10판/1. 측정

1-26 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 7. 21. 15:55

$$ \begin{aligned} \Ans &= \rho = ? \\ &= \frac{\mathrm{Mass}}{\mathrm{Volume}} \\ &= \frac{5.324\ut{g}}{2.5\ut{cm^3}} \\ &= \frac{5.324\ut{g}}{2.5\ut{cm^3}} \cdot \frac{1\ut{kg}}{1000\ut{g}}\cdot \left(\frac{100\ut{cm}}{1\ut{m}}\right)^3 \\ &= \frac{10648}{5}\ut{kg/m^3} \\ &= 2129.6\ut{kg/m^3} \\ &\approx 2.1\times 10^3\ut{kg/m^3} =2.1\ut{tonne/m^3} \\ &= 2.1\ut{kg/L}=2.1\ut{g/cm^3} \end{aligned} $$