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PirateFan308
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Homework Statement
A cylindrical bucket, open at the top, is 28.0 cm high and 10.0cm in diameter. A circular hole with a cross-sectional area 1.73cm2 is cut in the center of the bottom of the bucket. Water flows into the bucket from a tube above it at the rate of 2.11×10−4 m3/s. How high will the water in the bucket go (in centimeters)?
Homework Equations
[itex]P_1+1/2\rho v_1^{2}+\rho gy_1 = P_2+1/2\rho v_2^{2}+\rho gy_2[/itex]
[itex]A_1v_1=A_2v_2[/itex]
density of water is 1000kg/m3
The Attempt at a Solution
I let 1 be the top of the water, 2 be the hose flowing into the bucket and 3 be right above the hose but in the larger bucket. I know the pressure at 3 must equal the pressure at 2.
The area of bucket is π(r)(r) = π(0.05)(0.05) = 0.00785
[itex]P_1+1/2\rho v_1^{2}+\rho gy_1 = P_2+1/2\rho v_2^{2}+\rho gy_2[/itex]
[itex]P_1+1/2\rho v_1^{2}+\rho gy_1 = P_2+1/2\rho v_2^{2}[/itex]
[itex]P_3+\rho gh+1/2\rho \frac{A_2v_2}{A_1}^{2}+\rho gy_1 = P_2+1/2\rho v_2^{2}[/itex]
[itex]1/2\rho \frac{A_2v_2}{A_1}^2+2\rho gy_1 = 1/2\rho v_2^2[/itex]
[itex]1/2 \frac{A_2v_2}{A_1}^2+2 gh = 1/2v_2^2[/itex]
[itex]2gh = 1/2v_2^2-1/2 \frac{A_2v_2}{A_1}^2[/itex]
[itex]2gh = 1/2 (2.11*10^{-4})^2 - 1/2 \frac{(0.000173)(2.11*10^{-4})}{0.00785}^2[/itex]
[itex]2gh = 1.054999*10^{-4}[/itex]
[itex]h = 5.377*10^{-6} m = 5.377*10^{-4}cm[/itex]