Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 14 - Fluids and Elasticity - Exercises and Problems - Page 383: 4

Answer

The mass of the water in the pool is $1.44\times 10^5~m^3$

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: $M = \rho~V$ $M = (1.00\times 10^3~kg/m^3)(144.0~m^3)$ $M = 1.44\times 10^5~m^3$ The mass of the water in the pool is $1.44\times 10^5~m^3$.
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