrow {F_{friction}}=\mu_s\overrightarrow {F_{normal}}\\
+        \overrightarrow {W}=mg\\
+        \overrightarrow {F_{friction}}=\overrightarrow{F_{spring}}\\
+    \therefore\\
+        -kx=\mu_s((mg)\sin(270-\theta))\\
+    x=-\frac{\mu_s((mg)\sin(270-\theta))}{k}
+        $$
+        
+    3. Calculate the maximum amount of spring spring displacement given:
+        $$
+        m=2.00kg\\
+        \theta=65.0^\circ\\
+        k=360N/m\\
+        \mu_s=0.220
+        $$
+    
+        $$
+        x=-\frac{\mu_s((mg)\sin(270-\theta))}{k}\\
+        x=-\frac{0.220(((2.00kg)(-9.81m/s^2))|\sin(270-65.0^\circ)|)}{360N/m}\\
+        x=0.0050671930m\\
+        \overline{\underline{|x=5.07mm|}}
+        $$
+        
+    4. Does your answer to part 2.3 change if the mass of the box increases? Why?
+        The answer is proportional to the mass, so if the mass increases, the answer will also increase. 
+    
+3. A spring is attached to a rotating mass $m$ that is moving counterclockwise in a circle of radius $R$ at a constant speed $v$ on a smooth surface. During this motion, the spring is stretched a distance $x$ from its equilibrium length. ($x<R$). 
+
+    1. Draw a free body diagram.![image-20200922151737381](macdougall_skyler_week5.assets/image-20200922151737381.png)
+
+    2. Find an expression for the force constant of the spring. [^1]
+        $$
+        \overrightarrow{F_{net}}=ma_c=\frac{mv^2}{r}\\
+        \overrightarrow{F_{net}}=\overrightarrow{F_{spring}}=-kx\\
+        -kx=\frac{mv^2}{r}\\
+        k=-\frac{mv^2}{Rx}
+        $$
+        
+    3. Determine the value of the spring constant given:
+        $$
+        m=(200\pm1)g\\
+        v=(1.40\pm0.03)m/s\\
+        R=(13.0\pm0.4)\times10^{-3}m\\
+        x=(4.0\pm0.1)\times10^{-3}m
+        $$
+        
+   
+    $$
+        k=-\frac{mv^2}{Rx}\\
+        k=-\frac{((0.200\pm0.001)kg)*((1.40\pm0.03)m/s)^2}{((13.0\pm0.4)\times10^{-3}m)*((4.0\pm0.1)\times10^{-3}m)}\\
+        k=-\frac{((0.200\pm0.001))*((1.40\pm0.03))^2}{((13.0\pm0.4)\times10^{-3})*((4.0\pm0.1)\times10^{-3})}\frac{kgm^2/s^2}{m^2}\\
+        k=-\frac{(0.200\pm0.001)*(1.40\pm0.03)^2}{(13.0\pm0.4)*(4.0\pm0.1)*10^{-6}}{kg/s^2}\\
+        k=-\frac{(0.200\pm0.001)*(1.40\pm0.03)^2}{(13.0\pm0.4)*(4.0\pm0.1)*10^{-6}}N\\
+        k=-\frac{(0.200)*(1.40)^2}{(13.0)*(4.0)*10^{-6}}N=-7.5384615kN\\
+        k=-\frac{(0.201)*(1.43)^2}{(13.4)*(4.1)*10^{-6}}N=-7.4813415kN\\
+        \overline{\underline{|k=-(7.54\pm0.06)kN=-(7.54\pm0.06)\times10^3N|}}
+    $$
+
+
+# Circular Motion
+
+1. A car of mass $m$ drives through a curve on a horizontally flat road at a constant speed $v$. The curve has a radius of curvature $R$.
+
+    1. Draw a free body diagram for the car. (Side View)
+
+        ![freeBodyDiagram](macdougall_skyler_week5.assets/freeBodyQuestion6.png)
+
+    2. What is the magnitude of the frictional force that keeps the car moving in a circle of radius $R$ with a uniform speed $v$? [^1]
+        $$
+        \overrightarrow {F_{net}}=ma=\frac{mv^2}{r}\\
+        \overrightarrow {F_{net}}=\overrightarrow {F_{friction}}\\
+        \therefore\\
+        \overrightarrow {F_{friction}}=\frac{mv^2}{r}
+        $$
+        
+    3. If the car has a mass of $1500kg$, then what is the magnitude of the frictional force acting on the car that keeps it moving in a circle with a radius of $50.0m$ with a uniform speed of $15.0m/s$.
+        $$
+        \overrightarrow {F_{friction}}=\frac{mv^2}{r}\\
+            \overrightarrow {F_{friction}}=\frac{(1500kg)(15.0m/s)^2}{50.0m}\\
+            \overrightarrow {F_{friction}}=\frac{(1500)(15.0)^2}{50.0}\frac{kgm^2/s^2}{m}\\
+            \overrightarrow {F_{friction}}=\frac{(30\cancel{1500})(15.0)^2}{\cancel{50.0}}{kgm/s^2}\\
+            \overrightarrow {F_{friction}}={(30)(15.0)^2}N\\
+            \overrightarrow {F_{friction}}={(30)(15.0)^2}N\\
+            \overrightarrow {F_{friction}}=6750N\\
+            \overline{\underline{|\overrightarrow {F_{friction}}=6.75kN|}}
+        $$
+        ​    
+    
+    4. If the coefficient of static friction between the tires and the road is $\mu_s$, find the safe speed a car can pass through the curve without sliding. [^1]
+    
+    $$
+        \overrightarrow {F_{friction}}=\frac{mv^2}{r}\\
+        \overrightarrow {F_{friction}}=\mu_s\overrightarrow{F_{normal}}\\
+        \overrightarrow {F_{normal}}=mg\\
+        \mu_s(mg)=\frac{mv^2}{r}\\
+        \sqrt{\frac{\mu_s(mg)r}{m}}=v\\
+        \sqrt{{\mu_sgr}}=v\\
+    $$
+    
+    
+    5. Does the maximum safe speed increase or decrease on:
+            1. An icy road?
+                Decrease.
+                2. A tighter curve?
+                Decrease.
+                    3. A small car?
+                No change.
+    
+2. A conical pendulum is formed by attaching a mass $m$ to a string of length $l$, then allowing the mass to swing in a horizontal circle. The string makes a constant angle $\theta$ with respect to the vertical. As a result, the mass moves with constant speed in a horizontal circle.
+
+    1. Draw a free body diagram. 
+
+        ![freeBodyDiagram](macdougall_skyler_week5.assets/freeBodyQuestion7.png)
+
+    2. What forces in your free body diagram keep the mass moving in a circle at a constant speed?
+        The x component of tension.
+
+    3. Apply Newton's Second Law, and determine an expression for the magnitude of the tension in the string. [^1]
+        $$
+        \overrightarrow{F_{net}}=\overrightarrow{F_{tension}}\cos(90^\circ-\theta)\\
+        \overrightarrow{F_{tension}}=\frac{\overrightarrow{W}}{\sin(90^\circ-\theta)}\\
+        \overrightarrow{W}=mg\\
+        \overrightarrow{F_{net}}=mg\cot(90^\circ-\theta)
+        $$
+        
+    4. Determine an expression for the speed of the mass. [^1]
+        $$
+        \overrightarrow{F_{net}}=\frac{mv^2}{r}\\
+        \sqrt{\frac{\overrightarrow{F_{net}}r}{m}}=v\\
+        \sqrt{\frac{mg\cot(90^\circ-\theta)r}{m}}=v\\
+        \sqrt{g\cot(90^\circ-\theta)r}=v\\
+        r=l\cos(90^\circ-\theta)\\
+        \sqrt{g\cot(90^\circ-\theta)(l\cos(90^\circ-\theta))}=v
+        $$
+    
+    5. Find the tension and the speed, given the following:
+    
+    $$
+        m=500g\\
+        l=1.00m\\
+        \theta=11.5^\circ
+    $$
+    
+    $$
+    \sqrt{g\cot(90^\circ-\theta)(l\cos(90^\circ-\theta))}=v;\ \overrightarrow{F_{net}}=mg\cot(90^\circ-\theta)\\
+    \sqrt{(9.81m/s^2)\cot(90^\circ-11.5^\circ)((1.00m)\cos(90^\circ-11.5^\circ))}=v;\ \overrightarrow{F_{net}}=(0.500kg)(9.81m/s^2)\cot(90^\circ-11.5^\circ)\\
+    \sqrt{(9.81m/s^2)\cot(78.5^\circ)((1.00m)\cos(78.5^\circ))}=v;\ \overrightarrow{F_{net}}=(0.500kg)(9.81m/s^2)\cot(78.5^\circ)\\
+    0.63080258m/s=v;\ \overrightarrow{F_{net}}=0.99793353N\\
+    \overline{\underline{|v=63.1cm/s;\ \overrightarrow{F_{net}}=998.mN|}}
+    $$
+    
+    
+    
+3. In another type of amusement ride, passengers stand inside a cylinder with their backs against the wall. The cylinder has a diameter of $D$. The cylinder begins to rotate around the vertical axis. Once the cylinder reaches its operational rotational speed, the linear speed of a passenger is $v$ and the floor on which the passengers are standing drops away. If all goes well, the passengers will "stick" to the wall and not slide down. Let $m$ be the mass of the passenger. 
+
+    1. Draw a free body diagram.
+
+        ![freeBodyDiagram](macdougall_skyler_week5.assets/freeBodyLastQuestion.png)
+
+    2. Determine an expression for the minimum coefficient of static friction required to keep a passenger from sliding down the cylinder's wall when the floor drops away. [^1]
+        $$
+        \overrightarrow{F_{friction}}=\mu_s\overrightarrow{F_{normal}}\\
+        \overrightarrow{F_{friction}}=\overrightarrow{W}\\
+        \overrightarrow{W}=mg\\
+        \overrightarrow{F_{net}}=\frac{mv^2}r=\overrightarrow{F_{normal}}\\
+        \therefore\\
+        mg=\mu_s(\frac{mv^2}r)\\
+        \frac{\cancel{m}gr}{\cancel{m}v^2}=\mu_s\\
+        \frac{gr}{v^2}=\mu_s\\
+        r=\frac D2\\
+        \frac{gD}{2v^2}=\mu_s
+        $$
+        
+    3. Calculate a numerical value for the coefficient of static friction, given:
+        $$
+        D=15.0m\\
+        v=18.8m/s\\
+        m=70.0kg
+        $$
+        
+    
+    $$
+    \frac{gD}{2v^2}=\mu_s\\
+    \frac{(9.81m/s^2)(15.0m)}{2(18.8m/s)^2}=\mu_s\\
+    \frac{(9.81)(15.0)}{2(18.8)^2}=\mu_s\\
+    \mu_s=0.20816829\\
+    \overline{\underline{|\mu_s=0.208|}}
+    $$
+    
+    
+    
+    
+    
+
+
+
+[^1]: Write the result in terms of given quantities and constants.

