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 18 - A Macroscopic Description of Matter - Exercises and Problems - Page 511: 3


The diameter of the sphere is 8.4 cm

Work Step by Step

We can find the mass of the aluminum as: $m = \rho~V$ $m = (2.70\times 10^3~kg/m^3)(1000~cm^3)$ $m = (2.70\times 10^3~kg/m^3)(1000\times 10^{-6}~m^3)$ $m = 2.70~kg$ We then find the volume of copper that has this mass; $V = \frac{m}{\rho}$ $V = \frac{2.70~kg}{8.9\times 10^3~kg/m^3}$ $V = 3.034\times 10^{-4}~m^3$ We can find the radius of the copper sphere as: $\frac{4}{3}\pi~R^3 = V$ $R^3 = \frac{3V}{4\pi}$ $R = (\frac{3V}{4\pi})^{(1/3)}$ $R = [\frac{(3)(3.034\times 10^{-4}~m^3)}{4\pi}]^{(1/3)}$ $R = 0.042~m = 4.2~cm$ Since the diameter of the sphere is $2R$, the diameter of the sphere is 8.4 cm.
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