Question

a rectangular piece of aluminum foil measures 13.72 cm x 8.63 cm and has a mass of 3.1 g. Find how thick it is

Answer 1

The two dimensions of aluminum foil are given 13.72 cm and 8.63 cm respectively with mass 3.1 g.

The density of aluminum is
`2.7 g/cm^(3)`. It is defined as mass per unit volume thus, volume of aluminum can be calculated as follows:

`V=(m)/(d)`

Putting the values.

`V=(3.1 g)/(2.7 g/cm^(3))=1.148 cm^(3)`

The volume of cuboid is
`l* b* h`, the length and breadth are given, height can be calculated as follows;

`h=(V)/(l* b)`

Putting the values,

`h=(1.148 cm^(3))/((13.72 cm)(8.63 cm))=0.009695 cm`

Or, approximately 9.7\times 10^{-3}cm

Therefore, thickness of aluminum foil is
`9.7* 10^(-3)cm`