Answer:
[tex]h=1.60\mu m[/tex]
Explanation:
Hello,
In this case, given the density and the mass of the aluminum foil, we can compute the occupied volume as shown below:
[tex]\rho =\frac{m}{V}\\ \\V=\frac{m}{\rho}=\frac{2.07g}{2.7g/cm^3} =0.767cm^3[/tex]
Next, since the volume is defined as:
[tex]V=24cm*20cm*h[/tex]
Whereas [tex]h[/tex] accounts for its thickness, we can find it to be:
[tex]h=\frac{V}{24cm*20cm}=\frac{0.767cm^3}{20cm*24cm}\\ \\h=1.60x10^{-3}cm*\frac{10000\mu m}{1cm} \\\\h=1.60\mu m[/tex]
Regards.
