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Volume and Density Question

UrGrandpap

http://imgbox.com/L7qYzwCw

I tried working it out on the right but went all wrong.

Reply 1

mullac

It seemed to be going fine, why did you stop? (although 0.002^3 is not 0.0000335)

Reply 2

Kian Stevens

The equation for density is

\rho = \frac{m}{V}

so rearranging for mass gives

m = \rho V

Since the volume of a sphere is

V = \frac{4}{3} \pi r^{3}

, substitute this into the mass equation to give

m = \rho \frac{4}{3} \pi r^{3}

Hence, Sphere B has the greater mass - its mass is 1.13 kg, compared to Sphere A's mass of 0.637 kg.

(edited 5 years ago)

Reply 3

UrGrandpap

Original post by mullac

It seemed to be going fine, why did you stop? (although 0.002^3 is not 0.0000335)

Is it not? That's strange because it's what it says on my calculator. Anyways, would you mind working me throught this? I'm pretty confused

Reply 4

Kian Stevens

Original post by UrGrandpap

Is it not? That's strange because it's what it says on my calculator. Anyways, would you mind working me throught this? I'm pretty confused

2³ is 8. So (2x10^-3)³ is 8x10^-9...
You must've typed it in on your calculator wrong.

Reply 5

UrGrandpap

Original post by Kian Stevens

2³ is 8. So (2x10^-3)³ is 8x10^-9...
You must've typed it in on your calculator wrong.

Yeah I probably did.

Reply 6

UrGrandpap

Original post by Kian Stevens

2³ is 8. So (2x10^-3)³ is 8x10^-9...
You must've typed it in on your calculator wrong.

Oh wait I forgot to say that I converted the cm to m before I started hence why I calculated 4/3 * pi * 0.02^3

Reply 7

mullac

Original post by UrGrandpap

Yeah I probably did.

no disregard what i said, i didnt see that you had also multiplied it by 4/3 pi both your volumes are correct

Reply 8

UrGrandpap

Original post by mullac

no disregard what i said, i didnt see that you had also multiplied it by 4/3 pi both your volumes are correct

Oh wait I forgot to say that I converted the cm to m before I started hence why I calculated 4/3 * pi * 0.02^3

Reply 9

Kian Stevens

Original post by UrGrandpap

Oh wait I forgot to say that I converted the cm to m before I started hence why I calculated 4/3 * pi * 0.02^3

Yeah I've only just realised that too :biggrin:

No worries!

If you need help:
•

m = \rho V = \rho \frac{4}{3} \pi r^{3} \therefore

•

m_{SphereA} = \rho \frac{4}{3} \pi r^{3} = 19000 * \frac{4}{3} \pi * (\frac{2}{100})^{3}

= 0.637 kg
•

m_{SphereB} = \rho \frac{4}{3} \pi r^{3} = 10000 * \frac{4}{3} \pi * (\frac{3}{100})^{3}

= 1.13 kg

Hence, as I mentioned before, Sphere A has the greater mass.

(edited 5 years ago)

Reply 10

UrGrandpap

Original post by Kian Stevens

Yeah I've only just realised that too :biggrin:

No worries!

If you need help:
•

m = \rho V = \rho \frac{4}{3} \pi r^{3} \therefore

•

m_{SphereA} = \rho \frac{4}{3} \pi r^{3} = 19000 * \frac{4}{3} \pi * (\frac{2}{100})^{3}

= 0.637 kg
•

m_{SphereB} = \rho \frac{4}{3} \pi r^{3} = 10000 * \frac{4}{3} \pi * (\frac{3}{100})^{3}

= 1.13 kg

Hence, as I mentioned before, Sphere A has the greater mass.

Is it possible to do it the way I tried in the picture? If not then that's ok, I'm just wondering because I want to know if I was on the right track.

Reply 11

Kian Stevens

Original post by UrGrandpap

Is it possible to do it the way I tried in the picture? If not then that's ok, I'm just wondering because I want to know if I was on the right track.

Yeah, but it's a slower method of doing it. However, if you feel more confident if you take your time, then your way is obviously better.

The way I did it is essentially the way you were doing it in the picture. The only difference is that you were calculating volumes before calculating masses, whereas I just did it all in one step by substituting the volume equation into the rearranged density equation.

You'd arrive at the same answer regardless!

(edited 5 years ago)