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 18 - Exercises and Problems - Page 342: 40

Answer

$-6689 \ J$

Work Step by Step

We know the following equation for the work it will take: $ W_2=\frac{ln(c_2)}{ln(c_1)}(W_1)$ Where $c_1$ is the fraction that the volume is initially changed by and where $c_2$ is the factor by which the volume will be changed by. Thus, we find: $ W_2=\frac{ln(\frac{1}{22})}{ln(\frac{1}{2})}(-1500)=-6689 \ J$
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.