Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 6 - Applications of Newton's Laws - Problems and Conceptual Exercises - Page 183: 70

Answer

$1.4\times 10^5\frac{m}{s^2}$

Work Step by Step

We can find the centripetal acceleration as follows: $a_c=\frac{v^2}{r}$ We plug in the known values to obtain: $a_c=\frac{(77)^2}{0.042}$ $a_c=1.4\times 10^5\frac{m}{s^2}$
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