Explanation:
a) Using the provided information about the density of gold, the sample size, thickness, and the following equations and comersion factors, find the area of the gold leaf:
[tex]V=l \cdot w \cdot h=A \cdot h\\m=\rho \cdot V[/tex]
Gold [tex]_{\rho}=19.32 \mathrm{g} / \mathrm{cm}^{3}[/tex]
[tex]1 \mu=10^{-6} \mathrm{m}[/tex]
First, find the volume of the sample and then find the area of the sample.
[tex]V=\frac{m}{\rho}=\frac{27.6 \mathrm{g}}{19.32 \mathrm{g} / \mathrm{cm}^{3}} \cdot\left(\frac{0.01 \mathrm{m}}{1 \mathrm{cm}}\right)^{3}[/tex] [tex]=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}[/tex]
[tex]V=A \cdot h \rightarrow A=\frac{V}{h}=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{10^{-6} \mathrm{m}} \approx 1.429 \mathrm{m}^{2}[/tex]
b. Using the provided information from part [tex]a[/tex] ), the radius of the cylinder, and the following equation for the volume of a cylinder, find the length of the fiber :
[tex]V=\pi r^{2} h \rightarrow h=\frac{V}{\pi r^{2}}[/tex]
[tex]h=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{\pi \cdot\left(2.5 \times 10^{-6} \mathrm{m}\right)^{2}} \approx 72778 \mathrm{m}[/tex]
