Physics Technology Update (4th Edition)

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

Chapter 1 - Introduction to Physics - Problems and Conceptual Exercises - Page 17: 49

Answer

(a) 51 revolutions. (b) $8.7~ft$

Work Step by Step

See conversion factors (inside front cover). $1~yd=3~ft$, $1~ft=30.5~cm$ (a) $time=\frac{distance}{speed}=\frac{150~yd}{31~cm/s}(\frac{3~ft}{1~yd})(\frac{30.5~cm}{1~ft})(\frac{1~min}{60~s})=7.38~min$ Let N be the number of revolutions. $N=(7~revolutions/min)(7.38~min)=51.7$. That is, 51 revolutions completed plus 0.7 of a revolution. (b) $distance=(speed)(time)$ $distance=(31~cm/s)(\frac{1~revolution}{7~revolutions/min})(\frac{60~s}{1~min})(\frac{1~ft}{30.5~cm})=8.7~ft$
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