Given:
The coin wrapper volume = 27,480 mm³.
The coin diameter = 21.18 mm.
There are 50 such coins.
To find:
The thickness (height) of 1 coin.
Solution:
The 50 coins form a cylinder with a volume of 27,480 mm³.
If the diameter is 21.18 mm, the radius of a coin will be [tex]\frac{21.18}{2} = 10.59[/tex] mm.
The volume of a cylinder [tex]= \pi r^{2} h.[/tex]
The volume of all 50 coins is given as 27,480 mm³. We need to determine the height of a single coin.
[tex]50( \pi r^{2} h)=27,480.[/tex]
[tex]50(3.1415)(10.59^{2} )(h) = 27,480.[/tex]
[tex](17,615.662)(h)=27,480.[/tex]
[tex]h= \frac{27,480}{17,615.662} = 1.5599[/tex] mm.
So the thickness of 1 coin is 1.5599 mm.