Aug.-Sept./macdougall_skyler_week6.md

@@ -0,0 +1,221 @@
+# Week 6 Activities Problems
+
+## Skyler MacDougall
+
+## PHYS-111 2pm Lab
+
+### Group Members: Will Caliguri, Nina Chambliss
+
+#### Due: 9/25/2020
+
+# Forces And Newton's Second Law
+
+1. A helicopter is lifting two crates simultaneously. Crate A with mass $m_A$, is attached to the helicopter by a light cable. Crate B, with a mass of $m_B$, hangs below crate A and is attached to crate A by a second light cable. The cables are not connected to each other. The helicopter accelerates upward with an acceleration of magnitude $a$. Ignosre all frictional effects. Let $\overrightarrow {T_1}$ be the tension in cable 1, and $\overrightarrow {T_2}$ the tension in cable 2.
+
+    ![image-20201003165056892](macdougall_skyler_week6.assets/image-20201003165056892.png)
+
+    1. Which cable has greater tension? Don't do any math. Explain why.
+        Cable 1 will have more tension than cable 2. Cable 1 has to hold up both crates, while cable 2 only holds one crate.
+
+    2. Find $T_2$. [^1]
+        $$
+        T_2=m_B(a+g)
+        $$
+        
+3. Find $T_1$. [^1]
+        $$
+        T_1=(m_A+m_B)(a+g)
+        $$
+        
+    4. Find $T_1$ and $T_2$ given:
+    $$
+        m_A=200kg\\
+        m_B=250kg\\
+        a=1.00m/s^2
+        $$
+    
+        $$
+    T_2=m_B(a+g);\ T_1=(m_A+m_B)(a+g)\\
+        T_2=250kg(1.00m/s^2+(-9.81m/s^2))\\
+        T_2=-2.2025kN\\\rule{75pt}{0.4pt}\\
+        T_1=(m_A+m_B)(a+g)\\
+        T_1=(200kg+250kg)(1.00m/s^2+(-9.81m/s^2))\\
+        T_1=-3.9645kN\\
+        \overline{\underline{|T_1=-4kN;\ T_2=-2kN|}}
+        $$
+        
+        
+        
+    5. If the helicopter moves vertically upward with a constant velocity, which cable has greater tension? Why? Calculate the values.
+        The tension changes, but equally on each cable, so cable 1 will still have more tension.
+    
+2. Two blocks are connected by a light cord. The cord goes across a pulley, suspending Block 2. Block 1 is on the horizontal surface above Block 2. Block 1 has a weight $W_1=25.0N$, and block 2 has a weight $W_2=18.0N$. The pulley and cords are ideal. Block 1 is at rest.
+
+    ![image-20201003165108485](macdougall_skyler_week6.assets/image-20201003165108485.png)
+
+    1. What does rough surface mean?
+        There is friction.
+
+    2. Draw a free body diagram.
+
+    3. What is the magnitude of the tension in the cord?
+        $$
+        T=W_2\\
+        T=18.0N
+        $$
+        
+4. What is the magnitude and direction of the friction force on $W_1$?
+        $$
+        F_{friction}=T\\
+        F_{friction}=18.0N
+        $$
+        The friction force is pointed in the opposite direction of the tension force.
+    
+3. A Mass $m_1$ is connected by a light rope passing over a pulley to a second mass $m_2$, like before. Assume no friction and ideal components. 
+
+    ![image-20201003165120625](macdougall_skyler_week6.assets/image-20201003165120625.png)
+
+    1. What does the term "light rope" mean?
+        The rope does not have to be accounted for.
+
+    2. Find the magnitude of the acceleration of each mass. [^1]
+        $$
+        a_2=g\\
+        F=ma;\ a_1m_1=m_2a_2\\
+        \overline{\underline{|a_1=\frac{m_2g}{m_1};\ a_2=g|}}
+        $$
+        
+3. Find the tension on the rope.[^1]
+        $$
+        T=-m_2g
+        $$
+        
+    4. Check if your results are reasonable by evaluating numerical values for acceleration and tension at these limits:
+    1. What are the values for a and T as $m_1$ approaches 0? Does this make sense?
+            Tension and $a_2$ do not change as $m_1$ changes. $a_1$ increases as $m_1$ decreases. All of these make logical sense because it takes less effort to move a lighter object.
+        2. What are the values for a and T as $m_2$ approaches 0? Does this make sense?
+            $a_2$ does not change. This makes sense, because acceleration due to gravity is mass-independent. Tension and $a_1$ do change, both decreasing as $m_2$ decreases. This makes sense, because the lighter object can't output as much force as a heavier object.
+    
+4. Two objects are connected by a light string that passes over an ideal pulley. The $200g$ bunch of bananas is initially $140cm$ above the floor. The bananas are released from rest. How long does it take for the $200g$ banana bunch to hit the floor?
+
+    ![image-20201003172346963](macdougall_skyler_week6.assets/image-20201003172346963.png)
+    $$
+    F_{net_{L}}=m_Lg-m_sg\\
+    F_{net_L}=m_La\\
+    a_L=\frac{m_Lg-m_sg}{m_L}\\
+    x_f=x_i+v_it+\frac12at^2\\
+    x_f=x_i+v_it+\frac12\frac{m_Lg-m_sg}{m_L}t^2\\
+    x_f=x_i\cancel{+v_it}+\frac12\frac{m_Lg-m_sg}{m_L}t^2\\
+    x_f=x_i+\frac12\frac{m_Lg-m_sg}{m_L}t^2\\
+    x_f-x_i=\frac12\frac{m_Lg-m_sg}{m_L}t^2\\
+    2(x_f-x_i)=\frac{m_Lg-m_sg}{m_L}t^2\\
+    \frac{2(x_f-x_i)m_L}{m_Lg-m_sg}=t^2\\
+    \sqrt{\frac{2(x_f-x_i)m_L}{m_Lg-m_sg}}=t\\
+    \sqrt{\frac{2(0-140cm)(200g)}{(200g)(-9.81m/s^2)-(50g)(-9.81m/s^2)}}=t\\
+    t=35.2ms
+    $$
+
+# Work and Kinetic Energy
+
+1. The two ropes show n in a birds-eye view of the diagram are used to drag a crate $3.00m$ horizontally across a rough floor. The tensions in the two ropes are $T_1=402N;\ T_2=275N$. The magnitude of the kinzetic friction force is $f_k=500N$. 
+    1. What is the direction of the displacement vector $\overrightarrow d$ whose magnitude is $3.00m$? 
+        $$
+        F_{net}=f_k+\overrightarrow {T_1}+\overrightarrow{T_2}\\
+        F_{net}=500N\ang180^\circ+402N\ang20^\circ+275N\ang-30^\circ\\
+        F_{net}=115.92342\ang-0.0039061327^\circ\\
+    \overline{\underline{|F_{net}\approx0.1kN\ang0^\circ|}}\\
+        \overline{\underline{|F_{net}\approx0.1kN\ in\ the\ +x\ direction|}}
+        $$
+        
+    2. How much work is done by the rope $T_1$?
+        $$
+        W=Fd\cos(\theta)\\
+        W=(402N)(3.00m)\cos(20^\circ)\\
+        W=1.1332693kJ\\
+        \overline{\underline{|W_{T_1}=1.13kJ|}}
+        $$
+        
+    3. How much work is done by the rope $T_2$?
+        $$
+        W=Fd\cos(\theta)\\
+        W=(275N)(3.00m)\cos(-30^\circ)\\
+        W=kJ\\
+        \overline{\underline{|W_{T_2}=714.J|}}
+        $$
+        
+    4. How much work is done by kinetic friction?
+        $$
+        W=Fd\cos(\theta)\\
+        W=(500N)(3.00m)\cos(180^\circ)\\
+        W=-1.5kJ\\
+        \overline{\underline{|W_{f_k}=-2kJ|}}
+        $$
+        
+    5. What is the net work done on the crate?
+        $$
+        \sum W=W_{T_1}+W_{T_2}+W_{f_k}\\
+        \sum W=1.13kJ+714J-1.5kJ\\
+        \sum W=347.74026J\\
+        \overline{\underline{|\sum W=0.3kJ|}}
+        $$
+        
+    
+2. Sam's job at the amusement park is to slow down and bring to a stop the boats in a log ride. A boat and its riders have a total mass of $1.2Mg$ and drift with a speed of $1.20m/s$. 
+    1. How much total work is done on the boat to slow it from $1.20m/s$ to $0.6m/s$? 
+        $$
+        W_{total}=\Delta K;\ K=\frac12mv^2\\
+        W_{total}=\frac12m(v_f^2-v_i^2)\\
+        W_{total}=\frac12(1.2Mg){((0.6m/s)^2-(1.20m/s)^2)}\\
+        W_{total}=(0.6Mg){((0.6m/s)^2-(1.20m/s)^2)}\\
+        W_{total}=-648J\\
+        \overline{\underline{|W_{total}=-0.65kJ|}}
+        $$
+        
+    2. How much total work is done to bring the boat to rest?
+        $$
+        W_{total}=\Delta K;\ K=\frac12mv^2\\
+        W_{total}=\frac12m(v_f^2-v_i^2)\\
+        W_{total}=\frac12(1.2Mg){((0m/s)^2-(1.20m/s)^2)}\\
+        W_{total}=864\\
+        \overline{\underline{|W_{total}=-0.86kJ|}}
+        $$
+        
+    3. What average power is expended in bringing one boat to rest in $8.00s$?
+        $$
+        P=Fv;\ F=ma;\ a=\frac {\Delta v}t\\
+        P=\frac{m(\Delta v)^2}{t}\\
+        P=\frac{(1.2Mg){(0-1.20m/s)}^2}{8.00s}\\
+        P=216W\\
+        \overline{\underline{|P=0.22kW|}}
+        $$
+        
+    4. Instead of having Sam stop each boat by hand, a large spring is placed at the end of the ride such that each boat compresses the spring a distance $x$ from its equilibrium as it comes to a halt. The work $W$ required to stretch a spring from an initial position $x_i$ to a final position $x_f$ is $W=\frac12k(x_f^2-x_i^2)$. Assuming a boat is traveling as in part 2.2, find the spring constant required to stop the boat in no more than $2.00m$. 
+        $$
+        W=\frac12k(x_f^2-x_i^2)\\
+        k=\frac{2W}{x_f^2-x_i^2}\\
+        k=\frac{2(-0.86kJ)}{2.00m^2\cancel{-0^2}}\\
+        k=432N/m\\
+        \overline{\underline{|k=0.43kN/m|}}
+        $$
+    
+3. A catapult launcher on an aircraft carrier accelerates a jet from rest to $72.0m/s$. The work done by the catapult during the launch is $76.0MJ$. During the launch, the system is considered ideal, and all work done on the jet is done by the catapult.
+    1. What is the mass of the jet?
+        $$
+        W=\frac12mv^2\\
+        m=\frac{2W}{v^2}\\
+        m=\frac{2(76.0MJ)}{(72.0m/s)^2}\\
+        m=29.320988Mg\\
+        \overline{\underline{|m=29.3Mg|}}
+        $$
+        
+    2. If the jet is in contact with the catapult for $2.00s$, what is the power output of the catapult?
+        $$
+        P=\frac{m(\Delta v)^2}{t};\ m=\frac{2W}{\Delta v^2}\\
+        P=\frac {W}{t}\\
+        P=\frac{76.0MJ}{2.00s}\\
+        P=38.0MW\\
+        \overline{\underline{|P=38.0MW|}}
+        $$
+        
+    
+    

