Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 1 - Concepts of Motion - Exercises and Problems - Page 29: 24

Answer

a) $0.20 m$ b) $20 m/s$ c) $27 m/s$ d) $0.0090 m^{2}$

Work Step by Step

First identify the type of units the problem is describing, then find the corresponding SI unit. Round the converted number to the number of significant figures in the original unit of measurement. a) The inch is a unit of length, so we must convert it to meters. the conversion goes from inches to feet, and feet to meters. $$ 8.0 in * \frac{1 ft}{12 in} * \frac{.3048 m}{1 ft} = 0.2032m $$ b) Feet per second is a unit of length and a unit of time. Because seconds is already an SI unit, only distance must be converted. $$66ft/s * \frac{.3048m}{1ft} = 20.12ft/s$$ c) Miles per hour is a unit of distance per a unit of time, both which need to be converted to meters and seconds to become SI units. $$60mi/h * \frac{1.6093km}{1mi} * \frac{10^3m}{1km} * \frac{1h}{60min} * \frac{60s}{1min} = 62.82 m/s$$ d) Inches squared is a measurement of area and the SI unit counterpart is square meters. Each conversion needs to be done twice, one for each of the powers of the unit. $$14in^{2} * (\frac{1ft}{12in})^{2} * (\frac{.3048m}{1ft})^{2} = .009032 m^{2}$$
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