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: 6


The mass of the liquid is 2430 kg.

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 first find the density of the liquid. $P = P_0 + \rho~g~h$ $\rho~g~h = P- P_0$ $\rho = \frac{P- P_0}{g~h}$ $\rho = \frac{1.3~atm- 1.0~atm}{(9.80~m/s^2)(2.0~m)}$ $\rho = \frac{(0.3~atm)(\frac{1.01\times 10^5~N/m^2}{1~atm})}{(9.80~m/s^2)(2.0~m)}$ $\rho = 1546~kg/m^3$ We then find the volume of the vat. $V = \pi~r^2~h$ $V = \pi~(0.5~m)^2(2.0~m)$ $V = 1.57~m^3$ We then find the mass of the liquid. $M = \rho~V$ $M = (1546~kg/m^3)(1.57~m^3)$ $M = 2430~kg$ The mass of the liquid is 2430 kg.
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