1-11 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} 1\ut{acre} &= 4047\ut{m^2}\\ 1\ut{ft} &= 12\ut{in}\\ 1\ut{in} &= 2.54\ut{cm}\\ S &= 17\ut{km^2}\\ h &= 2.0\ut{in}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} V &= S \cdot h \\ &= (17\ut{km^2}) \cdot (2.0\ut{in})\\ &=\bra{17\ut{km^2} \cdot \(\cfrac{1000\ut{m}}{1\ut{km}}\)^2 \cdot \cfrac{1\ut{acre}}{4047\ut{m^2}}}\cdot\(2.0\ut{in}\cdot\cfrac{1\ut{ft}}{12\ut{in}}\)\\ &=\cfrac{17\cdot1000^2\cdot2.0}{4047\cdot12}\ut{acre-ft}\\ &=\cfrac{8500000}{12141}\ut{acre-ft}\\ &\approx700.1070752\ut{acre-ft}\\ &\approx7.0\times10^2\ut{acre-ft} \end{aligned} $$ $$\ab{b}$$ $$\rho=1.0\times10^3\ut{kg/m^3}$$ $$ \begin{aligned} V &= S \cdot h \\ &= (17\ut{km^2}) \cdot (2.0\ut{in})\\ &=\bra{17\ut{km^2} \cdot \(\cfrac{1000\ut{m}}{1\ut{km}}\)^2 }\cdot\(2.0\ut{in}\cdot\cfrac{2.54\ut{cm}}{1\ut{in}}\cdot\cfrac{1\ut{m}}{100\ut{cm}}\)\\ &=\cfrac{17\cdot1000^2\cdot2.0\cdot2.54}{100}\ut{m^3}\\ &=863600\ut{m^3} \end{aligned} $$ $$ \begin{aligned} m&=\rho \cdot V\\ &=1.0\times10^3\ut{kg/m^3}\cdot863600\ut{m^3}\\ &=863600000\ut{kg}\\ &\approx 8.6\times10^8\ut{kg}\\ &\approx 8.6\times10^5\ut{ton}\\ &\approx 0.86\ut{megaton}\\ \end{aligned} $$

풀이자주 : 톤이나 메가톤 단위는 SI비표준이므로 필수적인 표현은 아니나 풀이자가 좋아해서 넣었음.