5th semester files

5th-Semester-Fall-2023/ME-For-ECEs/Homework/Homework-01.md

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+# Aidan Sharpe - Homework 1
+
+1. $\vec{F}$ has magnitude $800$N and makes an angle of $35^\circ$ with the y-axis in the second quadrant.
+    
+    $\vec{F}_x = \lVert \vec{F} \rVert \sin(-35^\circ) = 800 \times -0.5736 = -458.861$
+
+    $\vec{F}_y = \lVert \vec{F} \rVert \cos(-35^\circ) = 800 \times 0.8192 = 655.322$
+
+    $\vec{F} = -458.861\hat{i} + 655.322\hat{j}$
+
+1. The force, $\vec{F}$, has magnitude $6.6$kN and slope is $-\frac{5}{12}$.
+    
+    $\theta = \arctan \left(-\frac{5}{12} \right) = -0.3948 = -22.620^\circ$
+
+    $\vec{F}_x = 6600\cos(-22.620^\circ) = 6600 \times 0.9231 = 6092.308$
+
+    $\vec{F}_y = 6600\sin(-22.620^\circ) = 6600 \times -0.3846 = -2538.462$
+
+    $\vec{F} = 6092.308\hat{i} - 2538.462\hat{j}$
+
+1. $F_1 = 500$N and $F_2 = 350$-N. $F_1$ is in the direction of the x-axis, and $F_2$ makes an angle $60^\circ$ with the x-axis.
+
+    $\vec{R} = \vec{F}_1 + \vec{F}_2$
+
+    $\vec{F}_1 = 500\hat{i} + 0\hat{j}$
+    
+    $\vec{F}_2 = 350\cos(60^\circ)\hat{i} + 350\sin(60^\circ)\hat{j} = 175\hat{i} + 303.109\hat{j}$
+
+    $\vec{R} = 675\hat{i} + 303.109\hat{j}$
+
+    $\theta_R = \arctan(\frac{303.109}{675}) = 0.4221 = 24.182^\circ$
+
+1. $F_y = 70$lbs. The slope of $\vec{F}$ is $\frac{12}{5}$.
+    
+    $\frac{70}{F_x} = \frac{12}{5}$
+
+    $70 \times 5 = 12 F_x$
+
+    $F_x = \frac{70 \times 5}{12} = 29.166$
+
+    $\vec{F} = F_x\hat{i} + F_y\hat{j}$
+
+    $\lVert \vec{F} \rVert = \sqrt{F_x^2 + F_y^2} = \sqrt{29.166^2 + 70^2} = 75.833$N
+
+1. $\vec{F}_1$ has magnitude $2$kN and makes an angle $30^\circ$ with the x-axis. $\vec{F}_2$ has magnitude $3$kN and has slope $-\frac{4}{3}$.
+
+    $\vec{F}_1 = 2000\cos(30^\circ)\hat{i} + 2000\sin(30^\circ)\hat{j} = 1732.051\hat{i} + 1000\hat{j}$
+
+    $\theta_{F_2} = \arctan(-\frac{4}{3}) = -0.9273 = -53.130^\circ$
+
+    $\vec{F}_2 = 3000\cos(\theta_{F_2})\hat{i} + 3000\sin(\theta_{F_2})\hat{j} = 1800\hat{i} - 2400\hat{j}$
+
+    $\vec{R} = \vec{F}_1 + \vec{F}_2 = (1732.051 + 1800)\hat{i} + (1000 - 2400)\hat{j} = 3532.051\hat{i} - 1400\hat{j}$
+
+    $\theta = \arctan(-\frac{1400}{3532.051}) = -0.3774 = -21.622^\circ$
+
+    $\lVert \vec{R} \rVert = \sqrt{3532.051^2 + (-1400)^2} = 3799.393$

5th-Semester-Fall-2023/ME-For-ECEs/Homework/Homework-02.md

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+# Aidan Sharpe - Homework 2
+
+## 1.
