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: 94

Answer

$F_a=9680N$

Work Step by Step

$P_i+\frac{1}{2}\rho V_i^2+\rho gy_i=P_o+\frac{1}{2}\rho V_o^2+\rho gy_o$ $P_i-P_o=\frac{1}{2}\rho V_o^2=\frac{F_a}{A_r}$ $F_a=\frac{1}{2}\rho(V_o^2)A_r$ $F_a=\frac{1}{2}(1.29\frac{kg}{m^3})(180\frac{km}{h}\frac{1h 1000m}{3600s 1km})(6.0m^2)=9680N$
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