1-17 할리데이 11판 솔루션 일반물리학

-	wey	chaldron	bag	pottle	gill
1 wey=	1	10/9	40/3	640	120240
1chaldron=	$\frac{9}{10}=0.900$	1	$\frac{9}{10}\cdot\frac{40}{3}=12.0$	$\frac{9}{10}\cdot640=576$	$\frac{9}{10}\cdot120240=1.08\times10^5$
1bag=	$\frac{3}{40}=0.0750$	$\frac{3}{40}\cdot\frac{10}{9}=8.33\times10^{-2}$	1	$\frac{3}{40}\cdot640=48.0$	$\frac{3}{40}\cdot120240=9.02\times10^3$
1pottle=	$\frac{1}{640}=1.56\times10^{-3}$	$\frac{1}{640}\cdot\frac{10}{9}=1.74\times10^{-3}$	$\frac{1}{640}\cdot\frac{40}{3}=2.08\times10^{-2}$	1	$\frac{1}{640}\cdot120240=188$

$$\ab{f}$$ $$\begin{cases} 1\ut{bag}&=0.1091\ut{m^3}\\ \end{cases}$$ $$\begin{aligned} V&=4.32\ut{chaldron}\\ &=4.32\ut{chaldron}\times\frac{1\ut{wey}}{\frac{10}{9}\ut{chaldron}} \times\frac{\frac{40}{3}\ut{bag}}{1\ut{wey}}\times\frac{0.1091\ut{m^3}}{1\ut{bag}}\\ &=4.32\cdot\frac{9}{10}\cdot\frac{40}{3}\cdot0.1091\ut{m^3}\\ &=\frac{88371}{15625}\ut{m^3}\\ &=5.655744\ut{m^3}\\ &\approx5.66\ut{m^3}\\ \end{aligned}$$