+a) Point $O$ exists at the origin, and point $B$ exists at $(16, 0)$. A $5$kN force, $\vec{F}$, makes an angle of $\frac{5\pi}{6}$ with the $\hat{x}$ direction at $(12, -15)$.
+
+### a) Find the moment, $M_O$, about point $O$.
+
+$$M_O = \vec{r} \times \vec{F}$$
+$$\vec{r} = (12 - 0)\hat{x} + (-15 - 0)\hat{y} = 12\hat{x} - 15\hat{y}$$
+$$\vec{F} = 5000\cos\left(\frac{5\pi}{6}\right)\hat{x} + 5000\sin\left(\frac{5\pi}{6}\right)\hat{y} = -4330.13\hat{x} + 2500\hat{y}$$
+$$\vec{r} \times \vec{F} = 
+\begin{vmatrix} 
+\hat{x} & \hat{y} & \hat{z} \\
+12 & -15 & 0 \\
+-4330.13 & 2500 & 0 \\
+\end{vmatrix} = -34951.91\hat{z}$$
+$$M_O = -34951.91\hat{z}\text{[N m]}$$
+
+### b) Find the moment, $M_B$, about point $B$
+
+$$M_B = \vec{r} \times \vec{F}$$
+$$\vec{r} = (12 - 16)\hat{x} + (-15 - 0)\hat{y} = -4\hat{x} - 15\hat{y}$$
+$$\vec{F} = 5000\cos\left(\frac{5\pi}{6}\right)\hat{x} + 5000\sin\left(\frac{5\pi}{6}\right)\hat{y} = -4330.13\hat{x} + 2500\hat{y}$$
+$$\vec{r} \times \vec{F} = 
+\begin{vmatrix} 
+\hat{x} & \hat{y} & \hat{z} \\
+-4 & -15 & 0 \\
+-4330.13 & 2500 & 0 \\
+\end{vmatrix} = -74951.91\hat{z}$$
+$$M_B = -74951.91\hat{z}\text{[N m]}$$
+
+## 2
+A $250$N force is applied at an angle $\frac{5\pi}{12}$ with respect to the $\hat{x}$ direction at a distance $30$mm above and $200$mm to the right of the center of a bolt. Find the moment, $M_B$, about the center of the bolt.
+
+$$M_B = \vec{r} \times \vec{F}$$
+Convert distance to meters:
+$$\vec{r} = 0.2\hat{x} + 0.03\hat{y}$$
+$$\vec{F} = 250\cos\left(\frac{5\pi}{12}\right)\hat{x} + 250\sin\left(\frac{5\pi}{12}\right)\hat{y} = 64.70\hat{x} + 241.48\hat{y}$$
+$$\vec{r} \times \vec{F} =
+\begin{vmatrix} 
+\hat{x} & \hat{y} & \hat{z} \\
+0.2 & 0.03 & 0 \\
+64.70 & 241.48 & 0 \\
+\end{vmatrix} = 46.36\hat{z}$$
+$$M_B = 46.36\hat{z}\text{[N m]}$$
+
+## 3
+Consider a bar made up of two segments, $\overline{AB}$ and $\overline{BC}$, each with length, $1.6$m. Segment $\overline{AB}$ makes an angle $\frac{\pi}{2}$ with the $\hat{x}$ direction, and segment $\overline{BC}$ makes an angle $\frac{3\pi}{4}$ with the $\hat{x}$ direction. A $30$N force, $\vec{P}$, is applied perpendicular to $\overline{BC}$. Determine the moment, $M_B$, about point $B$, and $M_A$ about point $A$.
