College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 10 - Problems - Page 397: 4


The wire stretches a distance of $2.2~cm$

Work Step by Step

(a) $Y = \frac{F/A}{\Delta L/L}$ $Y$ is Young's modulus $F$ is the force $A$ is the cross-sectional area $\Delta L$ is the change in length $L$ is the original length We can find the change in length: $\Delta L = \frac{F~L}{Y~A}$ $\Delta L = \frac{(5000~N)(2.0~m)}{(9.2\times 10^{10}~N/m^2)(5.0\times 10^{-6}~m^2)}$ $\Delta L = 2.2\times 10^{-2}~m = 2.2~cm$ The wire stretches a distance of $2.2~cm$.
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.