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# Show that the equation I=nAvq is homogeneous with respect to units?? watch

1. I have got the equation I=nAvq

Which when broken down into the units, is:
c/s = (n/m^3)(m^2)(m/s)(c)

How do I carry on from here to show that it is homogeneous/the same?

Upon googling it, I see someone on this forum asked the same thing, but after my c/s=..... step, they asked
I = nAvq
C/s = (n/m^3)(m^2)(m/s)(C)

(m^-3) x (m^-2) x m = m^0 = 1

so m's cancel

C/s = (n)C/s

How do you proceed from here?
Can you explain how I would go about after my first step? The other person said they used (m^-3) x (m^-2) x m = m^0 = 1, what does that mean and where does it come from? I am stuck after seperating it into the units of C/s = (n/m^3)(m^2)(m/s)(C)

Any help would be greatly appreciated,

Thanks
2. (Original post by blobbybill)
I have got the equation I=nAvq

Which when broken down into the units, is:
c/s = (n/m^3)(m^2)(m/s)(c)

How do I carry on from here to show that it is homogeneous/the same?

Upon googling it, I see someone on this forum asked the same thing, but after my c/s=..... step, they asked

Can you explain how I would go about after my first step? The other person said they used (m^-3) x (m^-2) x m = m^0 = 1, what does that mean and where does it come from? I am stuck after seperating it into the units of C/s = (n/m^3)(m^2)(m/s)(C)

Any help would be greatly appreciated,

Thanks

i.e. eliminate the denominators so that the whole equation is on one line.

Then tidy up.

Yu should be left with C = C.
3. (Original post by uberteknik)

i.e. eliminate the denominators so that the whole equation is on one line.

Then tidy up.

Yu should be left with C = C.
Why do you do 1/m^3 for example. Why do you do 1 divided by the units on the right hand side of the equation? And why do you not do the same to the left hand side?

Thanks
4. (Original post by blobbybill)
Why do you do 1/m^3 for example. Why do you do 1 divided by the units on the right hand side of the equation? And why do you not do the same to the left hand side?

Thanks
Have you not learned the basic maths rules for manipulating exponents, powers, roots etc?

You need to learn and practice these so that you can apply them to the problem.

http://www.bbc.co.uk/schools/gcsebit...tshirev1.shtml

http://www.mathcentre.ac.uk/resource...ers-2009-1.pdf
5. (Original post by uberteknik)
Have you not learned the basic maths rules for manipulating exponents, powers, roots etc?

You need to learn and practice these so that you can apply them to the problem.

http://www.bbc.co.uk/schools/gcsebit...tshirev1.shtml

http://www.mathcentre.ac.uk/resource...ers-2009-1.pdf
I have learnt those before, but I don't see why you need to say m^-2 rather than m^2
6. (Original post by blobbybill)
I have learnt those before, but I don't see why you need to say m^-2 rather than m^2
You don't need to write it out but you do need to perform the reduction nevertheless.
7. (Original post by uberteknik)
You don't need to write it out but you do need to perform the reduction nevertheless.
OK thanks. What is the reason as to why you need to perform the reduction with something as simple as m^2 to m^-2? I get that for longer equations it makes it much easier to reduce the powers, but I'm not sure why you need to in this case.
8. (Original post by blobbybill)
OK thanks. What is the reason as to why you need to perform the reduction with something as simple as m^2 to m^-2? I get that for longer equations it makes it much easier to reduce the powers, but I'm not sure why you need to in this case.
The units of I = C/s, n=m-3 , A=m2, v=m/s, q=C

so nAvq = m-3 x m2 x m/s x C

-3+2+1 = 0 so m has gone leaving C/s which is the same as the units of I.

n is the number per unit volume so has units of 1/m3 which is better written m-3
9. (Original post by GeddyBaby)
The units of I = C/s, n=m-3 , A=m2, v=m/s, q=C

so nAvq = m-3 x m2 x m/s x C

-3+2+1 = 0 so m has gone leaving C/s which is the same as the units of I.

n is the number per unit volume so has units of 1/m3 which is better written m-3
In this bit:
so nAvq = m-3 x m2 x m/s x C

-3+2+1 = 0 so m has gone leaving C/s which is the same as the units of I.
The equation at the bottom, the -3 comes from the m^-3, the +2 comes from the m^2 and the +1 comes from m/s, but why/how does the m/s end up being +1 in that equation at the bottom? And what happens to the s in m/s? WHere does that go?

Thanks, can you just explain how m/s ends up being +1 and where the s goes? I get and understand now the -3 and +2, just not the +1.
10. (Original post by blobbybill)
In this bit:

The equation at the bottom, the -3 comes from the m^-3, the +2 comes from the m^2 and the +1 comes from m/s, but why/how does the m/s end up being +1 in that equation at the bottom? And what happens to the s in m/s? WHere does that go?

Thanks, can you just explain how m/s ends up being +1 and where the s goes? I get and understand now the -3 and +2, just not the +1.
When a value has a power of 1 you don"t write that power down. Otherwise everything would have a power. For example when you learnt that 2 + 2 = 4 you didn't write it as
2¹ + 2¹ =4¹. You do have to practice the math of this.

The seconds don't "do" anything as there are no other seconds on the right hand side. So after the metres have cancelled you are left with C/s on both sides which is what the question asked you to show: it has the same units on both sides, it is homogeneous.

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