+
+### a) Find the moment, $M_B$, about point $B$
+$$M_B = \vec{r} \times \vec{F}$$
+$$\vec{r} = 1.6\cos\left(\frac{3\pi}{4}\right)\hat{x} + 1.6\sin\left(\frac{3\pi}{4}\right)\hat{y} =-1.13\hat{x} + 1.13\hat{y}$$
+$$\vec{F} = 30\cos\left(\frac{\pi}{4}\right)\hat{x} + 30\sin\left(\frac{\pi}{4}\right)\hat{y} = 21.21\hat{x} + 21.21\hat{y}$$
+$$\vec{r} \times \vec{F} =
+\begin{vmatrix} 
+\hat{x} & \hat{y} & \hat{z} \\
+-1.13 & 1.13 & 0 \\
+21.21 & 21.21 & 0 \\
+\end{vmatrix} = -47.93\hat{z}$$
+$$M_B = -47.93\hat{z}\text{[N m]}$$
+
+### b) Find the moment, $M_A$, about point $A$
+$$M_A = \vec{r} \times \vec{F}$$
+$$\vec{r} = 1.6\cos\left(\frac{3\pi}{4}\right)\hat{x} + \left[1.6 + 1.6\sin\left(\frac{3\pi}{4}\right)\right]\hat{y} = -1.13\hat{x} + 2.73\hat{y}$$
+$$\vec{F} = 30\cos\left(\frac{\pi}{4}\right)\hat{x} + 30\sin\left(\frac{\pi}{4}\right)\hat{y} = 21.21\hat{x} + 21.21\hat{y}$$
+$$\vec{r} \times \vec{F} =
+\begin{vmatrix} 
+\hat{x} & \hat{y} & \hat{z} \\
+-1.13 & 2.73 & 0 \\
+21.21 & 21.21 & 0 \\
+\end{vmatrix} = -81.87\hat{z}$$
+$$M_A = -81.87\hat{z}\text{[N m]}$$
+
+## 4
+Consider an arm holding a ball and the moments acting on the elbow. The forearm makes an angle, $-35^\circ$ with the $\hat{x}$ direction. The ball weighs 8lbs, and its center of gravity is a distance of $13$ inches away in the $\hat{x}$ direction. The forearm weighs $5$lbs, and its center of gravity is $6$ inches away along the forearm. A third, balancing force of tension acts $2$ inches down the forearm. Find the force of tension, $\vec{T}$, such that the moment about the elbow, $M_O$, is $0$.
+
+$$M_O = (\vec{r}_T \times \vec{T}) + (\vec{r}_G \times \vec{G}) + (\vec{r}_A \times \vec{A}) = 0$$
+Calculate radial vectors:
+$$\vec{r}_G = 6\cos(-35^\circ)\hat{x} + 6\sin(-35^\circ)\hat{y} = 4.91\hat{x} - 3.44\hat{y}$$
+$$\vec{r}_A = 13\hat{x} + 13\tan(-35^\circ)\hat{y} = 13\hat{x} - 9.10\hat{y}$$
+$$\vec{r}_T = 2\hat{x} + 2\tan(-35^\circ)\hat{y} = 2\hat{x} - 1.4\hat{y}$$
+
+Assign force vectors:
+$$\vec{G} = -5\hat{y}$$
+$$\vec{A} = -8\hat{y}$$
+
+Evaluate known cross products:
+$$\vec{r}_G \times \vec{G} = 
+\begin{vmatrix}
+\hat{x} & \hat{y} & \hat{z} \\
+4.91 & -3.44 & 0 \\
+0 & -5 & 0
+\end{vmatrix} = -24.55\hat{z}$$
+$$\vec{r}_A \times \vec{A} =
+\begin{vmatrix}
+\hat{x} & \hat{y} & \hat{z} \\
+13 & -9.10 & 0 \\
+0 & -8 & 0
+\end{vmatrix} = -104\hat{z}$$
+
+Combine all knowns:
+$$0 = (\vec{r}_T \times \vec{T}) - 24.55\hat{z} - 104\hat{z}$$
+$$\therefore \vec{r}_T \times \vec{T} = 128.55\hat{z}$$
+
+Since $\vec{T}$ only acts in the $\hat{y}$ direction:
+$$128.55 = r_{T_x} T_y - r_{T_y} T_x = (2)T_x-(-1.4)(0)$$
+$$\therefore 128.55 = (2)T_y$$
+$$\therefore T_y = 64.28$$
+$$\vec{T} = 64.28\hat{y} \text{[lb in]}$$

5th-Semester-Fall-2023/ME-For-ECEs/Homework/Homework-03.md

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+# Homework 3 - Aidan Sharpe
+
+## 1
+A 50[kg] homogeneous smooth sphere rests on a $30^\circ$ incline and against a vertical wall.
