Answer:
The thickness of the foil is 0.017 mm.
Explanation:
Given that,
Weight = 7.5 oz = 212.6175 gm
Density = 2.70 g/cm³
Area of aluminium = 50 ft² = 46451.52 cm²
We need to calculate the thickness of the foil
Using formula of density
[tex]\rho=\dfrac{m}{V}[/tex]
[tex]\rho=\dfrac{m}{A\times t}[/tex]
[tex]t=\dfrac{m}{A\times \rho}[/tex]
Where, A = area
t = thickness
m = mass
Put the value into the formula
[tex]t=\dfrac{212.6175}{46451.52\times2.70}[/tex]
[tex]t=0.00170\ cm[/tex]
[tex]t=0.017\ mm[/tex]
Hence, The thickness of the foil is 0.017 mm.
