1-1 할리데이 10판 솔루션 일반물리학 14.0[gal]=14.0[gal](231[inch3]1[gal])(2.540[cm]1[inch])3(1[L]103[cm3])=14.0⋅231⋅2.5403⋅11⋅1⋅103[L]=331223531162500000[L]≈52.995764976[L]≈53.0[L] \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} ===≈≈14.0[gal]14.0[gal](1[gal]231[inch3])(1[inch]2.540[cm])3(103[cm3]1[L])1⋅1⋅10314.0⋅231⋅2.5403⋅1[L]625000003312235311[L]52.995764976[L]53.0[L] 10판/1. 측정 2019.07.20