+
+Since the sphere is in static equilibrium:
+$$\sum \vec{F} = 0$$
+$$\therefore \sum \vec{F}_x = \sum \vec{F}_y = 0$$
+$$\sum \vec{F}_y = \vec{F}_g + \vec{F}_{A_y}$$
+$$\sum \vec{F}_x = \vec{F}_B + \vec{F}_{A_x}$$
+Since $\vec{F}_A$ is normal to the surface at $30^\circ$ to the $-\hat{x}$ direction:
+$$\vec{F}_{A_x} = F_A \cos(60^\circ)\hat{x}$$
+$$\vec{F}_{A_y} = F_A \sin(60^\circ)\hat{y}$$
+
+Find the force due to gravity:
+$$\vec{F}_g = (50)(-9.8)\hat{y} = -490\hat{y}$$
+
+Solve for $\|\vec{F}_A\|$:
+$$-490 + F_A \sin(60^\circ) = 0$$
+$$\therefore \|\vec{F}_A\| = {490 \over \sin(60^\circ)} = 565.8 \text{[N]}$$
+$$\vec{F}_B + 565.8\cos(60^\circ) = 0$$
+$$\therefore \vec{F}_B = -282.9\hat{x} \text{[N]}$$
+
+Since $F_{A_x} = -F_B$ and $F_{A_y} = -F_g$:
+$$\vec{F}_A = 282.9\hat{x} + 490\hat{y} \text{[N]}$$
+
+## 2
+A uniform 150[kg], 15[m] long pole is supported by vertical walls spaced 12[m] apart at points $A$ and $B$. A vertical tension force is applied 5[m] from point $A$ (10[m] from point $B$). Find the reactions at $A$ and $B$.
+
+#### Assumptions:
+The moment, $M_g$, acts at the center of mass. The pole pivots around the point where the cable is attached. The forces at points $A$ and $B$ act strictly in the horizontal direction. There is no friction between the pole and the walls.
+
+#### Find $F_A$ and $F_B$:
+$$\sum \vec{M} = 0 = \vec{M}_g + \vec{M}_A + \vec{M}_B$$
+Find the angle that the pole makes:
+$$\theta = \arccos\left({12 \over 15}\right) = 0.6435$$
+
+Find the about the cable due to gravity, $M_g$:
+$$\vec{M}_g = \vec{r}_g \times \vec{F}_g$$
+
+Since $\vec{r}_g$ is 2.5[m] along the bar from the point of tension:
+$$\vec{r}_g = 2.5 \cos(\theta)\hat{x} + 2.5 \sin(\theta)\hat{y} = 2\hat{x} + 1.5\hat{y}$$
+$$\vec{F}_g = mg = (150)(-9.8)\hat{y} = -1470\hat{y}$$
+
+Plug in and evaluate $\vec{M}_g$:
+$$\vec{M}_g =
+\begin{vmatrix}
+\hat{x} & \hat{y} & \hat{z} \\
+2 & 1.5 & 0 \\ 
+0 & -1470 & 0 
+\end{vmatrix} = -2940\hat{z}$$
+
+Find the moment about the cable attachment due to the support from point $A$, $\vec{M}_A$:
+$$\vec{M}_A = \vec{r}_A \times \vec{F}_A$$
+
+Since $\vec{r}_A$ is -5[m] along the pole from the point of tension:
+$$\vec{r}_A = -5\cos(\theta)\hat{x} - 5\sin(\theta)\hat{y} = -4\hat{x} - 3\hat{y}$$
+
+Since $\vec{F}_A$ is unknown, but known to only act in the $\hat{x}$ direction:
