5-61 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_{\max}&=22300\ut{kg}\\ V_0&=7682\ut{L}\\ \end{cases} $$ $$\ab{a,b}$$ $$k_A=1.77\ut{kg/L},$$ $$\ab{a}$$ $$ \begin{aligned} m_{A0}&=k_AV_0\\ &=1.77\ut{kg/L}\cdot7682\ut{L}\\ &=13597.14\ut{kg}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Delta V&=\frac{\Delta m}{k_A}\\ &=\frac{m_{\max}-m_{A0}}{k_A}\\ &=\frac{870286}{177}\ut{L}\\ &\approx 4916.870056497175\ut{L}\\ &\approx 4917\ut{L}\\ \end{aligned} $$ $$\ab{c,d,e}$$ $$1\ut{lb}=0.45359237\ut{kg},$$ $$ \begin{aligned} k_B&=1.771\ut{lb/L}\\ &=1.771\ut{lb/L}\cdot0.45359237\ut{kg/lb}\\ &=0.80331208727\ut{kg/L} \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} m_{B0}&=k_BV_0\\ &=0.80331208727\ut{kg/L}\cdot7682\ut{L}\\ &=6171.04345440814\ut{kg}\\ &\approx 6171\ut{kg}\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \Delta V&=\frac{\Delta m}{k_B}\\ &=\frac{m_{\max}-m_{B0}}{k_B}\\ &=\frac{806447827279593}{22679618500}\ut{L}\\ &\approx 35558.2624672277\ut{L}\\ &\approx 35558\ut{L}\\ \end{aligned} $$ $$\ab{e}$$ $$ \begin{aligned} \Ans&=\frac{V_0+\Delta V}{V_{\max}}\\ &=\frac{V_0+\Delta V}{\frac{m_{\max}}{k_B}}\\ &=\frac{7682\ut{L}+\frac{870286}{177}\ut{L}}{22300\ut{kg}}\cdot0.80331208727\ut{kg/L}\\ &=\frac{232647526573}{513300000000}\\ &\approx 0.45323889844730175\\ &\approx 45.323889844730175\% \\ &\approx 45.32\% \\ \end{aligned} $$