47 lines
2.1 KiB
Markdown
47 lines
2.1 KiB
Markdown
# Homework 6 - Aidan Sharpe
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## 1
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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.
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$$v_1 = 5.4 \text{[m/s]}$$
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$$h_1 = 0 \text{[m]}$$
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$$h_2 = 1.3 \text{[m]}$$
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$${1\over2}\rho v_1^2 + \rho g h_1 = {1\over2} \rho v_2^2 + \rho g h_2$$
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$${1\over2}(5.4^2) + 9.81 (0) = {1\over 2} v_2^2 + 9.81 (1.3)$$
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Solve for $v_2$ algebraically:
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$$v_2 = 1.911 \text{[m/s]}$$
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## 2
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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)
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$$P = \rho g h$$
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$$\rho = 1000 \left[{\text{kg} \over \text{m}^3}\right]$$
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$$h = 1.4 \text{[m]}$$
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$$P = 9.81(1000)(1.4) = 13720 \text{[Pa]}$$
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$$P_1 + {1\over2}\rho v_1^2 + \rho g h_1 = P_2 + {1\over2}\rho v_2^2 + \rho g h_2$$
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$${1\over2}\rho v_2^2 = P_1$$
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$$v_2 = 5.23 \text{[m/s]}$$
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## 3
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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.
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### a)
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What is the buoyancy force? Draw a Free Body Diagram and apply equations.
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$$F_b = mg = 4.9 \text{[N]}$$
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### b)
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What is the volume of water displaced by the ball? (consider it a sphere and use Archimedes principle)
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$$F_b = \rho v g$$
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$$v = {4.9 \over \rho g} = 5 \times 10^{-4} \text{[m}^3]$$
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### c)
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What is the average density of the basketball?
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$$d = {0.5 \over {4\over3}\pi r^3} = {0.5 \over {4\over3}\pi {0.11}^3} = 89.68 \left[{\text{kg} \over \text{m}^3}\right]$$
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## 4
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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?
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$$v_1 A_1 = v_2 A_2$$
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$$0.025 (\pi 0.02^2) = v_2 (\pi 0.0055^2)$$
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$$v_2 = 0.083 \text{[m/s]} = 8.3 \text{[cm/s]}$$
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