Essential University Physics: Volume 1 (4th Edition) Clone

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

Chapter 13 - Exercises and Problems - Page 248: 82


$2.1 \ m/s^2$

Work Step by Step

We know: $\omega^2 = \frac{g}{L}$ Thus, we find: $90^2 = \frac{g}{L}$ $1=\frac{g}{8100L}$ And $91^2 = \frac{\sqrt{g^2+a^2}}{L}$ $ 1 = \frac{g-a}{8281L}$ Setting these equal gives: $\frac{g}{8100L}= \frac{\sqrt{g^2+a^2}}{8281L}$ $a\approx2.1 \ m/s^2$
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