1-22 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \rho &= 19.32\ut{g/cm^3} \\ m &= 29.34\ut{g} \end{cases}$$
(a) S=? $$h = 1.000\ut{\mu m}$$ $$\begin{aligned} \\ \Ans &= S = \frac{\text{Volume}}{\text{Hight}} \\ &= \frac{m}{\rho}\cdot\frac{1}{h} \\ &= \frac{29.34\ut{g}}{19.32\ut{g/cm^3}}\cdot \frac{1}{1.000\ut{\mu m}} \\ &= \frac{489}{322}\ut{cm^3}\cdot \frac{1}{1.000\ut{\mu m}}\cdot \frac{10^6\ut{\mu m}}{1\ut{m}}\cdot \left(\frac{1\ut{m}}{100\ut{cm}}\right)^3 \\ &=\frac{489}{322}\ut{m^2} \\ &\approx 1.518633540372671\ut{m^2} \\ &\approx 1.519\ut{m^2} \end{aligned}$$
(b) L=? $$R = 2.500\ut{\mu m}$$ $$\begin{aligned} \\ \Ans &= L = \frac{\text{Volume}}{\text{Area}}= \frac{V}{\pi R^2} \\ &= \frac{489}{322}\ut{cm^3}\cdot \frac{1}{\pi(2.500\ut{\mu m})^2} \\ &= \frac{489}{322}\ut{cm^3}\cdot \frac{1}{\pi(2.500\ut{\mu m})^2}\cdot\left(\frac{1\ut{m}}{100\ut{cm}}\right)^3\cdot\left(\frac{10^6\ut{\mu m}}{1\ut{m}}\right)^2 \\ &= \frac{39120000}{161 \pi }\ut{m} \\ &\approx 77343.371102545907271\ut{m} \\ &\approx 7.734\times 10^4\ut{m} = 77.34\ut{km} \end{aligned}$$