Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 16 - Exercises and Problems: 55

Answer

Please see the work below.

Work Step by Step

We know that $(\frac{1}{2})Mv^2=4mc\Delta T$ This can be rearranged as: $\Delta T=\frac{mv^2}{8mc}$ We plug in the known values to obtain: $\Delta T=\frac{(1500)(\frac{40}{3.6})^2}{(8)(5.0)(502)}$ $\Delta T=9.2C^{\circ}$
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