1-27 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \rho&=40\ut{mi/gal}\\ 1\ut{U.K.~gal}&=4.546~090~0\ut{L}\\ 1\ut{U.S.~gal}&=3.785~411~8\ut{L}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \Ans&={1230\ut{mi}\over40\ut{mi/U.S.~gal}}\\ &={123\over4}\ut{U.S.~gal}\\ &={123\over4}\ut{U.S.~gal}\cdot{3.785~411~8\ut{L}\over1\ut{U.S.~gal}}\\ &=\frac{2~328~028~257}{20~000~000}\ut{L}\\ &\approx116.401~412~85\ut{L}\\ &\approx1.2\times10^2\ut{L} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Ans&={1230\ut{mi}\over40\ut{mi/U.K.~gal}}\\ &={123\over4}\ut{U.K.~gal}\\ &={123\over4}\ut{U.K.~gal}\cdot{4.546~090~0\ut{L}\over1\ut{U.K.~gal}}\\ &=\frac{55916907}{400000}\ut{L}\\ &=139.792~675\ut{L}\\ &\approx1.4\times10^2\ut{L} \end{aligned} $$