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

Answer

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.