College Physics (4th Edition)

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

Chapter 10 - Problems - Page 397: 5


The Young's modulus for this wire is $7.69\times 10^{10}~N/m^2$

Work Step by Step

$Y = \frac{F/A}{\Delta L/L}$ $Y$ 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 Young's modulus: $Y = \frac{F/A}{\Delta L/L}$ $Y = \frac{F~L}{A~\Delta L}$ $Y = \frac{(1000~N)(5.00~m)}{(0.100\times 10^{-4}~m^2)(6.50\times 10^{-3}~m)}$ $Y = 7.69\times 10^{10}~N/m^2$ The Young's modulus for this wire is $7.69\times 10^{10}~N/m^2$.
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