+$$\vec{M}_A =
+\begin{vmatrix}
+\hat{x} & \hat{y} & \hat{z} \\
+-4 & -3 & 0 \\
+F_A & 0 & 0
+\end{vmatrix} = 3\vec{F}_A \hat{z}$$
+
+Find the moment about the cable attachment due to the suport from point $B$, $\vec{M}_B$:
+$$\vec{M}_B = \vec{r}_B \times \vec{F}_B$$
+
+Since $\vec{r}_B$ is 10[m] along the pole from the point of tension:
+$$\vec{r}_B = 10\cos(\theta)\hat{x} + 10\sin(\theta)\hat{y} = 8\hat{x} + 6\hat{y}$$
+
+Since $\vec{F}_B$ is unknown, but known to only act in the $\hat{x}$ direction:
+$$\vec{M}_B =
+\begin{vmatrix}
+\hat{x} & \hat{y} & \hat{z} \\
+8 & 6 & 0 \\
+F_B & 0 & 0
+\end{vmatrix} = -6F_B \hat{z}$$
+
+To find the values $F_A$ and $F_B$:
+$$\sum \vec{M} = 0 = \vec{M}_A + \vec{M}_B + \vec{M}_g$$
+$$\therefore -2940 + 3F_A - 6F_B = 0$$
+$$\sum \vec{F}_x = 0 =\vec{F}_A + \vec{F}_B$$
+$$\therefore\vec{F}_A = -\vec{F}_B$$
+$$\therefore -2940 = 9\vec{F}_B$$
+$$\therefore \vec{F}_B = -326.6\hat{x}$$
+$$\therefore \vec{F}_A = 326.6\hat{x}$$
+
+## 4
+If the car accelerates at 2.75[$\text{m}/\text{s}^2$] for 3[m] and then maintains speed for 4[m], find the time it takes to travel the entire distance.
+
+$$v^2 = 2a\Delta p = 2(2.75)(3) = 16.5 \text{[m/s]}$$
+$$v = at$$
+$$t_\text{decline} = {16.5 \over 2.75} = 6\text{[s]}$$
+$$\Delta p = vt$$
+$$t_\text{coasting} =  {4 \over 16.5} = 0.242\text{[s]}$$
+$$t = t_\text{decline} + t_\text{coasting} = 6.242\text{[s]}$$
+
+## 5
+A ball is thrown upwards with an initial velocity of 30[m/s] at the edge of a 60[m] high cliff. Find the maximum height above the ground, $h$, and the total time, $t$, before the ball hits the ground.
+$$0^2 = (30)^2 + 2(-9.8)\Delta h$$
+$${900 \over (2)(9.8)} = \Delta h = 45.918 \text{[m]}$$
+$$h = 60 + \Delta h = 105.918 \text{[m]}$$
+$$\Delta h_\text{final} = 30 + {1 \over 2}(-9.8)t^2$$
+$$-60 = 30 - 4.9t^2$$
+$$\therefore t = +\sqrt{{-90 \over -4.9}} = 4.2857 \text{[s]}$$

5th-Semester-Fall-2023/ME-For-ECEs/Homework/Homework-04.md

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+# Homework 4 - Aidan Sharpe
+
+## 1
+The wall of a house is 7[m] wide, 6[m] high, and 0.3[m] thick with $k=0.6$[W/($m \cdot K$)]. The surface temperature on the inside of the wall is 16$^\circ$C and the temperature on the outside is 6$^\circ$. Find the heat flux through the wall and the total heat loss through it.
