Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 15 - Exercises and Problems - Page 290: 53

Answer

a) $49 \ kg$ b) $2.5 \times 10^3 \ kg$

Work Step by Step

We know the result from the last problem: $m_g = \frac{\rho_g M}{\rho_a-\rho_g}$ Thus, we find: a) $m = \frac{(280)(.18)}{1.22-.18}=49 \ kg$ b) $m = \frac{(280)(.9\rho_a)}{\rho_a-.9\rho_a}=2.5 \times 10^3 \ kg$
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