Aug.-Sept./week1ActivitiesProblems.md

@@ -0,0 +1,266 @@
+# Week 1 Activities Problems
+
+## Skyler MacDougall
+
+## PHYS-111 2pm Lab
+
+### Group: Group 2
+
+#### Due: 8/24/2020
+
+# Unit Conversion
+
+1. NASA's unpiloted, experimental X-43A "air breathing" test plane broke its own world aircraft speed record on its third and final flight in Nov. 16, 2004. The test plane, known as a "SCRAM jet", burns its own fuel - in this case, hydrogen - without the need to carry heavy tanks of oxidizer, as rockets must. Instead, it draws oxygen from the air, which is naturally compressed by the forward speed of the vehicle and the shape of the air inlet. Conventional jet engines have rotating blades to compress the air, SCRAM jets have no such moving parts. The test plane achieved a record speed of $11\times10^3km/hr$.
+    1. Convert the SCRAM jet's record breaking speed to miles per hour. **SHOW ALL WORK.**
+        $$
+        \frac{11\times10^3\cancel{km}}{1\ hour}*\frac{1 mile}{1.609344\cancel{km}}=6835.0831miles/hour\approx\overline{\underline{|6.8\times10^3miles/hour|}}
+        $$
+    
+    2. Convert this speed to nanometers per microsecond. **SHOW ALL WORK**
+        $$
+        \frac{11\times10^3\cancel{km}}{1\cancel{hour}}*\frac{1\times10^{12}nm}{1\cancel{km}}*\frac{1\ \cancel{hour}}{3600\ \cancel{second}}*\frac{1\ \cancel{second}}{1\times10^6\mu s}\\
+        \frac{(11\times10^3)*(1\times10^{\cancel{12}6})}{1*(3600)*\cancel{(1\times10^6)}}nm/sec\\
+        \frac{(11\cancel{\times10^3})*(1\times10^{6})}{\cancel{3600}3.6}nm/sec\\
+        \frac{11\times10^6}{3.6}nm/sec=3.0\overline5\times10^6nm/sec\\
+        \overline{\underline{|3.1\times10^6nm/sec|}}
+        $$
+    
+2. Determine the area of a CD in $cm^2$. Convert the area to $km^2$. Assume the CD is a uniform disk of diameter 12.0cm with a central hole diameter of 1.5cm. **SHOW ALL WORK.**
+    $$
+    A_{circle}=\pi (\frac {D_{outer}-D_{inner}} 2)^2\\
+    A_{circle}=\pi (\frac {12.0cm-1.5cm} 2)^2\\
+A_{circle}=\pi (\frac {10.5cm} 2)^2\\
+    \underline{\overline{|A_{circle}=86.59cm^2|}}\\
+    \rule{125pt}{0.4pt}\\
+    \frac{86.59cm^2}1*[\frac{1m}{100cm}]^2*[\frac{1km}{10^3m}]^2\\
+    \frac{86.59\cancel{cm^2}}1*\frac{1\cancel{m^2}}{10^4\cancel{cm^2}}*\frac{1km^2}{10^6\cancel{m^2}}\\
+    \frac{86.59}{10^{10}}km^2\\
+    \underline{\overline{|A_{circle}=8.6\times10^{-9}km^2|}}
+    $$
+    
+3. Density $\rho$ is defined as the ratio of mass to volume. The density of aluminum is $2702kg/m^3$. What is the density in $g/cm^3$? **SHOW ALL WORK.**
+    $$
+    \frac{2702\cancel{kg}}{1m^3}*\frac{10^3g}{1\cancel{kg}}*[\frac{1m}{100cm}]^3\\
+    \frac{2702\cancel{kg}}{1\cancel{m^3}}*\frac{10^3g}{1\cancel{kg}}*\frac{1\cancel{m^3}}{10^6cm^3}\\
+\frac{2072\cancel{*10^3}}{10^{\cancel{6}3}}g/cm^3\\
+    \underline{\overline{|2.072g/cm^3|}}
+    $$
+    
+4. The fuel efficiency of a car is described in the US by citing the number of ***miles per gallon*** (MPG) that it gets. In Europe they cite ***liters per 100 kilometers*** (LPK).
+    $$
+    Useful\ Unit\ Conversions\\
+    1mi=1609m=1.609km\\
+    1gal=3.785L
+    $$
+    
+    1. Does a more fuel efficient car have a larger or smaller MPG number? Why?
+        A more fuel efficient car has a larger MPG rating, because it can go a longer distance in the same amount of fuel.
+        
+    2. Does a more fuel efficient car have a larger or smaller LPK number? Why?
+        A fuel efficient car has a smaller LPK rating, because it can use less gas in the same amount of distance.
+        
+    3. You are considering buying a gas-electric hybrid from an upstart European manufacturer that claims its hybrid gets 4.6LPK on the highway. Is this reasonable, or are you being "taken for a ride"? **Support your conclusion with calculations**.
+       This is very fuel efficient.
+       $$
+       \frac{100km}{4.6L}*\frac{3.785L}{1gal}*\frac{1mi}{1.609km}\\
+       \frac{100\cancel{km}}{4.6\cancel{L}}*\frac{3.785\cancel{L}}{1gal}*\frac{1mi}{1.609\cancel{km}}\\
+       \frac{100*3.785}{4.6*1.609}mi/gal\\
+       51.139mi/gal\approx\underline{\overline{|51mi/gal|}}
+       $$
+       
+
+# Position-Time and Velocity-Time Graphs
+
+1. A dog walks along a straight line, first in one direction, then in the opposite direction. Shown below are two possible *Position-Time* graphs for a dog's motion.
+
+    ![image-20200824163606337](week1ActivitiesProblems.assets/image-20200824163606337.png)
+
+    2. Which graph (Graph A or Graph B) more closely reflects the dog's movement? **Why**?
+        Graph A represents the dogs positional movement, because it goes in one direction, then turns around.
+    2. Explain why someone might choose the other graph. Also, describe the dog's motion if it were following this "other" graph.
+        Someone would choose the other graph (Graph B) for the given explanation if they believed that it was a speed graph instead of a positional graph. Graph B actually represents a constant motion in one direction. 
+
+2. A turtle moves along a line; first in one direction, then shown in the opposite direction. Shown below are two possible *Velocity-Time* graphs for the turtle's motion.
+
+    ![image-20200824163638544](week1ActivitiesProblems.assets/image-20200824163638544.png)
+
+    1. Which graph (Graph A or Graph B) more closely reflects the turtle's motion? **Why**?
+        Graph B more closely reflects the turtle's motion, because the turtle has a negative speed first, then has a positive speed, so it changed directions.
+    2. Explain why someone might choose the other graph. Also, describe the turtle's motion if it were following this "other" graph. 
+        Someone would choose the other graph for the given explanation if they believed that it was a positional graph instead of a speed graph. Graph A represents a changing speed, but always positive. So, the turtle would start out slow, then speed up, then slow back down, but always travel in the same direction.
+
+# Basic Measurements and Uncertainties
+
+1. Suppose you measure the diameter of a sphere five times. Each of the values obtained is represented by the symbol $D_i$, where $D_1$ is the first measurement, $D_2$ is the second, etc. The average (mean) diameter is defined as:
+    $$
+    D_{AV}=\frac{\displaystyle\sum^{N}_{i=1}=D_i}{N}=\frac{D_1+D_2+D_3+D_4+D_5}{5}\\
+    \delta D_{AV}=\frac{\displaystyle\sum^{N}_{i=1}=|D_i-D_{AV}|}{N}=\frac{|D_1-D_{AV}|+|D_2-D_{AV}|+|D_3-D_{AV}|+|D_4-D_{AV}|+|D_5-D_{AV}|}{5}
+    $$
+    The diameter of a sphere is measured five times using a digital caliper with a least count of $0.1mm$. The results are: $25.8mm$, $26.2mm$, $24.8mm$, $24.9mm$, $25.8mm$.
+
+    1. Calculate the average and average deviation in the set of measurements.
+        $$
+        D_{AV}=\frac{\displaystyle\sum^{N}_{i=1}=D_i}{N}=\frac{D_1+D_2+D_3+D_4+D_5}{5}\\
+        D_{AV}=\frac{25.8mm+26.2mm+24.8mm+24.9mm+25.8mm}{5}\\
+        D_{AV}=\frac{127.5mm}5\\
+        \overline{\underline{|D_{AV}=25.5mm|}}
+        $$
+    
+        $$
+        \delta D_{AV}=\frac{|D_1-D_{AV}|+|D_2-D_{AV}|+|D_3-D_{AV}|+|D_4-D_{AV}|+|D_5-D_{AV}|}{5}\\
+        \delta D_{AV}=\frac{|25.8mm-25.5mm|+|26.2mm-25.5mm|+|24.8mm-25.5mm|+|24.9mm-25.5mm|+|25.8mm-25.5mm|}{5}\\
+        \delta D_{AV}=\frac{2.6mm}{5}\\
+        \overline{\underline{|\delta D_{AV}=0.5mm|}}
+        $$
+    
+        
+    
+    2. What is the correct uncertainty to report for the diameter of the sphere?
+        The correct uncertainty should be $\pm0.5mm$
+    
+    3. Write the diameter of the sphere in "proper form". 
+        $(25.5\pm0.5)mm$
+    
+    4. The formula for the volume of a sphere is $V=\frac 1 6\pi D^3$, where D is the average (best) diameter of the sphere. $\delta D$ is the uncertainty in D, in this case, the average deviation in D. Using the results obtained by $D_{best}$ and $\delta D$, calculate the best estimate for the volume of the sphere and the uncertainty in the volume.
+        $$
+        V=\frac16\pi (D)^3;\ D_{best}=25.5mm;\ D_{max}=26.0mm\\
+        V_{best}=\frac16\pi (25.5mm)^3;\ V_{max}=\frac16\pi (26.0mm)^3\\
+        V_{best}=8.681\mu m^3;V_{max}=9.202\mu m^3\\
+        \delta V=V_{best}-V_{max}=0.5207\mu m^3\\
+        \overline{\underline{|V=(9.20\pm0.52)\mu m^3|}}
+        $$
+        Recall that the uncertainty is the calculated quantity is $\delta V=V_{max}-V_{best}$.
+        Write the final result in "proper form".
+
+# One Dimensional Kinematics Problems
+
+1. The diagram below shows the position-time graph of a particle.
+
+    ![image-20200824163512980](week1ActivitiesProblems.assets/image-20200824163512980.png)
+
+    1. Draw the particle's velocity-time graph for the interval $0\le t\le4s$
+       
+        ![image-20200827150003322](week1ActivitiesProblems.assets/image-20200827150003322.png)
+    2. Does the particle have a turning point or turning points? If so, at what time or times? (A turning point is a point at which the particle reverses its direction of motion)
+       
+        It does have a singular turning point, at 2 seconds.
+
+2. A car starts from $x_i=10m,\ t_i=0$, and moves with the velocity time graph shown.
+
+    ![image-20200824163533019](week1ActivitiesProblems.assets/image-20200824163533019.png)
+
+    1. What is the object's position at $t=4s$? Show calculation.
+       
+        | Variable   | Value                            | Significance     |
+        | ---------- | -------------------------------- | ---------------- |
+        | $x_f,\ x$  | $90m$ (calculated)               | final position   |
+        | $x_i$      | $10m$ (given)                    | initial position |
+        | $v_{x_i}$  | $12m/s$ (given)                  | initial velocity |
+        | $\Delta t$ | $4s-0s$ (given)                  | time traveled    |
+        | $a_x$      | $4m/s^2$ (calculated from slope) | car acceleration |
+        
+        
+        $$
+        a=\frac{\Delta v}{\Delta t}=\frac{12\frac ms-(-4\frac ms)}{4s-0s}=4m/s^2\\
+        x_f=x_i+v_{x_i}\Delta t+\frac12a_x(\Delta t)^2\\
+        x=10m+(12\frac ms)(4s-0s)+\frac12(4\frac m{s^2})(4s-0s)^2\\
+        x=10m+(12\frac ms)(4s)+\frac12(4\frac m{s^2})((4s)^2)\\
+        x=10m+(48\frac {m\cancel s}{\cancel s})+\frac12(4\frac m{s^2})(16s^2)\\
+        x=10m+48m+\frac{4*16}{2}\frac {m\cancel{s^2}}{\cancel{s^2}}\\
+        x=10m+48m+{2*16}m\\
+        x=10m+48m+32m\\
+        x=90m\\
+        \overline{\underline{|x=9\times 10^1m|}}
+        $$
+    
+2. Does this car ever change direction, if so, at what time?
+        The car changes direction at $t=3s$
+    
+3. When you sneeze, it is claimed that the air droplets in your lungs accelerate from rest to 150km/hr in 0.500s. Assume the droplets accelerate uniformly.
+
+    1. What is the average acceleration of a droplet in $m/s^2$?
+       
+        | Variable     | Value                       | Significance                                  |
+        | ------------ | --------------------------- | --------------------------------------------- |
+        | $v^2_{x_f1}$ | $\approx14m/s$ (calculated) | Velocity upon entering the water              |
+        | $v^2_{x_f2}$ | $0m/s$ (given)              | Velocity upon touching the bottom of the pool |
+        | $v^2_{x_i1}$ | $0m/s$ (given)              | Initial velocity (from the platform)          |
+        
+        $$
+        a=\frac{\Delta v}{\Delta t}=\frac{150\frac{km}{hr}-0}{0.500s-0}\\
+    \frac{150\cancel{km}}{1\cancel{hr}}*\frac{10^3m}{1\cancel{km}}*\frac{1\cancel{hr}}{3600s}\\
+        \frac{150\cancel{*10^3}}{\cancel{3600}3.6}m/s=41.\overline6m/s\\
+        a=\frac{41.\overline6m/s}{0.5000s}\\
+        a=83.\overline3m/s^2\\
+        \overline{\underline{|a=83.3m/s^2|}}
+        $$
+        
+    2. Assuming the acceleration is constant, how far does the droplet travel in 0.500s?
+       
+        | Variable   | Value                                         | Significance                                  |
+        | ---------- | --------------------------------------------- | --------------------------------------------- |
+        | $x_f$      | $\approx14m/s$ (calculated)                   | Velocity upon entering the water              |
+        | $x_i$      | $0m/s$ (given)                                | Velocity upon touching the bottom of the pool |
+        | $v_{x_i}$  | $0m/s$ (given)                                | Initial velocity (from the platform)          |
+        | $a_x$      | $83.3m/s^2$ (calculated in previous question) | Acceleration of droplets                      |
+        | $\Delta t$ | $(0.500s-0s)$ (given)                         | given time period for travel                  |
+
+    $$
+    x_f=x_i+v_{x_i}\Delta t+\frac12a_x(\Delta t)^2\\
+    x_f=0m+(0m/s)(0.500s-0s)+\frac12(83.3m/s^2)(0.500s-0s)^2\\
+    x_f=\cancel{0m+(0m/s)(0.500s-0s)}+\frac12(83.3m/s^2)(0.500s)^2\\
+    x_f=\frac12(83.3m/s^2)(0.500s)^2\\
+    x_f=\frac{83.3}{2*(\frac 12^2)}\frac{\frac m{\cancel{s^2}}}{\cancel{s^2}}\\
+    x_f=\frac{83.3}{\cancel{2*}(\frac 12\cancel{^2})}m\\
+    x_f=\frac{83.3}{\frac12}m=x_f=2*83.3m\\
+    x_f=166.6m\\
+    \overline{\underline{|x_f=1.66\times10^2m|}}
+    $$
+
+4. Divers compete by diving in a 3.00m deep pool from a platform 10m above the water. What is the magnitude of the minimum acceleration in the water needed to keep a diver from hitting the bottom of a pool? Assume the diver drops from rest from the platform and the acceleration in the water is constant.
+
+| Variable     | Value                       | Significance                                             |
+| ------------ | --------------------------- | -------------------------------------------------------- |
+| $v^2_{x_f1}$ | $\approx14m/s$ (calculated) | Velocity upon entering the water                         |
+| $v^2_{x_f2}$ | $0m/s$ (given)              | Velocity upon touching the bottom of the pool            |
+| $v^2_{x_i1}$ | $0m/s$ (given)              | Initial velocity (from the platform)                     |
+| $v^2_{x_i2}$ | $v^2_{x_f1}$                | Velocity upon entering the water                         |
+| $a_{x1}$     | $-9.81m/s^2$ (given)        | Acceleration due to gravity                              |
+| $a_{x2}$     | $33m/s^2$ (calculated)      | Acceleration to avoid impact with the bottom of the pool |
+| $\Delta x_1$ | $0m-10m$ (given)            | Distance from the platform to the top of the water       |
+| $\Delta x_2$ | $0m-3.00m$ (given)          | distance from the top of the water to the bottom of th   |
+
+$$
+v^2_{x_f1}=v^2_{x_i1}+2a_{x1}\Delta x_1\\
+v^2_{x_f1}=\cancel{(0m)^2}+2(-9.81m/s^2)(\cancel{0m}-(10m))\\
+v^2_{x_f1}=((9.81m/s^2)(20m))\\
+\sqrt{v^2_{x_f1}}=\sqrt{196.2m^2/s^2}\\
+v_{x_f1}=14.007141m/s\\
+\rule{150pt}{0.4pt}\\
+v^2_{x_f2}=v^2_{x_i2}+2a_{x2}\Delta x_2\\
+0=(14.007141m/s)^2+2a_{x2}(0m-(-3.00m))\\
+(-14.007141m/s)^2=2a_{x2}(3.00m)\\
+\frac{(-14.007141m/s)^2}{2*3.00m}=a_{x2}\\
+\frac{(-14.007141)^2}{6}\frac{\frac {m\cancel{^2}}{s^2}}{\cancel m}=a_{x2}\\
+\frac{-196.2}{6}m/s^2=a_{x2}\\
+-32.7m/s^2=a_{x2}\\
+\overline{\underline{|3.3\times10^1m/s^2=a_{x2}|}}
+$$
+
+
+
+# In Class Questions
+
+​	1. Why are you in this class?
+
+I am in this class because it is required for my major, and I REALLY didn't want to take calc-based physics because I heard it was extremely difficult.
+
+2. What is the general process for an approach of a physics problem?
+    1. Read the question
+        1. Understand (or find) the units. 
+        2. Find the requested answer, and its associated units.
+    2. Use provided equations. (These will be taught in class.)
+    3. Convert units (may be optional)
+    4. Ensure the answer makes sense.
+    5. Reduce to required number of significant figures.
+    6. Change answer to scientific notation (or metric prefixes, if allowed). (May be optional, depending on length of value)

November/macdougall_skyler_week11.md

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+# Week 11 Activities Problems
+
+## Skyler MacDougall
+
+## PHYS-111 2pm Lab
+
+### Group Members: Will Caliguri, Nina Chambliss
+
+#### Due: 10/6/2020
+
+# Rotational Dynamics
+
+1. A low friction pulley, modeled as a cylinder with a mass $m=0.800kg$ and a radius $r=30.0cm$, has a rope going over it as shown. The tension in the rope is $T_1=12.0N;\ T_2=10.0N$, with $T_1$ on the left and $T_2$ on the right. Both tension forces are tangential. What is the angular acceleration? (Answer to 3 sig figs).
+    $$
+    I=\frac12mr^2;\ \sum\tau=I\alpha\\
+    \tau_1-\tau_2=\frac12mr^2\alpha\\
+    r\tau_1-r\tau_2=\frac12mr^2\alpha\\
+    T_1-T_2=\frac12mr\alpha\\
+    \alpha=\frac{2(T_1-T_2)}{mr}\\
+    \alpha=\frac{2(12.0N-10.0N)}{(0.800kg)(30.0cm)}\\
+    \alpha=16.\overline6rad/s^2\\
+    \overline{\underline{|\alpha=16.7rad/s^2|}}
+    $$
+
+2. The $m=2.00kg$ solid disk is spinning about its central axis at $300rpm=10\pi rad/s$. The radius of the disk is $r=30.0cm$. How much friction force mus the brake apply to the rim to bring the disk to rest in $t=3.00s$. The frictional braking force acts tangentially on the circumference of the wheel. 
+    $$
+    F=\frac{I\alpha}{r};\ I=\frac12mr^2;\ \alpha=\frac \omega t\\\\
+    F=\frac{\frac12mr^2*\frac{\omega}{t}}{r}\\
+    F=\frac{mr^2\omega}{2rt}\\
+    F=\frac{mr\omega}{2t}\\
+    F=\frac{(2.00kg)(0.300m)10\pi rad/s}{2(3.00s)}\\
+    F=\pi N\\
+    \overline{\underline{|F=\pi N|}}
+    $$
+    
+
+# Angular Momentum
+
+1.  Two flat disks are rotating about a common axis. The top disk is rotating clockwise. The bottom disk is rotating counterclockwise. The top disk is dropped on the bottom disk. After a moment of sliding, the two disks stick together and rotate with the same velocity. All values are shown below:
+    $$
+    \omega_{1i}=-7.20rad/s\\
+    \omega_{2i}=9.80rad/s\\
+    I_{1}=1.44\times10^{-4}kgm^2\\
+    I_2=3.55\times10^{-4}kgm^2\\
+    h=1.00cm
+    $$
+    
+
+    1. What is the common angular velocity?
+        $$
+        L_i=L_f;\ L=I\omega\\
+        I_1\omega_{1i}+I_2\omega_{2i}=I_{1+2}\omega_f\\
+        \omega_f=\frac{I_1\omega_{1i}+I_2\omega_{2i}}{I_{1+2}}\\
+        \omega_f=\frac{(1.44\times10^{-4}kgm^2)(-7.20rad/s)+(3.55\times10^{-4}kgm^2)(9.80rad/s)}{(3.55+1.44)\times10^{-4}kgm^2}\\
+        \omega_f=4.894188377rad/s\\
+        \overline{\underline{|\omega_f=4.89rad/s|}}
+        $$
+    
+2. What is the total kinetic energy before the collision?
+        $$
+        K=\frac12I\omega^2\\
+        K_i=\frac12I_1\omega_{1i}^2+\frac12I_2\omega_{2i}^2\\
+        K_i=\frac12(1.44\times10^{-4}kgm^2)(-7.20rad/s)^2+\frac12(3.55\times10^{-4}kgm^2)(9.80rad/s)^2\\
+        K_i=20.77958mJ\\
+        \overline{\underline{|K_i=20.8mJ|}}
+        $$
+        
+    
+3. ... after the collision?
+    
+2. A figure skater is spinnning at a rate of $10.0rad/s$ with her arms horizontally extended and her moment of inertia of $2.50kgm^2$. 
+
+    1. When she pulls her arms in close to her body, will her angular speed change?
+    2. What is her angular speed after she pulls her arms in and reduces her moment of inertia to $1.60kgm^2$?
+    3. Calculate her kinetic energy before and after she pulls her arms in. Is kinetic energy conserved? If kinetic energy was lost, where did it go? If it was gained, where did it come from?

