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

Published by Pearson

# Chapter 15 - Fluids and Elasticity - Exercises and Problems - Page 437: 39

#### Answer

(a) The pressure at point A is $1.057\times 10^5~N/m^2$ (b) The pressure difference between point A and point B is $4400~N/m^2$ The pressure difference between point A and point C is $4400~N/m^2$

#### 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 oil $h$ is the depth below the surface We can find the pressure $P_A$ at point A. $P_A = P_0 + \rho~g~h_A$ $P_A = (1.013\times 10^5~N/m^2) + (900~kg/m^3)(9.80~m/s^2)(0.50~m)$ $P_A = 1.057\times 10^5~N/m^2$ The pressure at point A is $1.057\times 10^5~N/m^2$. (b) The pressure difference between point A and point B is $\rho~g~\Delta h$. $\Delta P = \rho~g~\Delta h$ $\Delta P = (900~kg/m^3)(9.80~m/s^2)(0.50~m)$ $\Delta P = 4400~N/m^2$ The pressure difference between point A and point B is $4400~N/m^2$ We would find the pressure difference between point A and point C in exactly the same way, so the pressure difference between point A and point C is also $4400~N/m^2$.

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