# Volume calculation help

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Hello all,

I have had help on this forum before and it was very informative but I have been presented with a calculation I am not entirely sure how to work out so am looking for some help again please.

The question is shown below along with the answer I believed to have been correct. I understand the equation is V=M/P however I believe the bit that threw me off was the fact that the density has 2 powers, one in the sum 1.24x103  and one again in the unit m-3

I would be very grateful if somebody could break down how to work this out so I could give it another go.

I just realised some people may not be able to zoom in to see the image I uploaded … so, the question is: The planet Jupiter has a mass of 1.899 × 1027 kg and a density of 1.24 × 103 kg m–3.Use this information to calculate the volume of Jupiter

My incorrect answer was: 2.354 x 1030 kg m-3

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10 minutes ago, ScienceNoobie said:

I just realised some people may not be able to zoom in to see the image I uploaded … so, the question is: The planet Jupiter has a mass of 1.899 × 1027 kg and a density of 1.24 × 103 kg m–3.Use this information to calculate the volume of Jupiter

My incorrect answer was: 2.354 x 1030 kg m-3

Hello! A quick hint: the question is for Jupiter's volume, is the unit of your suggested answer a volume?

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1 hour ago, ScienceNoobie said:

density has 2 powers, one in the sum 1.24x103  and one again in the unit m-3

These two powers are not connected to one another. One is the value of the density and the other is the units of density.

The first one is just the value of the density being shown in scientific notation: 1.24x103  = 1,240

The second one is representing the fact that the units of density are mass / volume; in other words kgm3. Dividing by a power is the same as multiplying by the negative of the power, so the units of density can also be written as kg m-3.

You have one value measured in kg and another value measured in kg m-3. If your calculation is correct then, as Ghideon suggests, the result should have the units of volume (m3) because the units can be multiplied, divided or cancelled like numbers.. This process of analysing the units in an equation is called dimensional analysis and is a valuable way of checking that a calculation makes sense. It can also, as in this case, tell you which number needs to be divided by which.

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