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: 52

Answer

The proof is below.

Work Step by Step

We know that when the balloon is floating, it is necessary for the buoyancy force to at least equal to the force of gravity on the balloon (and the air in it). Thus, we find: $ \rho_a Vg = (m_g+M)g \\ \rho_a V = m_g + M$ We know that the volume is equal to $\frac{m_g}{\rho_g}$. This gives: $\rho_a \frac{m_g}{\rho_g}= m_g + M$ Simplifying gives: $m_g = \frac{\rho_g M}{\rho_a-\rho_g}$
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.