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: 12.3

Answer

The average density of this piece is $7.02\times 10^3~kg/m^3$. Since the actual density of gold is $19.3\times 10^3~kg/m^3$, it seems clear that we were cheated.

Work Step by Step

We can find the volume of the metal. $V = (5.0\times 10^{-3}~m)(15.0\times 10^{-3}~m)(30.0\times 10^{-3}~m)$ $V = 2.25\times 10^{-6}~m^3$ We can find the average density of the metal. $\rho = \frac{M}{V}$ $\rho = \frac{0.0158~kg}{2.25\times 10^{-6}~m^3}$ $\rho = 7.02\times 10^3~kg/m^3$ The average density of this piece is $7.02\times 10^3~kg/m^3$. Since the actual density of gold is $19.3\times 10^3~kg/m^3$, it seems clear that we were cheated.
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