University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 12 - Fluid Mechanics - Problems - Exercises - Page 392: 12.41



Work Step by Step

Using Bernoulli's equation, we can find the velocity of the water escaping through the small hole at the bottom of the tank: $P_1+\rho g y_1+{1 \over 2}\rho v_1^2=P_2+\rho g y_2+{1 \over 2}\rho v_2^2$ $y_1=11~m$ $y_2=0$ $\rho=1030~kg/m^3$ $v_1=0$ $P_1=3(1.013\times 10^5~Pa)=3.039\times 10^5~Pa$ $P_2=0$ $(3.039\times 10^5~Pa)+(1030~kg/m^3)(9.8~m/s^2)(11~m)={1 \over 2}(1030~kg/m^3)v_2^2$ $v_2=28.38~m/s$
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