November/macdougall_skyler_week12.md

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+# Week 12 Activities Problems
+
+## Skyler MacDougall
+
+## PHYS-111 2pm Lab
+
+### Group Members: Will Caliguri, Nina Chambliss
+
+#### Due: 11/13/2020
+
+# Simple Harmonic Motion (Oscillation)
+
+1. How long must a simple pendulum be in order to have a period of simple harmonic motion of $1.00s$ when operated on the Moon?
+    (note: $g_{moon}=1.6m/s^2$)
+    $$
+    T=2\pi\sqrt{\frac{L}{g}}\\
+    L=\frac{T^2g}{4\pi^2}\\
+    L=\frac{(1.00s)^2(1.6m/s^2)}{4\pi^2}\\
+    L=0.04052847346m\\
+    \overline{\underline{|L=0.04m|}}
+    $$
+
+2. A certain simple pendulum has a period of simple harmonic oscillation of $T_0$ when operated on Earth. If the length of the simple pendulum is tripled, what is the new period of the simple harmonic oscillations, assuming the system is still on earth? Express the answer as a multiple or fraction of $T_0$.
+    $$
+    T=2\pi\sqrt{\frac{L}{9.81m/s^2}}\\
+    T=19.67951414\sqrt L\\
+    answer=\frac{T_{3L}}{T_{0}}=\frac{19.67951414\sqrt{3m}}{19.67951414\sqrt{1m}}=\sqrt3\\
+    T_{3L}=\sqrt3T_0\\
+    \overline{\underline{|T_{3L}=1.73T_0|}}
+    $$
+
+3. Vanya is trying to design a metronome that will count our four beats per second. She decides to use a simple pendulum as her system where one beat corresponds to a full cycle of a simple harmonic motion. To build the metronome she attaches a small weight having a mass of $m=390g$ to a long string, of negligible mass. How long must the string be in order for the metronome to maintain cadence that Vanya requires?
+    $$
+    L=\frac{T^2g}{4\pi^2};\ T=\frac1{f}=\frac1{4Hz}=0.25s;\ g=9.81m/s^2\\
+    L=\frac{(0.25s)^2(9.81m/s^2)}{4\pi^2}\\
+    L=1.553063768cm\\
+    \overline{\underline{|L=1.55cm|}}
+    $$
+
+4. To measure the local acceleration due to gravity on planet Astroworld, Travis takes a string having a length of $L=2.40m$ and a small ball bearing having a mass of $m=102g$. When operated as a simple pendulum on planet Astroworld, the system has a period of oscillation of $T=3.53s$. What is the magnitude of the local acceleration due to gravity on planet Astroworld?
+    $$
+    g=\frac{4\pi^2L}{T^2};\ T=3.53s;\ L=2.40m\\
+    g=\frac{(2.40m)4\pi^2}{(3.53s)^2}\\
+    g=7.603640367m/s^2\\
+    \overline{\underline{|g=7.60m/s^2|}}
+    $$
+
+5. The position as a function of time $t$ of a $m=50.0g$ mass attached to a spring is given by
+    $$
+    x(t)=(2.00cm)\cos\{(10.0rad/s)t\}
+    $$
+    Given:
+    $$
+    x=A\cos(\omega t)\\
+    \omega=\frac{2\pi}{T}=2\pi f\\
+    T=period\\
+    A=amplitude
+    $$
+    
+
+    1. What is the amplitude (maximum displacement) of the motion?
+        $2.00m$
+
+    2. What is the period?
+        $$
+        T=\frac{2\pi}{\omega}\\
+        T=\frac{2\pi}{10.0rad/s}\\
+        T=0.6283185307s\\
+        \overline{\underline{|T=0.628s|}}
+        $$
+
+    3. What is the spring constant? 
+        ($T_{spring}=2\pi \sqrt{\frac mk}$)
+        $$
+        T=2\pi\sqrt{\frac{m}{k}}\\
+        \frac{T}{2\pi}=\sqrt{\frac{m}{k}}\\
+        \frac{T^2}{4\pi^2}=\frac{m}{k}\\
+        k=\frac{m4\pi^2}{T^2}\\
+        k=\frac{(50.0g)4\pi^2}{(0.628s)^2}\\
+        k=5.00N/m\\
+        \overline{\underline{|k=5.00N/m|}}
+        $$
+
+    4. What is the total mechanical energy? 
+        ($E=K_{max}=U_{max};\ U_{max}=\frac12kA^2$)
+        $$
+        U_{max}=\frac12kA^2\\
+        U_{max}=\frac12(5.00N/m)(2.00cm)^2\\
+        U_{max}=1.00mJ\\
+        \overline{\underline{|U_{max}=1.00mJ|}}
+        $$
+
+    5. What is the maximum speed?
+        ($K_{max}=\frac12mv_{max}^2$)
+        $$
+        K=\frac12mv^2;\ K_{max}=U_{max}\\
+        v_{max}=\sqrt{\frac{2U}{m}}\\
+        v_{max}=\sqrt{\frac{2(1.00mJ)}{50.0g}}\\
+        v_{max}=0.2m/s^2\\
+        \overline{\underline{|v_{max}=0.2m/s^2|}}
+        $$
+
+    6. Determine the position at $t=0.400s$.
+        $$
+        x(t)=(2.00cm)\cos\{(10.0rad/s)t\}\\
+        x(t)=(2.00cm)\cos(4rad)\\
+        x=-0.01307287242m\\
+        \overline{\underline{|x=-1.31cm|}}
+        $$
+
+6. A $m=0.500kg$ mass is attached to a spring and oscillates on a smooth horizontal surface. Ignore friction and air resistance. The velocity-time graph is shown below:
+    ![image-20201110144523829](macdougall_skyler_week12.assets/image-20201110144523829.png)
+
+    1. What is the period of motion?
+
+        $T=4.00s$
+
+    2. What is the maximum speed of the mass?
+        $v_{max}=2.0m/s$
+
+    3. What is the maximum displacement of the motion?
+        $$
+        A=\frac{-v_{max}}{\omega};\ \omega=\frac{2\pi}{T}\\
+        A=\frac{-v_{max}T}{2\pi}\\
+        A=\frac{-(2.0m/s)(4.00s)}{2\pi}\\
+        A=-1.273239545m\\
+        x_{max}=|A|\\
+        \overline{\underline{|x_{max}=1.27m|}}
+        $$
+
+    4. What is the spring constant?
+        ($T=2\pi\sqrt{\frac{m}{k}}$)
+        $$
+        k=\frac{m4\pi^2}{T^2}\\
+        k=\frac{(0.500kg)4\pi^2}{(4.00s)^2}\\
+        k=1.233700550N/m\\
+        \overline{\underline{|k=1.23N/m|}}
+        $$
+
+    5. What is the maximum elastic potential energy of the mass-spring system?
+        ($U_{max}=\frac12kx^2$)
+        $$
+        U_{max}=\frac12kx^2\\
+        U_{max}=\frac12(1.23N/m)(1.27m)^2\\
+        U_{max}=1.00J\\
+        \overline{\underline{|U_{max}=1.00J|}}
+        $$
+
+    6. What is the total energy of the system at time $t=3.50s$?
+        $$
+        E_{max}=U_{max}=1.00J
+        $$
+        
+
+
+
+# Introduction to Waves
+
+1. A metal guitar has a linear mass density of $\mu=3.20g/m$. What is the speed of transverse waves on this string when its tension is $F_{tension}=90.0N$?
+    $$
+    v=\sqrt{\frac{F_T}{\mu}}\\
+    v=\sqrt{\frac{90.0N}{3.20g/m}}\\
+    v=167.7050983m/s\\
+    \overline{\underline{|v=168.m/s|}}
+    $$
+    
+2. A transverse wave travels along a stretched horizontal rope. The vertical distance from crest to trough for this wave is $13cm$ and the horizontal distance from crest to trough is $28cm$.
+
+    1. Draw a sketch of one wavelength for this wave in the x-y plane.
+        ![image-20201110153110331](macdougall_skyler_week12.assets/image-20201110153110331.png)
+    2. What is the amplitude of the wave?
+        $7.5cm$
+    3. What is the wavelength of the wave?
+        $56cm$
+

November/macdougall_skyler_week13.md

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+# Week 13 Activities Problems
+
+## Skyler MacDougall
+
+## PHYS-111 2pm Lab
+
+### Group Members: Will Caliguri, Nina Chambliss
+
+#### Due: 11/20/2020
+
+# Beats
+
+1. The little brown bat is a common bas species in North America. It emits echolocation pulses at a frequency of $40kHz$, well above the range of human hearing. To allow observers to "hear" these bats, an electronic bat detector combines the bat's sound wave with a wave created by a tunable oscillator. The resulting beat frequency is isolated, amplified, then played through a speaker. What frequency should the oscillator be set to produce an audible beat frequency of $3.0kHz$?
+    $$
+    f_{beat}=|f_1-f_2|\\
+    f_1=f_{beat}\pm f_2\\
+    f_1=3.0kHz\pm40kHz\\
+    \overline{\underline{|f_1=43kHz,\ 37kHz|}}
+    $$
+    
+
+# Sound Waves Interference
+
+1. Two identical sources emit sound of frequency $2.00kHz$ and a speed of $v=343m/s$. They are separated by $d_{diff}=25.7cm$. Assume no loss in intensity with distance.
+    1. Calculate the interference heard at a distance of $5.00m$ away along the axis joining the speakers.
+        $$
+        d_{diff}=n\lambda\\
+        \lambda=\frac{d_{diff}}{n}\\
+        f=\frac{vn}{d_{diff}}
+        $$
+        If $2n$ is an even integer, than it is constructive interference. If $2n$ is an odd integer, than it is destructive interference. If $2n$ is not an integer, it is partial interference.
+        $$
+        f=\frac{vn}{d_{diff}}\\
+        n=\frac{fd_{diff}}{v}\\
+        n=\frac{(2.00kHz)(25.7cm)}{343m/s}\\
+        n=1.498542274\approx1.5\\
+        2n=3
+        $$
+        3 is odd, therefore, it is completely destructive interference.
+    
+    2. You now increase the frequency of both speakers. Calculate the next frequency for which you will hear destructive interference. 
+        $$
+        2n=5;\ n=2.5\\
+        f=\frac{vn}{d_{diff}}\\
+        f=\frac{(343m/s)(2.5)}{25.7cm}\\
+        f=3.336575875kHz\\
+        \overline{\underline{|f=3.34kHz|}}
+        $$
+    
+    3. Now return to $2.00kHz$. What can you do to get the opposite type of interference as in part 1.1?
+        Change the distance, either from the listener to the speakers, or between the speakers.
+    
+2. 2 isotropic point sources produce sound waves that are in phase with each other at the positions of the sources. Source 1 is $d_1=17.1m$ from the observer, and Source 2 is $d_2=41.6m$ from the observer. What are the 3 lowest frequencies in the audible range at which fully destructive interference will occur at the position of the observer? Ignore the loss of intensity. Use $v=343m/s$ for the speed of sound in air.
+    Waves are in complete destructive interference when they are one half wavelength out of phase. This means that the difference between $d_1$ and $d_2$ has to be equal to some half-integer multiple of the wavelength. Therefore:
+    $$
+    d_{diff}=d_2-d_1=41.6m-17.1m=24.5m\\
+    d_{diff}=\frac n2\lambda\\
+    \lambda=\frac{2d_{diff}}{n}
+    $$
+    Given $v$, we can choose $n$ such that it fits within the human hearing range $20Hz-20kHz$.
+    $$
+    f=\frac v\lambda\\
+    f=\frac{vn}{2d_{diff}}\\
+    20Hz\le f\le20kHz
+    $$
+    Inputing $f=20Hz$ does not give us an integer $n$, but rather gives us guidance as to where our first value will be. The first integer above $n_{20Hz}=2.857...$ is 3. Thus, the 3 lowest frequencies where destructive interference will be calculated using $f=\frac{vn}{2d_{diff}}$, where $n=3,4,5$.
+    $$
+    \overline{\underline{|f=21.0Hz,\ 35.0Hz,\ 49.0Hz|}}
+    $$
+    
+
+# Standing Waves
+
+1. A string of length $L=2.00m$ is fixed at both ends and tightened until the wavespeed is $40.0m/s$.
+    1. What is the wavelength of the standing wave, given that the wave has 6 antinodes?
+        $$
+        \lambda=\frac{2L}{n}\\
+        \lambda=\frac{2(2.00m)}{6}\\
+        \lambda=0.\overline6m\\
+        \overline{\underline{|\lambda=0.67m|}}
+        $$
+    
+    2. What is the frequency of the standing wave stated above?
+        $$
+        f=\frac v\lambda\\
+        f=\frac{40.0m/s}{0.67m}\\
+        \overline{\underline{|f=60Hz|}}
+        $$
+    
+    3. What is the fundamental frequency of the stretched string?
+        $f_0=10Hz$
+    
+    4. If tension in the string is $T=3.60N$, what is the mass per unit length of the string?
+        $$
+        v=\sqrt{\frac{T}{\mu}}\\
+        \mu=\frac{T}{v^2}\\
+        \mu=\frac{3.60N}{(40m/s)^2}\\
+        \overline{\underline{|\mu=2.25\times10^{-3}kg/m|}}
+        $$
+        
+2. A $m=12.5g$ clothesline is stretched with a tension $T=22.1N$ between 2 poles $L=7.66m$ apart.
+    1. What is the fundamental frequency?
+        $$
+        f=\frac{v}{\lambda};\ v=\sqrt{\frac{T}{\mu}};\ \lambda=2L;\ \mu=\frac mL\\
+        f=\frac{\sqrt{\frac{T}{\frac{m}L{}}}}{2L}\\
+        f=\frac{\sqrt{\frac{22.1N}{\frac{12.5g}{7.66m}}}}{2(7.66m)}\\
+        f=7.596206281Hz\\
+        \overline{\underline{|f=7.60Hz|}}
+        $$
+    
+    2. What is the frequency of the second harmonic?
+        $$
+        f_n=nf_0\\
+        f_n=2(7.60Hz)\\
+        \overline{\underline{|f_n=15.20Hz|}}
+        $$
+        
+    
+    3. Make a simple sketch of the standing wave pattern corresponding to the second harmonic.
+        ![image-20201117162410368](macdougall_skyler_week13.assets/image-20201117162410368.png)
+    
+    4. If the tension in the clothesline is increased, does the frequency of the second harmonic change? 
+    
+        The frequency will also increase. This is because $f=\sqrt{\frac{T}{4mL}};\ f\propto T^{\frac12}$.
+    
+    5. If a heavier rope is used, but is stretched the same distance under the same tension, does the frequency of the second harmonic change?
+        The frequency will decrease. This is because $f=\sqrt{\frac{T}{4mL}};\ f\propto \frac1{m^{\frac12}}$.
+

