1-6 할리데이 11판 솔루션 일반물리학

$$\begin{cases} 1\ut{diplacement~ton}&=7\ut{barrel~bluk}\\ 1\ut{freight~ton}&=8\ut{barrel~bluk}\\ 1\ut{register~ton}&=20\ut{barrel~bluk}\\ 1\ut{barrel~bluk}&=0.1415\ut{m^3}\\ 1\ut{m^3}&=28.378\ut{U.S~bushel} \end{cases}$$ $$A = 73\ut{diplacement~ton}$$ $$\ab{a}$$ $$\begin{aligned} \Ans &= 73\ut{freight~ton}-73\ut{diplacement~ton}\\ &=73\cdot\(1\ut{freight~ton}-1\ut{diplacement~ton}\)\\ &=73\cdot\(8\ut{barrel~bluk}-7\ut{barrel~bluk}\)\\ &=73\ut{barrel~bluk}\\ &=73\ut{barrel~bluk}\times\cfrac{0.1415\ut{m^3}}{1\ut{barrel~bluk}}\times\cfrac{28.378\ut{U.S~bushel}}{1\ut{m^3}}\\ &=293.130551\ut{U.S~bushel}\\ &\approx2.9\times10^2\ut{U.S~bushel} \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \Ans &= 73\ut{register~ton}-73\ut{diplacement~ton}\\ &=73\cdot\(1\ut{register~ton}-1\ut{diplacement~ton}\)\\ &=73\cdot\(20\ut{barrel~bluk}-7\ut{barrel~bluk}\)\\ &=73\cdot13\ut{barrel~bluk}\\ &=73\cdot13\ut{barrel~bluk}\times\cfrac{0.1415\ut{m^3}}{1\ut{barrel~bluk}}\times\cfrac{28.378\ut{U.S~bushel}}{1\ut{m^3}}\\ &=3810.697163\ut{U.S~bushel}\\ &\approx3.8\times10^3\ut{U.S~bushel} \end{aligned}$$