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Often I am not sure when to make use of significant figures in a calculation.

For example at the moment. I have to calculate a certain equilibrium point which is given by:

$m^* = \frac{\phi_{\mathrm o}^{*2} A}{g K^2 A_{\mathrm v}^{*2}}$

And the values are: $\phi_{\mathrm o}^* = 0.001$, $A_{\mathrm v}^* = 0.001$, $\rho = 980$, $A = 1$, $K = 0.01$, $g = 9.81$. Now $m^* = 1.019367991845056e+03$

When applying the rules of significant figures the correct answer should be $m^* = 1e+03$, I believe. Since the amount of significant figures of the answer is equal to the number with the least amount of significant figures, which are $\phi_{\mathrm o}^*$, $A_{\mathrm{v}}^*$, $A$ and $K$, they all have 1 significant figure.

I am really questioning myself whether if I should use significant figures or not.

I hope some people can give some idea's/views when to use significant figures or not to use them.

Because $m^*$ is a equilibrium point I would say I would rather use $m^* \approx 1019.37$ rather then $1000$. But still one could argue also that I might use more digits, 1019.368.

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    WG- apologies: I did not read your post carefully enough!2012-12-14

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Are the values of $\phi_{\mathrm o}^*$, $A_{\mathrm{v}}^*$, $A$ and $K$ measurements or are they exact values? If the later, then I would claim that you shouldn't use them in determining the number of significant figures. This is because you can consider exact values as having infinite significant figures (i.e. you can add as many zeros as you wish after the decimal place) and therefore will never be chosen as having the minimum number of significant figures.

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    @WG- Technically, the later is correct. However, in a physics class $=$ is almost always understood to mean $\approx$.2012-12-15