Strategy to finish this problem, the volume of the gold needs to be decisive using the gold’s mass and density. Half of that volume is assigned on each face of the coin, and, for each face, the gold can be represented as a cylinder that is 5″ in diameter with a height equal to the thickness.
Operation the volume formula for a cylinder to complete the thickness.
Solution
The mass of the gold is given by the formula
m=ρV=25×10^−3g
m=ρV=25×10^−3g:
where
ρ=19.3g/cc
and V is the volume. Solving for the volume gives
V=mρ=(25×10^−3g):19.3g/cc≅12.9×10−4cc.
If t is the thickness, the volume corresponding to half the gold is
1/2(12.9×10^−4)=πr^2t=π(2.54)2t
where the 1″ radius has been converted to cm. Solving for the thickness gives
t=(6.45×10^−4)/π(2.54)^2 ≅3.2×10^−5cm=0.00039 mm.
The amount of gold used is stated to be 25 mg, which is identical
to a thickness of about 0.00039 mm. The mass figure may cause the amount of gold to sound bigger, both because the number is much bigger (25 versus 0.00039) and because people may have a more intuitive feel for how much a millimeter is than for how much a milligram is. A simple breakdown of this sort can clarify the significance of claims made by advertisers.
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