University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 12 - Fluid Mechanics - Problems - Exercises - Page 390: 12.4


The height of the cube would be 10.8 cm

Work Step by Step

We can find the mass of the cube. $M = \frac{\$1,000,000}{\$1282/troy~ounce} = 780~troy~ounces$ We can convert the mass to units of kilograms. $M = (780~troy~ounces)(\frac{31.1035~g}{1.0000~troy~ounces})(\frac{1~kg}{1000~g})$ $M = 24.26~kg$ We can find the volume of the cube. $V = \frac{M}{\rho}$ $V = \frac{24.26~kg}{19.3\times 10^3~kg/m^3}$ $V = 1.257\times 10^{-3}~m^3$ We can find the length of each side of the cube. $L = (V)^{1/3}$ $L = (1.257\times 10^{-3}~m^3)^{1/3}$ $L = 0.108~m = 10.8~cm$ The height of the cube would be 10.8 cm
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