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 15 - Fluids and Elasticity - Exercises and Problems: 27

Answer

Young's modulus for the rope is $5.5\times 10^9~N/m^2$

Work Step by Step

We can find Young's modulus for the rope as: $Y = \frac{F/A}{\Delta~L/L}$ $Y = \frac{F~L}{A~\Delta L}$ $Y = \frac{(70~kg)(9.80~m/s^2)(50~m)}{(\pi)(5.0\times 10^{-3}~m)^2(0.080~m)}$ $Y = 5.5\times 10^9~N/m^2$ Young's modulus for the rope is $5.5\times 10^9~N/m^2$.
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