1-10 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} x_A &= 12\ut{m}\\ y1_{A} &= 6.0\ut{m}\\ y2_{A} &= 3.0\ut{m}\\ z_A &= 20\ut{m}\\ L_A : L_B &= 12:1\\ L_A : L_C &= 144:1\\ \end{cases} $$ $$ \begin{aligned} V_A &= S_{xy} \times L_z\\ &= (S1_{xy} + S2_{xy})\times L_z\\ &=\(6.0\times12+\cfrac{1}{2}\times12\times3.0\)\times20\\ &=1800\ut{m^3} \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} V_A : V_B &= (L_A)^3 : (L_B)^3 \\ &= 12^3:1^3\\ &= 1728:1\\ \end{aligned} $$ $$ \begin{aligned} V_B &= \cfrac{1}{1728}\times V_A\\ &= \cfrac{1}{1728}\times 1800\ut{m^3}\\ &= \cfrac{25}{24}\ut{m^3}\\ &\approx 1.04166666667\ut{m^3}\\ &\approx 1.0\ut{m^3}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} V_A : V_C &= (L_A)^3 : (L_C)^3 \\ &= 144^3:1^3\\ \end{aligned} $$ $$ \begin{aligned} V_B &= \cfrac{1}{144^3}\times V_A\\ &= \cfrac{1}{144^3}\times 1800\ut{m^3}\\ &= \cfrac{25}{41472}\ut{m^3}\\ &\approx 0.000602816358025\ut{m^3}\\ &\approx 6.0\times10^{-4}\ut{m^3}\\ \end{aligned} $$