Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 11 - Work - Exercises and Problems: 23

Answer

$v = 1360~m/s$

Work Step by Step

We can find the kinetic energy of one helium atom. $KE = \frac{3700~J}{6.02\times 10^{23}~atoms}$ $KE = 6.146\times 10^{-21}~J$ Using this, we can find the speed of each helium atom. $\frac{1}{2}mv^2 = KE$ $v^2 = \frac{2~KE}{m}$ $v = \sqrt{\frac{2~KE}{m}}$ $v = \sqrt{\frac{(2)(6.146\times 10^{-21}~J)}{6.68\times 10^{-27}~kg}}$ $v = 1360~m/s$
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.