Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 3 - Exercises and Problems: 74

Answer

Please see the work below.

Work Step by Step

We know that $a_r=\frac{v^2}{r}$ $r=\frac{v^2}{a_r}$ but $a_r=g$ Hence $r=\frac{v^2}{g}$
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