Respuesta :
We use formula, the volume of aluminium
[tex]Volume = \frac{mass}{density} \\\\ Area \times thickness = \frac{mass}{density} \\\\ thickness = \frac{mass}{ Area \times density}[/tex]
Given, [tex]m = 1275 \ g[/tex], [tex]Area = 18.5 m^2 =18.5 \times 10^4 mm^2[/tex] and [tex]density = 2.7 \ g/cm^3 = 2.7 \times 10^{-3} g/mm^3[/tex]
Substituting these values, we get
[tex]thickness = \frac{1275 \ g}{ 18.5 \times 10^4 mm^2 \times 2.7 \times 10^{-3} g/mm^3} = 2.55 \ mm[/tex].
Thus, the thickness of the aluminium foil is 2.55 mm.
Answer:
The thickness in millimeters of the is 0.02552.
Explanation:
Area of an aluminum foil = A =[tex]18.5 m^2=185,000 cm^2[/tex]
[tex]1 m^2=10000 cm^2[/tex]
Mass of the assuming foil = 1275 g
Volume of the aluminum foil = V
Thickness of the assuming foil = h
Density of the aluminum foil = [tex]2.7 g/cm^3[/tex]
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]2.7 g/cm^3=\frac{1275 g}{V}[/tex]
[tex]V = 472.222 cm^3[/tex]
Volume = Area × Depth
[tex]V=A\times h[/tex]
[tex]h=\frac{472.222 cm^3}{185,000 cm^2}=0.002552 cm[/tex]
1 cm = 10 mm
h = 0.02552 mm
The thickness in millimeters of the is 0.02552.
