1-9 할리데이 10판 솔루션 일반물리학

$$\begin{aligned} & \Ans \\ =& 1.00\ut{cm} \cdot \sqrt[3]{1\ut{mol}} \\ =& 1.00\ut{cm}\cdot \frac{1\ut{m}}{100\ut{cm}} \cdot \sqrt[3]{1\ut{mol}\cdot \frac{6.02\times10^{23}\ut{EA}}{1\ut{mol}}} \\ =& \frac{1.00\cdot1\cdot\sqrt[3]{6.02\times10^{23}}}{100}\ut{m} \\ =& 100000\sqrt[3]{602}\ut{m} \\ \approx& 844368.773356757\ut{m} \\ \approx& 8.44\times 10^{5}\ut{m} = 844\ut{km} \end{aligned}$$