1-2 할리데이 10판 솔루션 일반물리학

$$ \begin{aligned} & 0.75\ut{gry^2} \\ =& 0.75\ut{gry^2} \(\frac{1\ut{line}}{10\ut{gry}}\)^2 \(\frac{1\ut{inch}}{12\ut{line}}\)^2 \(\frac{72\ut{Point}}{1\ut{inch}}\)^2 \\ =& \frac{0.75\cdot 1^2\cdot 1^2\cdot 72^2}{10^2\cdot12^2\cdot 1^2}\ut{Point^2} \\ =& \frac{27}{100}\ut{Point^2} \\ =& 0.27\ut{Point^2} \end{aligned} $$