If pi*r^2*h= 1.69x10^-6m3 what would the density be if a coin weighed 14.1 g
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If pi*r^2*h= 1.69x10^-6m3 what would the density be if a coin weighed 14.1 g

[From: ] [author: ] [Date: 12-04-20] [Hit: ]
The density of a material is its mass divided by its volume. Therefore,= (14.1 g)/(1.≈ 8343 g/m^3.I hope this helps!......
Assuming that r is the radius of the coin and h is its height, then πr^2*h represents the volume of the coin (assuming that the coin is perfectly cylindrical).

The density of a material is its mass divided by its volume. Therefore, the required density is:
d = m/V
= (14.1 g)/(1.69 x 10^-3 m^3)
≈ 8343 g/m^3.

I hope this helps!
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