Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 13 - Exercises and Problems: 28

Answer

a) .0099 meters b) 6.2 meters c) 3575 meters

Work Step by Step

We simplify the equation for the period of the pendulum so that length is isolated on one side of the equation: $L = \frac{T^2g}{4\pi^2}$ We plug in each of the values to find: a) $L = \frac{(200\times10^-3)^2(9.81)}{4\pi^2}=\fbox{.0099 meters}$ b) $L = \frac{(5)^2(9.81)}{4\pi^2}=\fbox{6.2 meters}$ c) $L = \frac{(120)^2(9.81)}{4\pi^2}=\fbox{3575 meters}$
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