1-30 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} 43\ut{cm}<1\ut{cubit}<53\ut{cm}\\ L=11\ut{cubit}\\ 2R=2.5\ut{cubit}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} 43\ut{cm}<1&\ut{cubit}<53\ut{cm}\\ 11\cdot43\ut{cm}<11&\ut{cubit}<11\cdot53\ut{cm}\\ 473\ut{cm}<11&\ut{cubit}<583\ut{cm}\\ 4.73\ut{m}<11&\ut{cubit}<5.83\ut{m}\\ \end{aligned} $$ $$ \begin{cases} \text{Max}&\approx5.8\ut{m}\\ \text{Min}&\approx4.7\ut{m}\\ \end{cases} $$ $$\ab{b}$$ $$ \begin{cases} \text{Max}&\approx5.8\ut{m}=5.8\times10^3\ut{mm}\\ \text{Min}&\approx4.7\ut{m}=4.7\times10^3\ut{mm}\\ \end{cases} $$ $$\ab{c}$$ $$ \begin{aligned} 43\ut{cm}<1&\ut{cubit}<53\ut{cm}\\ (43\ut{cm})^3<1&\ut{cubit^3}<(53\ut{cm})^3\\ 79507\ut{cm^3}<1&\ut{cubit^3}<148877\ut{cm^3}\\ 0.079507\ut{m^3}<1&\ut{cubit^3}<0.148877\ut{m^3}\\ \end{aligned} $$ $$ \begin{aligned} V&=\pi R^2L\\ &=\pi \(\frac{25}{20}\ut{cubit}\)^2\cdot11\ut{cubit}\\ &=\frac{275\pi}{16}\ut{cubit^3}\\ &=\frac{275\pi}{16}\ut{cubit^3} \end{aligned} $$ $$ \begin{aligned} \frac{275\pi}{16}\times0.079507\ut{m^3}< V&<\frac{275\pi}{16}\times0.148877\ut{m^3}\\ \frac{874577\pi}{640000}\ut{m^3}< V&<\frac{1637647\pi}{640000}\ut{m^3}\\ 4.293069809685314\ut{m^3}\lesssim V&\lesssim8.03878091308338\ut{m^3}\\ 4.3\ut{m^3}\lesssim V&\lesssim8.0\ut{m^3}\\ \end{aligned} $$