I have got two final questions that are really stumping me. I cant figure how to do these. Any help would be great..!

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- Jun 8th 2008, 06:36 AMmoolimanjDimensional Analysis
I have got two final questions that are really stumping me. I cant figure how to do these. Any help would be great..!

- Jun 8th 2008, 07:27 AMTheEmptySeta)
Write out an equation using only the units and what you want to solve for.

Note: Since the webber constant is "unitless" we can write its units as 1.

remember that $\displaystyle \rho=\frac{m}{V} \\\ V=m^3$

So the equation looks like.

$\displaystyle 1=\frac{\frac{m}{V}\left( \frac{m}{s}\right)^2m}{\sigma}$

Solving for sigma and multiplying the numerator we get

$\displaystyle \sigma=\frac{m^4}{Vs^2}=\frac{m^4}{m^3s^2}=\frac{m }{s^2}$

So sigma must have units distance over time squared. - Jun 8th 2008, 07:30 AMmoolimanj
Thanks - sort of makes sense. Can you help with the second part as well?

- Jun 10th 2008, 06:38 AMmoolimanj
For part (c) I get the following (can someone check this please:

v-ms^-1

p=Nm^-2

N= gms^-2 (i.e. Force = mass x acceleration)

so p = gms^2/m^2 = g/s^2m

V=m^3

Density (theta) = g/m^3

so

p/density = g/ms^2/g/m^3 which simplifies to m/s^2

Hence,

V is proportional to square root of pressure/Density

Does this sound right? - Aug 2nd 2008, 06:59 AMCoomast
moolimanj, that is indeed correct. The relation you derived indicates that the speed of sound is independent of the volume as can be expected.

There is a more general theory on this method of working. You were given the dependent parameters, but in more complicated cases you don't know which are important and this is were the theorem of Buckingham comes in the picture. It is a very important theorem in applied physics for looking at parameters which form dimensionless groups that define the problem completely. If you want to read more on them look at the following link.

Buckingham ? theorem - Wikipedia, the free encyclopedia

best regards, Coomast