Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 13 - Exercises and Problems - Page 248: 83


$.378\sqrt h$

Work Step by Step

We know the following value for k: $k = \frac{mg\sqrt h}{200\pi} $ *Note, it is $200\pi$ instead of $2\pi$ because h has to be converted to meters. Thus, we can find: $I = \frac{T^2k}{4\pi^2}$ $I = \frac{T^2(\frac{mg\sqrt h}{200\pi} )}{4\pi^2}$ $I = \frac{6.91^2(\frac{(20)(9.81)\sqrt h}{200\pi} )}{4\pi^2}=.378\sqrt h$
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