# Chapter 13 - Exercises and Problems - Page 245: 20

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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