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 - Page 436: 5


The pressure 11 km below sea level is $1.10\times 10^3~atm$

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 11 km below sea level. $P = P_0 + \rho~g~h$ $P = (1.01\times 10^5~Pa) + (1030~kg/m^3)(9.80~m/s^2)(11,000~m)$ $P = 1.11\times 10^8~Pa$ We can convert this pressure to units of atmospheres. $P = (1.11\times 10^8~Pa)(\frac{1~atm}{1.01\times 10^5~Pa})$ $P = 1.10\times 10^3~atm$ The pressure 11 km below sea level is $1.10\times 10^3~atm$.
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