October/invertedLecture.md

@@ -0,0 +1,133 @@
+# Inverted Lecture : Gravitation
+
+## Skyler MacDougall
+
+## PHYS-111 1pm Lecture
+
+#### Due: 10/9/2020
+
+1. In one short sentence, explain why we call the force of gravity an attractive force.
+    Gravity is an attractive force because it is a force that "pulls" one object to another.
+
+2. Does gravity exist between any two objects?
+
+    1. How does the force of gravity depend on the distance between the two objects?
+        The force of gravity is proportional to the inverse square of the distance between the two objects.
+    2. Is the gravitational attraction between the two objects the same magnitude for each object? Is it the same direction for each object?
+        The magnitude between the two objects is the same. The direction is directly opposite of the other's direction. Or, in other words, the gravitational attraction force vectors point towards each other.
+
+3. If there's more than one gravitational force acting on a body, can you just add the magnitudes of the gravitational forces acting on that body? If not, state why you cannot, and what mathematical tools you would then have to use to find the total gravitational force acting on a body.
+    You cannot simply add the magnitudes of the gravitational forces acting on that body. The gravitational force vectors must be added to properly represent the gravitational forces acting on an object.
+
+4. Write the formula for the magnitude of the gravitational force for two bodies of masses $m$ and $M$ whose centers are a distance $r$ apart.
+    $$
+    F_g=G\frac{mM}{r^2}
+    $$
+
+5. As you already know, the force of gravity acting on a body of mass $m$ is also given by the formula $F_g=w=mg$, where $g$ is the acceleration due to gravity. Put the two formulas for the force of gravity on earth equal to each other to derive an equation for the acceleration due to gravity on any planet of mass $M_p$ and radius $R_p$.
+    $$
+    ma=G\frac{mM}{r^2}\\
+    a=G\frac{M_p}{R_p^2}\\
+    $$
+
+6. Describe in one short sentence how an object in orbit experiences weightlessness even though the force of gravity is always acting on it.
+   
+7. In one short sentence, explain how circular space stations in sci-fi movies create artificial gravity.
+    In space, there is (effectively) no forces acting on you. As such, you are "weightless". Circular space stations create artificial gravity by spinning the station. This spin can be kept going essentially indefinitely, due to the lack of air resistance, and it causes the objects within the space ship to accelerate out towards the outer radius of the spaceship. When the spaceship is spinning at the correct speed, the force due to this acceleration is equal to the force due to gravity on earth.
+
+8. Using the concepts of a projectile's trajectory, of the planet's curvature, and of free fall, answer in one or two short sentences the following question:  "If gravity is always pulling on the moon, why doesn't the moon fall into the Earth?"
+    The moon is in an orbit. As such, it can be thought of as a circular motion problem, with the force of gravity being the acceleration on the moon. The force of gravity does not have enough of an acceleration to modify the velocity vector of the moon so that the velocity vector falls in line with the earth.
+
+9. What are Kepler's 3 laws of planetary motion? State Laws 1 and 2, and write the formula for Law 3.
+    Law 1: A planet orbits its star about an eclipse, with the sun at one focus of the eclipse.
+    Law 2: A line connecting a planet to its star sweeps out equal areas in equal times as the planet moves around its orbit. That is, the planet moves faster when it is closer to its star.
+    Law 3:
+    $$
+    t_{orbit}=Ar^{\frac32}
+    $$
+    , where $A$ is some constant.
+
+10.  
+
+    1. In one short sentence, describe "orbital speed".
+        Orbital speed is the speed necessary to achieve a stable orbit around an object. 
+
+    2. Derive the formula for orbital speed by starting with the general formula for the force of gravity and the formula for newton's second law of motion.
+        $$
+        F_g=G\frac{Mm}{r^2};\ F_{net}=m\frac {v^2}{r}\\
+        G\frac{Mm}{r^2}=m\frac{v^2}r\\
+        G\frac{M}{r}={v^2}\\
+        v=\sqrt{\frac{GM}{r}}
+        $$
+        
+3. What is the condition for this formula of orbital speed to reduce it to $v=\sqrt{rg}$?
+        When the orbit is very close to the surface of the object being orbited.
+    
+11. In one sentence, describe what geosynchronous satellites and geostationary satellites are.
+    Both geosynchronous satellites and geostationary satellites are satellites that orbit an object at a synchronous speed. Geostationary satellites orbit around the same pole as the object they're orbiting, while geosynchronous satellites can orbit with an inclination. That is, the pole of the geosynchronous orbit is not necessarily the same as the pole of the object's rotation. 
+
+12.  
+
+    1. Write the general equation for the Earth's gravitational potential energy (it includes the dependence on the distance from the center of the earth, $r$).
+        $$
+        U=-G\frac{mM_E}{r}
+        $$
+        
+2. With respect to the surface of the earth, where is the approximation $U=mgh$ applied?
+    
+    When the difference between the radius of the earth and the object's distance from the center of the earth is close to the radius of the earth.
+    
+13.  
+
+    1. Describe in one short sentence what "Escape speed" is.
+        Escape speed is the speed necessary to escape the pull of an object's gravity.
+
+    2. Starting wit the conservation of energy equation, $\Delta K+\Delta U=0$, derive an equation for the escape speed, $v_e$, of any object of mass $m$ from any planet of mass $M_p$ and radius $R_p$. 
+        $$
+        K+U=0\\
+        \frac12mv_e^2+\frac{-GMm}{r}=0\\
+        v_e=\sqrt{\frac{2GM}{r}}
+        $$
+        
+3. Now, as an example, calculate the escape speed from Earth, given the following:
+        $$
+        M_E=5.98\times10^{24}kg\\
+        R_E=6380km\\
+        G=6.67\times10^{-11}Nm^2kg^{-2}
+        $$
+    
+$$
+    v_e=\sqrt{\frac{2GM}{r}}\\
+    v_e=\sqrt{\frac{2(6.67\times10^{-11}Nm^2kg^{-2})(5.98\times10^{24}kg)}{6380km}}\\
+    v_e=11.19\times10^3m/s\\
+    \overline{\underline{|v_e=1.12\times10^4m/s|}}
+    $$
+    
+
+    
+14. Put the "orbital speed" and "escape speed" formulae next to each other, and circle the differences.
+    $$
+    v_{orbit}=\sqrt{\frac{GM}{r}}\\
+    v_e=\sqrt{\frac{\underline{\overline{|2|}}GM}{r}}
+    $$
+    
+15.  
+
+    1. In one short sentence, describe what a "black hole" is.
+        A black hole is an object with (theoretically) infinite mass and an infinitely small diameter. 
+
+    2. Use plausible numbers for the mass and radius of a black hole like the one at the center of our Galaxy and calculate the escape speed from your black hole. Use this to show why, "Nothing escapes a black hole, not even light".
+        According to [*Astronomy & Astrophysics*](https://www.aanda.org/articles/aa/full_html/2019/05/aa35656-19/aa35656-19.html), the mass of the black hole at the center of our galaxy is $(4.154\pm0.014)\times10^6M_\odot$. $M_\odot$ is the mass of our Sun, at $9.9891\times10^{30}kg$, according to [NASA](https://web.archive.org/web/20100715200549/http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html). The radius of a black hole, named the Schwarzschild radius, is calculated based off the mass, using the following formula:
+        $$
+        r=\frac{2GM}{c^2}
+        $$
+        Therefore.
+        $$
+        v_e=\sqrt{\frac{2GM}{r}};\ r=\frac{2GM}{c^2}\\
+        v_e=\sqrt{\frac{2GM}{\frac{2GM}{c^2}}}\\
+        v_e=\sqrt{\frac{1}{\frac{1}{c^2}}}=c
+        $$
+        The velocity required to escape a black hole is the speed of light. 
+
+16. Briefly describe a concept that surprised or interested you the most in this chapter.
+    $GM$ is often abbreviated in the astronomy/astrophysics realm to $\mu$, which kinda makes sense, because you don't tend to use micro- when you're working at the scale of the universe.

