College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 1 - Problems: 78

Answer

(a) $166~\mu m/s$ (b) $1.44\times 10^{-2}~km/day$

Work Step by Step

(a) We can convert 1 furlong per fortnight to units of $\mu m/s$: $\frac{220~yd}{14~days}\times \frac{1~day}{(24)(3600~s)}\times \frac{0.9144~m}{1~yd}\times \frac{10^6~\mu m}{1~m} = 166~\mu m/s$ (b) We can convert 1 furlong per fortnight to units of $km/day$: $\frac{220~yd}{14~days}\times \frac{0.9144~m}{1~yd}\times \frac{1~km}{1000~m} = 1.44\times 10^{-2}~km/day$
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