p=mv - issue with the 'v' equation.

Hi all,

I'm looking at density etc and I'm doing okay so far, I'm having a few questions about the equations though.

I know that to work out density the equation is p=m/v (p=mv).

I've been give the radius of an object and it's mass.

So all I need to do now is figure out it's Volume.

For doing that I've been told this equation: V= 4/3 TTr3.

For working that out I've been told to cube the radius, then multiply that answer by 4, them multiply that answer by TT (pi), then divide the answer by 3.

Why is that? Why do we multiply by 4, then by pi then divide by 3?

Do we multiply by 4 because it's the top number of the fraction and then divide by 3 as it's the lower part of the fraction?

I'm just looking to learn the 'rules' for this kind of equation.

Thanks for your help,

Hyrulian

---Edit---

Also, if r= 2403 km, 2403,000km would it be acceptable to say that it is 2.40 x 10⁶ or even 2.403 x 10⁵ (not sure on that one).

Why only use two significant figures instead of three?

(edited 10 years ago)

Reply 1

ghostwalker

17

Original post by Hyrulian

Hi all,

I'm looking at density etc and I'm doing okay so far, I'm having a few questions about the equations though.

I know that to work out density the equation is p=m/v (p=mv).

I've been give the radius of an object and it's mass.

So all I need to do now is figure out it's Volume.

For doing that I've been told this equation: V= 4/3 TTr3.

For working that out I've been told to cube the radius, then multiply that answer by 4, them multiply that answer by TT (pi), then divide the answer by 3.

Why is that? Why do we multiply by 4, then by pi then divide by 3?

Do we multiply by 4 because it's the top number of the fraction and then divide by 3 as it's the lower part of the fraction?

Yes.

4/3 = 4 x (1/3)

So, multiply by 4; and multiply by 1/3, or putting it another way divide by 3

I'm just looking to learn the 'rules' for this kind of equation.

Thanks for your help,

Hyrulian

Anything on top of the fraction is multiply by.

And the bottom is divide by.

I truely hope this isn't "undergraduate level" for you, else you've a long learning curve for basic arithmetic.

Reply 2

ghostwalker

17

Original post by Hyrulian

---Edit---

Also, if r= 2403 km, 2403,000km would it be acceptable to say that it is 2.40 x 10⁶ or even 2.403 x 10⁵ (not sure on that one).

Why only use two significant figures instead of three?

2403 km is 2,403,000 m

In scientific notation, it would be

2.403\times 10^6

m, which is the same thing.

I wouldn't truncate it to 2 or 3 sig.fig. unless you'd been told to.

Reply 3

Hyrulian

OP

Original post by ghostwalker

Yes.

4/3 = 4 x (1/3)

So, multiply by 4; and multiply by 1/3, or putting it another way divide by 3

Anything on top of the fraction is multiply by.

And the bottom is divide by.

I truely hope this isn't "undergraduate level" for you, else you've a long learning curve for basic arithmetic.

Thank you for the answer, a brief explanation is that I am taking an undergrad course through the OU.

I was never that great at Maths at school but pretty good with Science.
I'm taking a Mathematics and Physics degree and am really trying hard to brush up in my Maths.
---

Ah okay, thanks.

I had seen an example where the r = 1140km (3 sig.figs/ well it said that the zero's can be considered to not be significant) so it was written as 1.14 x 10^6.

If it was't truncated it'd have been 1.140 x 10^6, or if it was truncated to 2 sig.figs it'd be 1.1 x 10^6.

Thanks again for your reply.

(edited 10 years ago)

Reply 4

ghostwalker

17

Original post by Hyrulian

Thank you for the answer, a brief explanation is that I am taking an undergrad course through the OU.

I was never that great at Maths at school but pretty good with Science.
I'm taking a Mathematics and Physics degree and am really trying hard to brush up in my Maths.

Go for it!

Even though you're doing an undergrad course, you may find it better to label, as appropriate, threads as either secondary school or sixth form. You'll have a greater chance of response, particularly by people who have recently covered material that's relates to what you're asking.

Good luck.

Reply 5

ghostwalker

17

Original post by Hyrulian

I had seen an example where the r = 1140km (3 sig.figs/ well it said that the zero's can be considered to not be significant) so it was written as 1.14 x 10^6.

As it's quoted to 3sig.fig you know that last 0 is an approximation, and is required as a place holder - 114km would be clearly wrong.

In scientific notation 1.14 x 10^6 m would be correct, quoting to 3 sig fig.
Note we have converted km to m.

If it was't truncated it'd have been 1.140 x 10^6, or if it was truncated to 2 sig.figs it'd be 1.1 x 10^6.

Thanks again for your reply.

1.140 x 10^6 is quoting 4 figures (that 0 is no longer required as a place holder) and you'd need to state 3 sig.fig. if you meant that, otherwise it's implicit that it is to 4 sig.fig.

1.1 x 10^6 would be correct ot 2 sig.figs, if you wanted to make an approximation.

Reply 6

Hyrulian

OP

Thanks Ghostwalker.

I have a quick question about how to write out an equation like that in full, plus why certain parts are written the way that they are.

If we had a radius of 1140km and a mass of 130 X 10²⁰ kg.

When it comes to writing out the equation for the volume, I could write it out simply as
V= 4/3TTr³ = 6.207 X 10¹⁸ m³

I've seen an example where the Volume equation was written out in full, like this:

V = 4/3 TT r³

= 4/3 TTX (1.14 X 10⁶ m)³

= 4/3 TT X 1.482 X 10¹⁸ m³

= 6.207 X 10¹⁸ m³

To calculate the density:

p=mv = 130 X 10²⁰ kg / 6.207 X 10¹⁸ m³= 2094 kg m^-3

= 2100 kg m^-3(to 2 sig.figs)

My question is, with the second line why is that sum in brackets?

m³ That's basically a kilometer, I assume the ³ stands for number of 0's?

Thanks,

Hyrulian

Reply 7