Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 10 - Fluids - General Problems - Page 290: 89

Answer

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Work Step by Step

a. The water starts and ends at ground level. Use the projectile motion range equation. $$R=\frac{v_0^2sin 2 \theta}{g}$$ $$v_0=\sqrt{\frac{Rg}{sin 2 \theta}}=\sqrt{\frac{(6.0m)(9.80m/s^2)}{sin 70^{\circ}}}=7.9m/s$$ b. The volume rate of flow is the area of the flow, multiplied by the speed of the flow. There are four sprinkler heads. $$Flow=Av=4\pi r^2v=4 \pi (0.0015m)^2(7.910m/s)$$ $$=2.237\times10^{-4}m^3/s=0.22\;liters/s$$ c. Apply the equation of continuity. $$(Av)_{supply}=(Av)_{heads}$$ Calculate the flow speed in the pipe. $$v_{supply}=\frac{(Av)_{heads}}{A_{supply}}$$ $$v_{supply}=\frac{2.237\times10^{-4}m^3/s }{\pi (0.0095m)^2}=0.79m/s$$
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