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: 63

Answer

$19\frac{m}{s}$

Work Step by Step

We know that $N-mg=-ma_c$ $\implies 0-mg=-\frac{mv^2}{r}$ This simplifies to: $v=\sqrt{rg}$ We plug in the known values to obtain: $v=\sqrt{(35)(9.81)}$ $v=19\frac{m}{s}$
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