+
+Heat flux is given by
+$$q''_x = -k{dT \over dx}$$
+$\Delta T = 10$, $k = 0.6$, $\Delta x$ = 0.3
+$$q''_x= -0.6 {10 \over 0.3} = -20\text{[W/m$^2$]}$$
+
+Total heat loss is the product of heat flux and the area:
+$$-20 (6 \times 7) = -840\text{[W]}$$
+
+## 2
+A 20[mm] diameter copper pipe is used to carry heated water. The external surface of the pipe is subjected to a convective heat transfer coefficient of $h = 6$[W/(m$^2 \cdot$ K)]. Find the heat loss by convection per meter length of the pipe when the external surface temperature is 80$^\circ$C and the surroundings are at 20$^\circ$C. Assuming black body radiation, what is the heat loss by radiation?
+
+$$\dot{Q} = hA(T-T_f)$$
+$h=6$, $A=2\pi 0.02 l$, $T = 80$, $T_f = 20$
+$$\dot{Q} = 6 (2 \pi 0.02 x) (80 - 20) = 45.2389x$$
+Heat loss due to convection is $45.2389$[W/m].
+
+$$q'' = \sigma T_s^4 = 5.67 \times 10^{-8} (80 + 273.15) = 2 \times 10^{-5} \text{[W/m$^2$]}$$
+
+The total heat loss is the product of surface area and thermal flux:
+$$2\times 10^{-5} (2\pi 0.02x) = 2.516 x \text{[W]}$$
+
+## 3
+A plate 0.3[m] long and 0.1[m] wide, with a thickness of 12[mm] is made from stainless steel ($k = 16$[W/(m $\cdot$ K)]), the top surface is exposed to a 20$^\circ$C airstream. In an experiment, the plate is heated by an electrical heater (also 0.3[m] by 0.1[m]) positioned on the underside of the plate and the temperature of the plate adjacent to the heater is maintained at 100$^\circ$C. A voltmeter and ammeter are connected to the heater and these read 200[V] and 0.25[A] respectively. Assuming that the plate is perfectly insulated on all sides except the top surface, what s the convective heat transfer coefficient?
+
+Heat transfer from heater to plate:
+$$P = I V = 200 \times 0.25 = 50\text{W}]$$
+$$q'' = {50 \over 0.3 \times 0.1} = 1666.667 \text{[W/m$^2$]}$$
+
+Heat transfer from the plate to the air:
+$$q'' = 1666.667 = {d\dot{Q} \over dA}$$
+$$\dot{Q} = h A(T_\text{plate} - T_\text{air})$$
+$$h = {q'' \over T_\text{plate} - T_\text{air}} = {1666.667 \over 100 - 20} = 20.833\text{[W/(m$^2 \cdot$K)]}$$

5th-Semester-Fall-2023/ME-For-ECEs/Homework/Homework-05.md

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+# Homework 5 - Aidan Sharpe
+
+## 1
+If the specific weight, $\gamma$, of a substance is 8.2[kN/m$^3$], what is its mass density, $\rho$, in kg/m$^3$?
+
+On Earth, 9810[kg] weighs 1[kN], so an object that weighs 8.2[kN] will have a mass of 80442[kg]. Therefore, $\rho = 80442$[kg/m$^3$].
+
+## 2
+A fluid flowing between two parallel plates has a viscosity, $\mu = 0.62$[Ns/m$^2$], and density, $\rho = 1250$[kg/m^3]. Calculate the intensity of shear stress, $\tau$, in pascals at $y = 3$[cm], assuming a straight-line viscosity distribution, given that the top plate has a velocity of 100[cm/s] and the fluid is 6[cm] thick.
