Physics Technology Update (4th Edition)

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

Chapter 10 - Rotational Kinematics and Energy - Problems and Conceptual Exercises - Page 326: 48

Answer

$v=0.609m/s$

Work Step by Step

We know that $\omega=0.373rev/s=0.373\times 2\pi rad/s$ Now $v=r\omega$ We plug in the known values to obtain: $v=(0.373\times 2\pi \space rad/s)(0.260m)=0.609m/s$
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