## 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 435: 3

#### Answer

The mass of the water in the pool is $1.44\times 10^5~kg$

#### Work Step by Step

Note that the average depth of the water in the pool is 2.0 meters. We can find the volume of the water in the pool. $V = (6.0~m)(12.0~m)(2.0~m)$ $V = 144.0~m^3$ We can find the mass of the water in the pool. $M = \rho~V$ $M = (1.00\times 10^3~kg/m^3)(144.0~m^3)$ $M = 1.44\times 10^5~kg$ The mass of the water in the pool is $1.44\times 10^5~kg$.

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