1-1 할리데이 10판 솔루션 일반물리학

$$ \begin{aligned} & 14.0\ut{gal} \\ =& 14.0\ut{gal}\(\frac{231\ut{inch^3}}{1\ut{gal}}\)\(\frac{2.540\ut{cm}}{1\ut{inch}}\)^{3}\( \frac{1\ut{L}}{10^{3}\ut{cm^3}} \) \\ =& \frac{14.0\cdot231\cdot2.540^{3}\cdot1}{1\cdot1\cdot10^{3}}\ut{L} \\ =& \frac{3312235311}{62500000}\ut{L} \\ \approx& 52.995764976\ut{L} \\ \approx& 53.0\ut{L} \end{aligned} $$