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 16: 31

Answer

(a) $67~mutchkins$ (b) $1~noggin=0.031~gal$

Work Step by Step

See conversion factors (inside front cover). $1~m=3.281~ft$, $1~mutchkin=0.42~L$, $1~m^{3}=10^{3}~L$ (a) Let V be the volume of a container ($a=1~ft)$. $V=a^{3}=(1~ft)^{3}=(1~ft^{3})(\frac{1~m}{3.281~ft})^{3}=2.83\times10^{-2}~m^{3}$ $V=(2.83\times10^{-2}~m^{3})(\frac{10^{3}~L}{1~m^{3}})(\frac{1~mutchkin}{0.42~L})=67~mutchkin$ (b) $1~noggin=0.28~mutchkin$, $1~gal=3.785~L$, $1~mutchkin=0.42~L$ $1~noggin=(1~noggin)(\frac{0.28~mutchkin}{1~noggin})(\frac{0.42~L}{1~mutchkin})(\frac{1~gal}{3.785~L})=0.031~gal$
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