Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 15 - Fluids and Elasticity - Exercises and Problems: 8

Answer

The pressure at the bottom of the water layer is $1.17\times 10^5~N/m^2$

Work Step by Step

$P = P_0 + \rho~g~h$ $P$ is the pressure $P_0$ is the atmospheric pressure $\rho$ is the density of the liquid $h$ is the depth below the surface We can find the pressure $P_1$ at the bottom of the oil layer. $P_1 = P_0 + \rho_o~g~h_1$ $P_1 = (1.013\times 10^5~N/m^2) + (900~kg/m^3)(9.80~m/s^2)(0.50~m)$ $P_1 = 1.057\times 10^5~N/m^2$ We can find the pressure $P_2$ at the bottom of the water layer. $P_2 = P_1 + \rho_w~g~h_2$ $P_2 = (1.057\times 10^5~N/m^2) + (1000~kg/m^3)(9.80~m/s^2)(1.20~m)$ $P_2 = 1.17\times 10^5~N/m^2$ The pressure at the bottom of the water layer is $1.17\times 10^5~N/m^2$
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