October/macdougall_skyler_week10.md

@@ -0,0 +1,229 @@
+# Week 10 Activities Problems
+
+## Skyler MacDougall
+
+## PHYS-111 2pm Lab
+
+### Group Members: Will Caliguri, Nina Chambliss
+
+#### Due: 10/30/2020
+
+# Work, Rotational Energy and Energy Conservation
+
+1. A $12.0g$ CD with a radius of $6.00cm$ rotates with an angular speed of $34.0rad/s$. Treat the CD as a uniform disk.
+
+    1. What is its kinetic energy?
+        $$
+        K=\frac12I\omega^2;\ I=mr^2\\
+        K=\frac12mr^2\omega^2\\
+        K=\frac12(12.0g)(6.00cm)^2(34.0rad/s)^2\\
+        K=0.0249696J\\
+        \overline{\underline{|K=25.0mJ|}}
+        $$
+
+    2. What angular speed must the CD have if its kinetic energy is doubled?
+        $$
+        \omega=\sqrt{\frac{2K}{mr^2}}\\
+        \omega=\sqrt{\frac{2(50.0mJ)}{(12.0g)(6.00cm)^2}}\\
+        \omega=48.083261rad/s\\
+        \overline{\underline{|\omega=48.1rad/s|}}
+        $$
+
+    3. What average power is required to bring it to rest from its original speed in $2.50s$?
+        $$
+        P=\tau\omega;\ \tau=I\alpha;\ \alpha=\frac{\omega}{t};\ I=mr^2\\
+        P=\frac{mr^2\omega^2}{t}\\
+        P=\frac{(12.0g)(6.00cm)^2(34.0rad/s)^2}{2.50s}\\
+        P=0.01997568W\\
+        \overline{\underline{|P=20.0mW|}}
+        $$
+
+2. When a baseball pitcher throws a curve ball, the ball is given a fairly rapid spin. If a $0.150kg$ baseball with a radius of $3.70cm$ is thrown with a linear center-of-mass speed of $48.0m/s$ and an angular speed of $42.0rad/s$ around its center-of-mass, how much of its total kinetic energy is translational kinetic energy of its center-of-mass and how much is rotational kinetic energy around its center-of-mass? Assume the ball is a uniform solid sphere.
+    $$
+    K_{rot}=\frac12I\omega^2;\ K_{trans}=\frac12mv^2;\ I=mr^2\\
+    K_{rot}=\frac12mr^2\omega^2|K_{trans}=\frac12mv^2\\
+    K_{rot}=\frac12(0.150kg)(3.70cm)^2(42.0rad/s)^2|K_{trans}=\frac12(0.150kg)(48.0m/s)^2\\
+    K_{rot}=0.1811187J|K_{trans}=172.8J\\
+    K_{total}=172.98112J\\
+    K_{rot_\%}=\frac{K_{rot}}{K_{total}}=\frac{0.1811187J}{172.98112J}=0.10470432\%\\
+    K_{trans_\%}=\frac{K_{trans}}{K_{total}}=\frac{172.8J}{172.98112J}=99.895296\%\\
+    \overline{|K_{rot_\%}=0.10\%|}\\
+    \underline{|K_{trans_\%}=99.90\%|}
+    $$
+
+3. A wheel of radius $R$ is free to rotate about a frictionless axle. The moment of inertia of the wheel with respect to its axis of rotation is $I$. A light rope is wrapped around the fixed wheel. The other end of the rope is attached to a box of mass $m$. The box is released at rest. The rope does not stretch or slip over the surface of the wheel. $I$ will be given as an intertial value, do not break it down.
+
+    1. Use conservation of energy to derive an expression of the **speed** $v$, of the box after it has descended a vertical distance $h$. [^1]
+        $$
+        U=mgh;\ \Delta K=\frac12I\omega^2;\ \omega=\frac{v_t}{r}\\
+        h_i=0;\ v_{t_i}=0\\
+        U_i+K_i=U_f+K_f\\
+        mgh_i+\frac{Iv_i^2}{2r^2}+\frac12mv^2=mgh_f+\frac{Iv_f^2}{2r^2}\\
+        \cancel{mgh_i}^0+\cancel{\frac{Iv_i^2}{2r^2}}^0+\cancel{\frac12mv^2}^2=mgh_f+\frac{Iv_f^2}{2r^2}+\frac12mv^2\\
+        -mgh_f=\frac{Iv_f^2}{2r^2}+\frac12mv^2\\
+        v^2(\frac{I+mr^2}{2r^2})=-mgh_f\\
+        v=\sqrt{\frac{-2mgh_fr^2}{I+mr^2}}
+        $$
+        
+2. Determine the speed of the box given the following:
+        $$
+        h=1.50m\\
+        R=0.200m\\
+        I=0.400kg\times m^2\\
+        m=6.00kg
+        $$
+    
+    $$
+        v_f=\sqrt{\frac{-2mgh_fr^2}{I+mr^2}}\\
+        v_f=\sqrt{\frac{-2(6.00kg)(9.81m/s^2)(-1.50m)(0.200m)^2}{0.400kg\times m^2+(6.00kg)(0.200m)^2}}\\
+        v_f=3.3220852m/s\\
+        \overline{\underline{|v_f=3.32m/s|}}
+        $$
+    
+3. Using your result for part 3.1, write the speed of the box after it has descended a distance $h$ assuming the pulley is "*light*". Does this result make sense? What is the numerical value of the speed of the box in the limiting case of a "*light*" pulley?
+        $$
+        v_f=\sqrt{\frac{-2mgh_fr^2}{I+mr^2}};\ I=0\\
+        v_f=\sqrt{\frac{-2mgh_fr^2}{\cancel{I}^0+mr^2}}\\
+        v_f=\sqrt{\frac{-2\cancel{m}gh_f\cancel{r^2}}{\cancel{mr^2}}}\\
+        v_f=\sqrt{-2gh_f}\\
+        v_f=\sqrt{-2(9.81m/s^2)(1.50m)}\\
+        v_f=5.4249424m/s\\
+        \overline{\underline{|v_f=5.42m/s|}}
+        $$
+
+# Torque
+
+1. A $1.40m$ long bar on a horizontal surface is free to rotate about an axis perpendicular to the page and passing through its center. A force $\overrightarrow F$ of magnitude $35.0N$ is applied to the left end of the bar at different angles as shown in the overhead views below. Determine the torque on the bar produced by each of the forces shown below. 
+
+    1.  ![image-20201027160607756](macdougall_skyler_week10.assets/image-20201027160607756.png)
+        $$
+        \tau=-rF\sin\theta\\
+        \sin\theta=\sin(0^\circ)=0\\
+        \tau=0
+        $$
+
+    2. ![image-20201027160635126](macdougall_skyler_week10.assets/image-20201027160635126.png)
+        $$
+        \tau=-rF\sin\theta\\
+        \tau=-(1.40m)(35.0N)\sin(90^\circ)\\
+        \overline{\underline{|\tau=-49.0J|}}
+        $$
+        
+3.  ![image-20201027160703612](macdougall_skyler_week10.assets/image-20201027160703612.png)
+        $$
+        \tau=-rF\sin\theta\\
+        \overrightarrow{F_{1_y}}=-\overrightarrow{F_{2_y}}\\
+        \tau=0
+        $$
+    
+4.  ![image-20201027160731415](macdougall_skyler_week10.assets/image-20201027160731415.png)
+        $$
+        \tau=-rF\sin\theta\\
+        \tau=-(1.40m)(35.0N)\sin(30.0^\circ)\\
+        \tau=-24.5J
+        $$
+    
+2. A rod is being used as a lever as shown in the diagram. The fulcrum is $1.2m$ from the load and $2.4m$ from the applied force. The rod makes a $25.0^\circ$ angle with respect to the horizontal. The applied force $\overrightarrow{F_A}$ has a magnitude of $128N$ and is directly down. What is the torque due to the applied force relative to the fulcrum?
+    $$
+    \tau=-rF\sin\theta\\
+    \tau=-(2.4m)(128N)\sin(90^\circ+25^\circ)\\
+    \tau=-278.41775J\\
+    \overline{\underline{|\tau=278.J,\ clockwise|}}
+    $$
+    
+3. A house painter stands $3.00m$ above the ground on a $5.00m$ long ladder that leans against the wall at a point $4.70m$ above the ground. The painter weighs $680N$ and stands vertically on the ladder. The ladder weighs $120N$. The ladder makes an angle $\theta$ with the horizontal ground. Choose an axis of rotation perpendicular to the page and passing down through the bottom of the ladder. 
+
+    1. Create an extended free body diagram for the ladder using the diagram below. 
+
+    2. Find the torque due to the ladder.
+        $r=2.5m$ because that is the distance between the center of mass of the ladder and the point of rotation.
+        $$
+        \tau=-rF\sin\theta;\ \theta=\sin^{-1}(\frac{h_1}{l_l})_+90^\circ\\
+        \tau=-rF\sin(\sin^{-1}(\frac{h_1}{l_l})+90^\circ)\\
+        \tau=-(2.50m)(120N)(\sin(\sin^{-1}(\frac{4.70m}{5.00m})+90^\circ))\\
+        \tau=-102.3523327J\\
+        \overline{\underline{|\tau=102.J,\ clockwise|}}
+        $$
+
+    3. Find the torque due to the painter.
+        $$
+        \tau=-rF\sin\theta;\ r=\frac{3.00m}{\sin(\theta-90^\circ)}\\
+        \tau=-(\frac{3.00m}{\sin(\sin^{-1}(\frac{4.70m}{5.00m}))})*680N*\sin(\sin^{-1}(\frac{4.70m}{5.00m})+90^\circ)\\
+        \tau=-740.4211298J\\
+        \overline{\underline{|\tau=740.J,\ clockwise|}}
+        $$
+        
+
+# Equilibrium
+
+1. Your job is to ensure the safety of the house painter shown in the diagram below. You quickly realize that preventing the ladder from slipping requires a friction force one the bottom of the ladder. The house painter stands $3.0m$ above the ground on the $5.0m$ long ladder that leans at rest against a smooth vertical wall at a point $4.7m$ above the ground. The painter weighs $680N$ and the uniform ladder weighs $120N$. Assume no friction between the house and the upper end of the ladder.
+
+    1. What is the direction of the static frictional force exerted on the ladder by the ground?
+        Towards the wall.
+
+    2. Draw a free body diagram.
+
+    3. Calculate the magnitude of the static frictional force exerted on the ladder by the ground. 
+        $$
+        \sum\overrightarrow F=0\\
+        \therefore\\
+        \overrightarrow {n_w}=\overrightarrow{F_{static}}\\
+        \overrightarrow {n_g}=\overrightarrow{W_L}+\overrightarrow{W_P}\\
+        \tau_{n_w}=\tau_{w_L}+\tau_{w_P}\\
+        \tau=rF\sin\theta\\
+        \overrightarrow{F_{static}}=\frac{\tau_{n_w}}{r\sin\theta}\\
+        \overrightarrow{F_{static}}=\frac{\tau_{w_P}+\tau_{w_L}}{r\sin\theta}\\
+        r=l_l;\ \theta=\cos^{-1}(\frac{h_l}{l_l})\\
+        \overrightarrow{F_{static}}=\frac{740.J+102.J}{5.00m(\sin(\sin^{-1}(\frac{4.70m}{5.00m})))}\\
+        \overrightarrow {F_{static}}=179.3135027N\\
+        \overline{\underline{|\overrightarrow {F_{static}}=179.N|}}
+        $$
+
+    4. What is the minimum coefficient of static friction required to keep the ladder from slipping?
+        $$
+        F_{static}=\mu N\\
+        N=n_g=\overrightarrow{W_L}+\overrightarrow{W_P}\\
+        179.N=\mu (680.N+120.N)\\
+        \mu=0.2241418783\\
+        \overline{\underline{|\mu=0.224|}}
+        $$
+
+2. A $40.0kg,\ 5.00m$ long uniform horizontal beam is supported by, but not attached to, the two vertical posts. The vertical posts are separated by $3.00m$. A $25.0kg$ girl starts walking along the beam to the right.
+
+    1. Determine an algebraic expression for the normal force $N_1$ exerted on the beam by the left vertical post when the girl is a distance $x$ to the right of the vertical post. The result should be $N_1$ written as a function of $x$.
+        $$
+        F=0;\ \tau=0\\
+        N_1+N_2=W_b+W_g\\
+        \tau_{N_1}+\tau_{N_2}+\tau_{W_b}+\tau_{W_g}=0\\
+        \tau=rF\sin\theta;\ r_{\tau_{N_1}}=0;\ \tau_{N_1}=0\\
+        \tau_{N_2}=\tau_{W_b}+\tau_{W_g}\\
+        \tau_{W_b}=1.5W_b;\ \tau_{W_g}=(3+X)W_g;\ \tau_{N_2}=3N_2\\
+        N_2=\frac{1.5W_b+(3+X)W_g}{3}\\
+        N_1+\frac{1.5W_b+(3+X)W_g}{3}=W_b+W_g\\
+        N_1+\frac12W_b+(1+\frac13X)W_g=W_b+W_g\\
+        N_1=\frac12W_b+(1-(1+\frac13X))W_g\\
+        N_1=\frac12W_b-\frac13XW_g
+        $$
+
+    2. What is the maximum distance to the right of the right vertical post $x_{max}$ she can walk before the beam begins to tip?
+        $$
+        
+        $$
+
+3. A man is attempting to raise a uniform $34.0kg$ flagpole that is hinged at the base by pulling on a rope to the top of the pole, as shown. Let the length of the flagpole be $L$.
+
+    1. Draw a free body diagram.
+
+    2. With what force does the person have to pull on the rope to hold the pole motionless in the position shown?
+        $$
+        \tau=rF\sin\theta;\ \sum\tau=0\\
+        \tau_{pole}=\tau_{man}\\
+        \frac L2mg\sin(120^\circ)=LF\sin(170^\circ)\\
+        F=\frac{mg\sin(120^\circ)}{2\sin(170^\circ)}\\
+        $$
+        
+3. What are the horizontal and vertical components of the force exerted on the bottom of the pole by the hinge?
+        $$
+        asdf
+        $$
+        