+
+Velocity at $y = 3$[cm]:
+$$\tau = \mu{du \over dy} = 0.62\left(0.03 \times {1 \over 0.06}\right) = 0.31\text{[Pa]}$$
+
+## 3
+A cube with side length 10[cm] is placed into two different liquids. In the first liquid, the top of the cube is $h_1 = 1$[cm] above the surface. In the second liquid, the top of the cube is $h_2$ above the surface. If the densities of the liquids are known to be $\rho_1 = 1000$[kg/m$^3$], and $\rho_2 = 1300$[kg/m$^3$] respectively, find $h_2$.
+
+Displaced mass for liquid 1, same as mass of cube:
+$$0.1 \times 0.1 \times (0.1 - 0.01) \times 1000 = 0.9\text{[kg]}$$
+
+Displaced mass for liquid 2:
+$$0.1 \times 0.1 \times (0.1 - h_2) \times 1300 = 0.9$$
+$$\therefore h_2 = 0.031\text{[m]} = 3.1\text{[cm]}$$

5th-Semester-Fall-2023/ME-For-ECEs/Homework/Homework-06.md

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+# Homework 6 - Aidan Sharpe
+
+## 1
+A hose lying on the ground has water coming out of it at a speed of 5.4 meters/sec. You lift the nozzle of the hose to a height of 1.3 meters above the ground. At what speed does the water now come out of the hose? (Note: Atmospheric pressure acts on the fluid at both points). List any assumptions.
+
+$$v_1 = 5.4 \text{[m/s]}$$
+$$h_1 = 0 \text{[m]}$$
+$$h_2 = 1.3 \text{[m]}$$
+$${1\over2}\rho v_1^2 + \rho g h_1 = {1\over2} \rho v_2^2 + \rho g h_2$$
+$${1\over2}(5.4^2) + 9.81 (0) = {1\over 2} v_2^2 + 9.81 (1.3)$$
+Solve for $v_2$ algebraically:
+$$v_2 = 1.911 \text{[m/s]}$$
+
+## 2
+A dam is holding back the water in a lake. There is a small hole 1.4 meters below the surface of the lake. At what speed does water exit the hole? List any assumptions. (Note: if the hole is small, what is a safe assumption for the particles at the top of the lake)
+$$P = \rho g h$$
+$$\rho = 1000 \left[{\text{kg} \over \text{m}^3}\right]$$
+$$h = 1.4 \text{[m]}$$
+$$P = 9.81(1000)(1.4) = 13720 \text{[Pa]}$$
+$$P_1 + {1\over2}\rho v_1^2 + \rho g h_1 = P_2 + {1\over2}\rho v_2^2 + \rho g h_2$$
+$${1\over2}\rho v_2^2 = P_1$$
+$$v_2 = 5.23 \text{[m/s]}$$
+
+## 3
+A basketball is floating in a bathtub full of water. The basketball has a mass of 0.5 kg and a diameter of 22 cm.
+
+### a)
+What is the buoyancy force? Draw a Free Body Diagram and apply equations.
+$$F_b = mg = 4.9 \text{[N]}$$
+
+### b)
+What is the volume of water displaced by the ball? (consider it a sphere and use Archimedes principle)
+
+$$F_b = \rho v g$$
+$$v = {4.9 \over \rho g} = 5 \times 10^{-4} \text{[m}^3]$$
+
+### c)
+What is the average density of the basketball?
+$$d = {0.5 \over {4\over3}\pi r^3} = {0.5 \over {4\over3}\pi {0.11}^3} = 89.68 \left[{\text{kg} \over \text{m}^3}\right]$$
+
+## 4
+You need to extend a 2 cm diameter pipe, but you have only a 1.10 cm diameter pipe on hand. You make a fitting to connect these pipes end to end. If the water is flowing at 2.50 cm/sec in the wide pipe, how fast will it be flowing through the narrow one?
+
+$$v_1 A_1 = v_2 A_2$$
+$$0.025 (\pi 0.02^2) = v_2 (\pi 0.0055^2)$$
+$$v_2 = 0.083 \text{[m/s]} = 8.3 \text{[cm/s]}$$