1-26 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} 1\ut{ken}&=1.97\ut{m}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \frac{1\ut{ken^2}}{1\ut{m^2}}&=\(\frac{1\ut{ken}}{1\ut{m}}\)^2\\ &=\(\frac{1.97\ut{m}}{1\ut{m}}\)^2\\ &=\frac{38809}{10000}\\ &\approx \frac{388}{100}\\&\approx \frac{97}{25}\\ \end{aligned} $$ $$1\ut{ken^2}:1\ut{m^2}\approx 97:25 $$ $$\ab{b}$$ $$ \begin{aligned} \frac{1\ut{ken^3}}{1\ut{m^3}}&=\(\frac{1\ut{ken}}{1\ut{m}}\)^3\\ &=\(\frac{1.97\ut{m}}{1\ut{m}}\)^3\\ &=\frac{7645373}{1000000}\\ \end{aligned} $$ $$1\ut{ken^3}:1\ut{m^3}= 7645373:1000000 $$ $$\ab{c}$$ $$ \begin{aligned} \Ans&=6.90\ut{ken}\times\pi\cdot(3.80\ut{ken})^2\\ &=\frac{24909 }{250}\pi \ut{ken^3}\\ &\approx313.01572563307263\ut{ken^3}\\ &\approx313\ut{ken^3}\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \Ans&=\frac{24909 }{250}\pi \ut{ken^3}\\ &=\frac{24909 }{250}\pi \ut{ken^3}\cdot\(\frac{1.97\ut{m}}{1\ut{ken}}\)^3\\ &=\frac{190438596057 }{250000000}\pi\ut{m^3}\\ &\approx2393.1219773305015\ut{m^3}\\ &\approx2.39\times10^3\ut{m^3}\\ \end{aligned} $$