October/macdougall_skyler_week7.md

@@ -0,0 +1,359 @@
+# Week 7 Activities Problems
+
+## Skyler MacDougall
+
+## PHYS-111 2pm Lab
+
+### Group Members: Will Caliguri, Nina Chambliss
+
+#### Due: 10/9/2020
+
+# Conservative and Non-Conservative Forces
+
+$$
+W=Fd\cos\theta\\
+\theta=\ang d-\ang F
+$$
+
+
+
+1. Write the work-kinetic energy theorem in words and as a formula.
+
+    The total work done equals the change in kinetic energy.
+2. A book of mass $m$ moves along a horizontal table where a kinetic frictional force of constant magnitude of $5.0N$ acts on the book. Point $B$ is located $2.0m$ north and $1.5m$ east of point $A$. The book is motionless at the start and at the end. 
+    1. Determine the work done by the force of friction along the path $A\rightarrow C\rightarrow B$. Make a sketch of $\overrightarrow {f_k}$, $\overrightarrow d$, and $\theta$ for each calculation of work.
+        $$
+        A\rightarrow C\\
+        \overrightarrow d=2.0m\ang90^\circ;\ \overrightarrow{f_k}=5.0N\ang-90^\circ;\\
+        W=Fd\cos\theta\\
+        W_1=5.0N*2.0m*\cos(90^\circ-(-90^\circ))\\
+        W_1=10.0Nm*-1\\
+        W_1=-10.0J\\\rule{70pt}{0.4pt}\\
+        C\rightarrow B\\
+        \overrightarrow d=1.5m\ang0^\circ;\ \overrightarrow{f_k}=5.0N\ang180^\circ;\\
+        W_2=5.0N*1.5m*\cos(0-180^\circ)\\
+        W=-7.5J\\\rule{70pt}{0.4pt}\\
+        W_{total}=\sum W\\
+        W_{total}=(-10.0J)+(-7.5J)\\
+        W_{total}=-17.5J
+        $$
+        
+    2. Determine the work done by the force of friction along the path $A\rightarrow B$. Make a sketch of $\overrightarrow {f_k}$, $\overrightarrow d$, and $\theta$ for the calculation of work.
+        $$
+        \overrightarrow d=(\sqrt{(2.0m)^2+(1.5m)^2})\ang(\tan^{-1}(\frac{1.5m}{2.0m}))^\circ;\ F=5.0N\ang(\tan^{-1}(\frac{20.m}{1.5m}))^\circ\\
+        \overrightarrow d=2.5m\ang53.13010235^\circ;\ F=5.0N\ang-126.8698976^\circ\\
+        W=5.0N*2.5m*\cos(180^\circ)\\
+        W=-12.5J
+        $$
+        
+    3. Are your answers the same or different for the two paths?
+        The answers are different for the two paths.
+    
+    4. Using the work-kinetic energy theorem, find the total work done on the book in going from point $A$ to point $B$. This is the work done by ALL  forces acting on the book.
+        $$
+        W=K_f-K_i\\
+        K=\frac12mv^2\\
+        v_i=0=v_f\\
+        \therefore\\
+        W=0
+        $$
+        
+    5. Is the work done by the friction force equal to the total work? Why or why not?
+        The friction force is not equal to the total work. There is another force that caused the movement of the book, which is counteracted by the friction, equating to zero work.
+3. Now consider a ball of mass $m$ that moves in the vertical plane. Points $A,\ B$, and $C$ are the same as before. You move from $A$ to $B$ along two different paths. The ball has a mass of $4.0kg$ and is motionless at the start and end (points $A$ and $B$ respectively.)
+    1. Determine the work done by the weight force along the path $A\rightarrow C\rightarrow B$. Make a sketch of $\overrightarrow {f_k}$, $\overrightarrow d$, and $\theta$ for each calculation of work.
+        $$
+        A\rightarrow C\\
+        \overrightarrow d=2.0m\ang90^\circ;\ \overrightarrow{f_k}=39.24N\ang-90^\circ;\\
+        W=Fd\cos\theta\\
+        W_1=39.24N*2.0m*\cos(90^\circ-(-90^\circ))\\
+        W_1=78.48J*-1\\
+        W_1=-78.48J\\\rule{70pt}{0.4pt}\\
+        C\rightarrow B\\
+        \overrightarrow d=1.5m\ang0^\circ;\ \overrightarrow{f_k}=39.24N\ang-90^\circ;\\
+        W_2=39.24*1.5m*\cos(0-90^\circ)\\
+        W=0J\\\rule{70pt}{0.4pt}\\
+        W_{total}=\sum W\\
+        W_{total}=(-78.48J)+(0J)\\
+        W_{total}=-78.48J
+        $$
+        
+    2. Determine the work done by the weight force along the path $A\rightarrow B$. Make a sketch of $\overrightarrow {f_k}$, $\overrightarrow d$, and $\theta$ for the calculation of work.
+        $$
+        \overrightarrow d=(\sqrt{(2.0m)^2+(1.5m)^2})\ang(\tan^{-1}(\frac{1.5m}{2.0m}))^\circ;\ F=39.24N\ang-90^\circ\\
+        \overrightarrow d=2.5m\ang53.13010235^\circ;\ F=39.24N\ang-90^\circ\\
+        W=39.24N*2.5m*\cos(53.13010235^\circ-(-90^\circ))\\
+        W=-78.48J
+        $$
+        
+    3. Are your answers the same or different for the two paths?
+        The answer is the same for both paths.
+    
+    4. Using the work-kinetic energy theorem, find the total work done on the book in going from point $A$ to point $B$. This is the work done by ALL  forces acting on the book.
+        $$
+        W=K_f-K_i\\
+        K=\frac12mv^2\\
+        v_i=0=v_f\\
+        \therefore\\
+        W=0
+        $$
+        
+    5. Is the work done by the weight force equal to the total work? Why or why not?
+        The friction force is not equal to the total work. There is another force that caused the movement of the book, which is counteracted by the friction, equating to zero work.
+
+# Energy Conservation
+
+1. A slingshot fires a pebble from the top of a building at a speed of $14.0m/s$. The bulding is $31.0m$ tall. Assume an ideal system. Find the speed with which the pebble strikes the ground when the pebble is fired:
+
+    1. Horizontally.
+    2. Straight up.
+    3. Straight down.
+    4. At an angle of $\theta=27.5^\circ$ above the horizontal.
+
+    All of these will be the same, at:
+    $$
+    W=\Delta K+\Delta U\\
+    K=\frac12mv^2;\ U=mgy\\
+    W=(\frac12m(v_f^2-v_i^2))+(mg\Delta y)\\
+    W=0\\
+    \therefore\\
+    \frac12m(v_f^2-v_i^2)=mg\Delta y\\
+    \frac12(v_f^2-v_i^2)=g\Delta y\\
+    v_f^2-v_i^2=2g\Delta y\\
+    v_f^2=2g\Delta y+v_i^2\\
+    v_f=\sqrt{2g\Delta y+v_i^2}\\
+    v_f=\sqrt{2(9.81m/s^2)(31.0m)+(14.0m/s)^2}\\
+    v_f=28.35677289m/s\\
+    \overline{\underline{|v_f=28.4m/s|}}
+    $$
+
+2. A cart slides to the right along a smooth surface and passes position 1 ($x_i$) with a speed of $15.0m/s=v_i$. Ignore all frictional effects. The vertical height at each position is labeled.
+
+    1. What is the speed of the cart as it passes position 2 (+8.00m vertical)?
+        Smooth surface = Conservative
+        $$
+        K_1+U_1=K_2+U_2\\
+        K=\frac12mv^2;\ U=mgy\\
+        \frac12mv_1^2+mgy_1=\frac12mv_2^2+mgy_2\\
+        \frac12mv_1^2+mg(0)=\frac12mv_22^2+mgy_2\\
+        \frac12\cancel{m}v_1^2+\cancel{mg(0)}^0=\frac12\cancel{m}v_2^2+\cancel{m}gy_2\\
+        \frac12v_2^2=\frac12v_2^2-gy_2\\
+        v_2^2=v_1^2-2gy_2\\
+        v_2=\sqrt{v_1^2-2gy_2}\\
+        v_2=\sqrt{(15.0m/s)^2-2(9.81m/s^2)(8.00m)}\\
+        v_2=8.248636251m/s\\
+        \overline{\underline{|v_2=8.25m/s|}}
+        $$
+
+    2. Will the cart reach position 4? If so, what is the speed as it passes position 4?
+        Potential energy at most must be equal to kinetic energy, because of conservation of energy. Therefore:
+        $$
+        \frac12mv^2=mgy\\
+        \frac12v^2=gy\\
+        \frac{v^2}{2g}=y\\
+        y=11.46788991m\\
+        11.46788991m\cancel{\ge}15.0m\\
+        \therefore
+        $$
+        The cart will not reach position 4.
+
+3. In an emergency, the passengers on an airplane slide down a curved escape ramp to safety. The ramp has a radius of curvature $R=10.0m$ and the passengers exit the plane $2.50m$ above the ground. 
+
+    1. What is the change in gravitational potential energy of a $72.0kg$ passenger due to sliding down the ramp?
+        $$
+        \Delta U=mg\Delta y\\
+        \Delta U=(72.0kg)(9.81m/s^2)(2.50m)\\
+        \Delta U=1.7658kJ\\
+        \overline{\underline{|\Delta U=1.77kJ|}}
+        $$
+    
+2. What is the passenger's speed at the bottom of the slide, assuming they start at rest at the top of the slide?
+        Assuming an energy-conservative system:
+    $$
+        \Delta U=\Delta K\\
+        K=\frac12mv^2\\
+        \Delta U=\frac12mv^2\\
+        \sqrt\frac{2\Delta U}{m}=v\\
+        \sqrt\frac{2(1.77kJ)}{72.0kg}=v\\
+        v=7.003 570 518 m/s\\
+        \overline{\underline{|v=7.00m/s|}}
+    $$
+    
+    3. If the passenger reaches the bottom of the slide with a speed of $3.75m/s$, how much work was done on the passenger by non-conservative forces? What forces are responsible for the non-conservative work?
+    $$
+        K=\frac12mv^2\\
+        K=\frac12(72.0kg)(3.75m/s)^2\\
+        K=506.25J\\
+        K_{ideal}-K_{real}\\
+        1.77kJ-506.25J\\
+        1.25955kJ
+    $$
+    $1.30kJ$ of work is done by non-conservative forces. The force most likely to have done this work is friction.
+    
+4. A vertical spring with a spring constant of $k$ is mounted on the floor. From directly above the spring, which is unstrained, a block of mass $m$ is dropped from rest. It collides with and sticks to the spring, which compresses a maximum amount $x_{max}$ in bringing the block to a momentary halt. Assume an ideal system.
+
+    1. From what height $H$ above the compressed spring was the block dropped? [^1]
+        $$
+        U_g=U_{spring}\\
+        U_g=mgy;\ U_{spring}=\frac12kx^2;\ y=H\\
+        mgH=\frac12kx^2\\
+        H=\frac{kx^2}{2mg}
+        $$
+
+    2. Calculate $H$ given:
+        $$
+        x_{max}=2.50cm\\
+        m=0.300kg\\
+        k=450N/m
+        $$
+        
+
+    $$
+    H=\frac{kx^2}{2mg}\\
+    H=\frac{(450N/m)(2.50cm)^2}{2(0.300kg)(9.81m/s)}\\
+    H=47.782 874 62 mm\\
+    \overline{\underline{|H=4.78cm|}}
+    $$
+
+    
+
+5. A spring has a constant $k$. The spring is near the base of a ramp inclined at an angle $\theta$ above the horizontal. A block of mass $m$ is pressed against the spring, compressing it a distance $x_i$ from its equilibrium position. The block is then released. The horizontal and inclined surfaces are smooth. The block is not attached to the spring.
+
+    1. What is the maximum distance it travels up the inclined plane? [^1]
+        $$
+        K=\frac12mv^2\\
+        U_g=mgy\\
+        mgy=\frac12mv^2\\
+        y=\frac{\frac12v^2}{g}\\
+        \sin(\theta)=\frac OH\\
+        H=\frac{O}{\sin(\theta)}\\
+        d=\frac{\frac{\frac12v^2}{g}}{\sin\theta}\\
+        F=ma=-kx\\
+        a=\frac{-kx}m\\
+        v_f^2=v_i^2+2ax\\
+        v_i=0\\
+        v_f^2=2\frac{-kx}mx\\
+        d=\frac{\frac{\frac12(2\frac{-kx^2}m)}{g}}{\sin\theta}\\
+        d=\frac{\frac{\frac{-kx^2}m}{g}}{\sin\theta}\\
+        d=\frac{{\frac{-kx^2}{mg}}}{\sin\theta}\\
+        d=\frac{-kx^2}{mg\sin\theta}
+        $$
+    
+2. Calculate the numerical value for the maximum distance it travels up the inclined plane given the following:
+        $$
+        k=40.0N/m\\
+        \theta=30.0^\circ\\
+        m=0.500kg\\
+        x_i=20.0cm
+        $$
+    
+    $$
+        d=\frac{-kx^2}{mg\sin\theta}\\
+        d=\frac{-(40.0N/m)(20.0cm)^2}{(0.500kg)(-9.81m/s^2)\sin(30.0^\circ)}\\
+        d=32.619776mm\\
+        d=32.6mm
+    $$
+    
+    
+    
+3. When the block slides down the inclined plane and collides with the spring , what is the maximum compression of the spring?
+        Due to conservation of energy, the spring will be compressed exactly as much as the initial compression ($20.0cm$).
+    
+6. A piece of wood of mass $m$ slides on the surface shown below. All parts of the surface are frictionless, except a $30.0m$ long rough flat segment in the bottom, where the coefficient of kinetic friction with the wood is $\mu_k$. The wood starts at rest a vertical distance $y_i$ above the bottom.
+
+    1. How far to the right of the left edge of the rough segment will the wood eventually come to rest? [^1]
+    $$
+        U=K\\
+        mg\Delta y=\frac12mv^2\\
+        v^2={2g\Delta y}\\
+        v_f^2=v_i^2+2a{\Delta x}\\
+        0=2g\Delta y+2a{\Delta x}\\
+        \Delta y=0-y_i
+    F=ma;\ F=\mu_kmg;\ ma=\mu_kmg;\ a=\mu_kg\\
+        0=2gy+2(\mu_kg){\Delta x}\\
+        \frac{-2g\Delta y}{2\mu_kg}=x\\
+        \frac{-\Delta y}{\mu_k}=x\\
+        \frac{y_i}{\mu_k}=x\\
+    $$
+    
+    
+    2. Calculate a numerical value for the distance to the right of the left edge of the rough segment that the wood eventually comes to rest given the following:
+        $$
+        m=2.00kg\\
+        \mu_k=0.200\\
+        y_i=4.00m
+        $$
+    
+        $$
+        \frac{y_i}{\mu_k}=x\\
+        \frac{4.00m}{0.200}=x\\
+        x=20.0m
+        $$
+        
+        
+        
+    3. How much work is done by friction by the time the wood stops?
+        $$
+        U=K=W_{friction}\\
+        \therefore\\
+        W_{friction}=mg\Delta y\\
+        W_{friction}=(2.00kg)(9.81m/s^2)(-4.00m)\\
+        W_{friction}=78.48J\\
+        \overline{\underline{|W_{friction}=78.5J|}}
+        $$
+        
+
+# Energy Graphs
+
+1. In the graph below, a cart is released from rest at point A and slides without friction along the vertical track. The cart is always in contact with the track.
+![image-20201008154437925](macdougall_skyler_week7.assets/image-20201008154437925.png)
+   
+    1. What is the kinetic energy of the cart at point A?
+        If the cart is at rest, then the kinetic energy of the cart is zero.
+    2. How is the total mechanical energy at point A related to the gravitational potential energy at point A?
+        They are the same.
+    3. Draw a dashed horizontal line representing the total mechanical energy $E$ on the graph.
+    4. Label with the letter $B$ the point at which the potential energy has its lowest value.
+    5. At point B, show by using arrows how the potential energy and the kinetic energy add to give the constant mechanical energy $E$. Make sure you identify each arrow appropriately. 
+    6. What is the value of the kinetic energy at a turning point?
+        The kinetic energy is not changing at a turning point, and either growing more positive or more negative. This is when the velocity is changing from positive to negative, or vice versa.
+    7. Identify the turning points on the graph with a letter T.
+    8. identify the point at which the cart will reach its maximum height with the letter M.
+2. A cart slides without friction along the vertical track. The total energy $E$ of the cart is indicated in the diagram. The cart does not leave the track.
+    ![image-20201008154448127](macdougall_skyler_week7.assets/image-20201008154448127.png)
+    
+    1. Is the object at rest at point A?
+        No.
+    2. Identify the point at which the object will reach it maximum height with the letter M. How does this point compare to the maximum location in Question 1?
+        The point at which the object will reach its maximum height is higher for question 2 than in question 1. This is because the object has a higher total energy in question 1, leading to more potential energy at its peak.
+3. A $1.60kg$ object in a conservative system moves along the x-axis, where the potential energy as a function of position is shown in the graph. The value for $U\ @\ x=0,2$ are shown as $9.35J$ and $4.15J$. 
+    ![image-20201008154455947](macdougall_skyler_week7.assets/image-20201008154455947.png)
+    
+    1. If the object's speed at $x=0$ is $2.30m/s$, what is its speed at $2.00m$?
+        $$
+        U_1+K_1=U_2+K_2\\
+        U_1= 9.35J;\ U_2=4.15J;\\ v_1=2.30m/s;\ K=\frac12mv^2\\
+        U_1+\frac12v_1^2=U_2+\frac12v_2^2\\
+        \frac{2(U_1-U_2)}m+v_1^2=v_2^2\\
+        \sqrt{\frac{2(U_1-U_2)}m+v_1^2}=v_2\\
+        \sqrt{\frac{2(9.35J-4.15J)}{1.60kg}+(2.30m/s)^2}=v_2\\
+        v_2=3.433 656 943 m/s\\
+        \overline{\underline{|v_2=3.43m/s|}}
+        $$
+    
+    2. What is the value for the mechanical energy in the system?
+        $$
+        Mechanical\ Energy=U+K\\
+        U=9.35J;\ K=\frac12mv^2;\ m=1.60kg;\ v=2.30m/s\\
+        U+\frac12mv^2\\
+        9.35J+\frac12(1.60kg)(2.30m/s)^2\\
+        Mechanical\ Energy=13.582J\\
+        \overline{\underline{|Mechanical\ Energy=13.6J|}}
+        $$
+        
+    3. What is the value of the potential energy at the turning points?
+    
+        At turning points, the potential energy is the same as the mechanical energy, at $13.6J$.
+    
+        
+

