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 182: 56

Answer

$196\frac{m}{s}$

Work Step by Step

We know that $a_c=\frac{v^2}{r}$, where $a_c$ shows centripetal acceleration This simplifies to: $v=\sqrt{ra_c}$ We plug in the known values to obtain: $v=\sqrt{(0.075m)(52000)(9.81\frac{m}{s^2})}$ $v=196\frac{m}{s}$
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