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We need to make $1.15\times10^5$ coins with a thickness of $2.0\times10^{-3}m(2mm)$ from a $0.12m^3$ block of silver. I'm trying to calculate what diameter the coins will end up being.

Did I set up the equation correctly where I'd get my answer from solving for D? I tried to draw it out to help understand what needs to be done. $0.12m^3 = (1.15\times10^5) \times ((2.0\times10^{-3}m)\times \frac{\pi D}{2})$

My solution: $0.12m^3 = (1.15\times10^5) \times ((2.0\times10^{-3}m)\times \pi r^2)$ $\frac{0.12}{(1.15\times10^5)} = (2.0\times10^{-3})\times \pi r^2$ $\pi r^2 = \frac{(0.12)}{(2.0\times10^{-3})(1.15\times10^5)}$ $r^2 = \frac{(0.12)}{\pi(2.0\times10^{-3})(1.15\times10^5)}$ $ 2r = 2\sqrt{\frac{(0.12)}{\pi(2.0\times10^{-3})(1.15\times10^5)}} = 0.025774m$ $ D = 2.5774cm $

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    One way to find the error is to check the units. The left side has dimensions of volume ($m^3$) while the right has dimensions of $m^2$ because $D$ has dimensions of $m$.2012-08-30

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The term $\dfrac{\pi D}{2}$ needs to be changed to $\dfrac{\pi D^2}{4}$.

This is because the area of a circle of radius $r$ is $\pi r^2$, so the area of a circle of diameter $D$ is $\dfrac{\pi D^2}{4}$. Thus if the coin has thickness $t$, it has volume $\dfrac{\pi D^2}{4}t$. If we want to make $n$ coins, and the amount of silver available is $V$, then $V=\frac{\pi D^2}{4}tn.$

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    Oh that makes sense I was wondering if I did that conversion correctly. Thanks.2012-08-30
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Ok so you know that $\pi r^2(2*10^{-3})*1.18*10^5=0.12$

So $r^2=\frac {0.12}{2\pi10^{-3}*1.18*10^5}$ and

$\huge2r=2\sqrt {\frac{0.12}{2\pi10^{-31.18*10^5}}} \approx 0.0254443$

The problem in what you did is that $r=d/2$ so $r^2=d^2/4$ as André ponted out.