October/macdougall_skyler_week8.md

@@ -0,0 +1,230 @@
+# Week 8 Activities Problems
+
+## Skyler MacDougall
+
+## PHYS-111 2pm Lab
+
+### Group Members: Will Caliguri, Nina Chambliss
+
+#### Due: 10/9/2020
+
+# Law of Universal Gravitation
+
+1. At new moon, the earth, moon, and sun are in a straight line.
+    $$
+    Givens:\\
+    M_{earth}=5.67\times 10^{24}kg\\
+    M_{moon}=7.35\times10^{22}kg\\
+    M_{sun}=2.00\times10^{30}kg\\
+    d_{moon-earth}=3.84\times10^5km\\
+    AU=1.50\times10^8km
+    $$
+    
+
+    1. What is the net gravitational force exerted on the earth by the moon and the sun?
+        $$
+        F_g=G\frac{mM}{r^2}\\
+        F_{net}=F_{g1}+F_{g2}\\
+        F_g=G\frac{mM}{r^2}+G\frac{mM}{r^2}\\
+        F_g=Gm(\frac{M}{r^2}+\frac{M}{r^2})\\
+        F_g=G(5.67\times 10^{24}kg)(\frac{2.00\times10^{30}kg}{(1.50\times10^8km)^2} +\frac{7.35\times10^{22}kg} {(3.84\times10^5km)^2})\\
+        F_g=G(5.67\times 10^{24}kg)(8.938734266\times10^7 kg ∕ m^2)\\
+        F_g=3.383\times10^{22} N\\
+        \overline{\underline{|F_g=3.38\times10^{22}N\ towards\ the\ sun|}}
+        $$
+
+    2. ... on the moon by the sun and the earth?
+        $$
+        F_g=Gm(\frac{M}{r^2}-\frac{M}{r^2})\\
+        F_g=G(7.35\times10^{22}kg)(\frac{2.00\times10^{30}kg}{(1.50\times10^8km-3.84\times10^5km)^2} -\frac{5.67\times 10^{24}kg} {(3.84\times10^5km)^2})\\
+        F_g=G(5.67\times 10^{24}kg)(5.089360517\times10^7 kg ∕ m^2)\\
+        F_g=2.497\times10^{20} N\\
+        \overline{\underline{|F_g=2.497\times10^{20} N\ towards\ the\ sun|}}
+        $$
+        
+    3. ... on the sun by the moon and the earth?
+    
+    $$
+        F_g=Gm(\frac{M}{r^2}+\frac{M}{r^2})\\
+        F_g=G(2.00\times10^{30}kg)(\frac{7.35\times10^{22}kg}{(1.50\times10^8km-3.84\times10^5km)^2} +\frac{5.67\times 10^{24}kg} {(3.84\times10^5km)^2})\\
+        F_g=G(5.67\times 10^{24}kg)(3.845215172\times10^7 kg ∕ m^2)\\
+        F_g=5.133\times10^{27} N\\
+        \overline{\underline{|F_g=5.133\times10^{27} N\ towards\ the\ sun|}}
+    $$
+    
+2. When the earth, moon, and sun form a right triangle, with the moon located at the  right angle, the moon is in its third quarter phase. What is the net gravitational force on the moon?
+    $$
+    F_{g}=G\frac{mM}{r^2}+G\frac{mM}{r^2}\\
+    F_{g}=G\frac{(7.35\times10^{22}kg)(2.00\times10^{30}kg)}{(1.50\times10^8km)^2-(3.84\times10^5km)^2}\ang180^\circ+G\frac{(7.35\times10^{22}kg)(5.67\times10^{24}kg)}{(3.84\times10^5km)^2}\ang270^\circ\\
+    F_{g}=1.910534752\times10^{20}N\ang260.8078552^\circ\\
+    \overline{\underline{|F_g=1.91\times10^{20}N\ang261^\circ|}}
+    $$
+    
+4. An asteroid has a mass of $3.35\times10^{15}kg$ and a radius of $19.0km$. Assume the asteroid is a uniform sphere.
+
+    1. What is the magnitude of gravitational acceleration on the surface of the asteroid?
+        $$
+        F_g=G\frac{mM}{r^2}\\
+        a=\frac{GM}{r^2}\\
+        a=\frac{G(3.35\times10^{15}kg)}{(19.0km)^2}\\
+        \overline{\underline{|a=6.19\times10^{-4}m/s^2|}}
+        $$
+        
+    2. With what speed must you launch a rocket of mass $m$ so that it completely escapes the asteroid's gravitational pull?
+        $$
+        v_e=\sqrt{\frac{2GM}{r}}\\
+        v_e=\sqrt{\frac{2G(3.35\times10^{15}kg)}{19.0km}}\\
+        v_e=4.851m/s\\
+        \overline{\underline{|v_e=4.85m/s|}}
+        $$
+        
+
+5. The largest moon in the solar system is Ganymede, a moon of Jupiter. Ganymede orbits at a distance of $1.07\times10^9m$ from the center of Jupiter, with an orbital period of $6.18\times10^5s$. Using this information, calculate the mass of Jupiter.
+    $$
+    v_o=\sqrt{\frac{GM}{r}}\\
+    v=\frac dt=\frac{2\pi r}t=\frac{2\pi (1.07\times10^9m)}{6.18\times10^5s}\\
+    v_o=1.527949235\times10^4m/s\\
+    \frac{v_o^2r}{G}=M\\
+    M=\frac{(1.527949235\times10^4m/s)^2(1.07\times10^9m)}{G}\\
+    M=3.743\times10^{27}kg\\
+    \overline{\underline{|M=3.74\times10^{27}kg|}}
+    $$
+    
+
+# Linear Momentum
+
+1. A puck of mass $m_1=0.151kg$ slides along a frictionless horizontal surface with a velocity of $\overrightarrow{v_{1i}}=0.450m/s$ directed horizontally to the right. This puck makes a glancing collision with a second puck of mass $m_2=0.250kg$ sliding with a velocity $\overrightarrow{v_{2i}}=0.135m/s$ directed horizontally to the left. After the collision, $m_1$ is observed to be moving with a speed $\overrightarrow{v_{1vf}}=0.355m/s$ at an angle of $\theta_{1f}=17.1^\circ$.
+
+    1. What is the velocity of puck $m_2$ after the collision?
+        $$
+        p_i=p_f\\
+        m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}\\
+        v_{2f}=\frac{m_1v_{1i}+m_2v_{2i}-m_1v_{1f}}{m_2}\\
+        v_{2f}=\frac{(0.151kg)(0.450m/s\ang0^\circ)+(0.250kg)(0.135m/s\ang180^\circ)-(0.151kg)(0.355m/s\ang17.1^\circ)}{0.250kg}\\
+        v_{2f}=0.009283469707m/s\ang-137.2232125^\circ\\
+        \overline{|0.0928m/s\ang-137.^\circ|}\\\underline{|0.0928m/s\ang137.^\circ clockwise|}
+        $$
+        
+    2. Is the collision elastic? If not, calculate the fractional change in the kinetic energy ($\frac{\Delta K}{K_i}=\frac{K_f-K_i}{K_i}$) .
+        $$
+        Elastic:\Delta K=0\\
+        \Delta K=K_f-K_i\\
+        \Delta K=\frac12(m_{1}v_{1i}^2+m_{2}v_{2i}^2-(m_{1}v_{1f}^2+m_{2}v_{2f}^2))\\
+        \Delta K=\frac12((0.151kg)(0.450m/s\ang0^\circ)^2+(0.250kg)(0.135m/s\ang180^\circ)^2-...\\
+        ((0.151kg)(0.355m/s\ang17.1^\circ)^2+(0.250kg)(0.0928m/s\ang-137^\circ)^2))\\
+        \Delta K=0.01156154894J\\
+        \Delta K\ne0J\\\therefore\\inelastic\\
+        \frac{\Delta K}{K_i}=\frac{\frac12(m_{1}v_{1i}^2+m_{2}v_{2i}^2-(m_{1}v_{1f}^2+m_{2}v_{2f}^2))}{\frac12(m_{1}v_{1f}^2+m_{2}v_{2f}^2)}\\
+        \frac{\Delta K}{K_i}=\frac{0.01156154894J}{((0.151kg)(0.355m/s\ang17.1^\circ)^2+(0.250kg)(0.0928m/s\ang-137^\circ)^2))}\\
+        \frac{\Delta K}{K_i}=\frac{0.01156154894J}{0.01022235025J}\\
+        \frac{\Delta K}{K_i}=1.131006927\\
+        \overline{\underline{|\frac{\Delta K}{K_i}=1.13|}}
+        $$
+        
+    
+2. A block of mass $m_1$ is attached to a horizontal spring that is at equilibrium. The block rests on a frictionless horizontal surface. A wad of putty mass $m_2$ is thrown horizontally at the block, hitting it with a speed of $v_{2i}$ and sticking to the block.
+
+    1. Derive an expression for the maximum amount that the putty-block system compresses the spring after the collision.
+        $$
+        U_{spring}=\frac12kx^2;\ K=\frac12mv^2;\ p_i=p_f\\
+        m_2v_{2i}=m_{1+2}v_{f}\\
+        v_f=\frac{m_2v_{2i}}{m_{1+2}}\\
+        kx^2=mv^2\\
+        x^2=\frac{mv^2}{k}\\
+        x=\sqrt{\frac{m_{1+2}(\frac{m_2v_{2i}}{m_{1+2}})^2}{k}}
+        $$
+        
+2. Determine the compression distance given the following:
+        $$
+        m_1=0.430kg\\
+        m_2=0.0500kg\\
+        k=20.0N/m\\
+        v_{2i}=2.330m/s
+        $$
+    
+$$
+    x=\sqrt{\frac{m_{1+2}(\frac{m_2v_{2i}}{m_{1+2}})^2}{k}}\\
+    x=\sqrt{\frac{(0.430kg+0.500kg)(\frac{(0.500kg)(2.330m/s)}{0.430kg+0.500kg})^2}{20.0N/m}}\\
+    x=0.2701279068m\\
+    \overline{\underline{|x=0.270m|}}
+    $$
+    
+
+    
+3. Block A in the figure below has a mass $m_A=1.50kg$, and block B has a mass $m_B=3.00kg$. The blocks are forced together, compressing a spring between them. The system is released from rest on a level, frictionless surface. The spring which has negligible mass, is not attached to either block, and it drops to the surface after it has expanded and pushed on both blocks. Block B acquires a speed of $1.25m/s$. 
+
+    1. What is the final speed of block A?
+        $$
+        p_i=p_f\\
+        p_i=0\\
+        m_Av_A=-m_Bv_B\\
+        v_A=\frac{m_Bv_B}{m_A}\\
+        \frac{m_B}{m_A}=2\\
+        \overline{\underline{|v_A=2.50m/s|}}
+        $$
+        
+2. How much potential energy was stored in the compressed spring?
+        $$
+        P=K\\
+        K=\frac12(m_Av_A^2+m_Bv_B^2)\\
+        P=\frac12((1.50kg)(2.50m/s)^2+(3.00kg)(1.25m/s)^2)\\
+        P=63.28125J\\
+        \overline{\underline{|P=63.3J|}}
+        $$
+        
+    
+4. A chunk of ice of mass $m_1$ is sliding with a horizontal velocity of magnitude $v_{1i}$ on the floor of an ice-covered valley when it collides with and sticks to another chunk of ice mass $m_2$ that is at rest at the base of a hill. The two blocks stick together and move up the hill together. Since the valley and the hill are icy, there is no friction between the chunks and the ground.
+
+    1. Which conservation laws may be applied to understand the motion of the ice chunks in this problem?
+        Conservation of energy and conservation of momentum.
+
+    2. What is the maximum vertical distance $H$ that the combined ice chunks will go up the hill after the collision? [^1]
+        $$
+        p_f=p_i\\
+        m_fv_f=m_iv_{1i}\\
+        v_f=\frac{m_iv_{1i}}{m_f}\\
+        v_f=\frac{m_1v_{1i}}{m_1+m_2}\\
+        U=K\\
+        mgh=\frac12mv^2\\
+        H=\frac{(\frac{m_1v_{1i}}{m_1+m_2})^2}{2g}
+        $$
+        
+3. Calculate the numerical value $H$ given:
+        $$
+        m_1=5.00kg\\
+        m_2=7.50kg\\
+        v_{1i}=12.0m/s
+        $$
+    
+    $$
+        H=\frac{(\frac{m_1v_{1i}}{m_1+m_2})^2}{2g}\\
+        H=\frac{(\frac{5.00kg(12.0m/s)}{(5.00kg)+(7.50kg)})^2}{2(9.81m/s^2)}\\
+        H=1.174311927m\\
+        \overline{\underline{|H=1.17m|}}
+        $$
+        
+        
+        
+    4. How much kinetic energy is lost due to the collision?
+        $$
+        \Delta K=K_f-K_i;\ K_f=U_f\\
+        \Delta K=\frac12mv^2-mgh\\
+        \Delta K=\frac12m_1v_{1i}^2-m_{1+2}gH\\
+        \Delta K=\frac12(5.00kg)(12.0m/s)^2-(5.00kg+7.50kg)(9.81m/s^2)(1.17m)\\
+        \Delta K=216J\\
+        \overline{\underline{|\Delta K=216.J|}}
+        $$
+        
+    
+    
+    
+5. A Prius of mass $m_1=1.61\times10^3kg$ is traveling north with an unknown initial speed $v_{1i}$. A RAV4 of mass $m_2=1.61\times10^3kg$ is travelling with an initial velocity of $\overrightarrow{v_{2i}}=17.8m/s\ang 160^\circ$. The two vehicles collide annd stick together. Immediately after the collision, the combined vehicles are moving with a velocity of $\overrightarrow{v_f}=13.1m/s\ang133.4^\circ$. What is the initial speed of the prius before the collision?
+    $$
+    p_i=p_f\\
+    m_1v_{1i}+m_2v_{2i}=m_{1+2}v_{f}\\
+    v_{1i}=\frac{m_{1+2}v_{f}-m_2v_{2i}}{m_1}\\
+    v_{1i}=\frac{((1.38+1.61)\times10^3kg)(13.1m/s\ang133.4^\circ)-(1.61\times10^3kg)(17.8m/s\ang160^\circ)}{1.38\times10^3kg}\\
+    v_{1i}=13.51999850m/s\ang89.94724050^\circ\\
+    \overline{\underline{|v_{1i}=13.5m/s|}}
+    $$
+