Showing the Terminal Velocity equation is dimensionally correct.

[Plebert]

Hey guys, this is my er...first post.
It's a first year university physics assignment that I'm having a bit of trouble with, any help will be rewarded with kind words!(bit of an empty gift, but it's all I have)

Ok, here's the problem.

The terminal velocity of a mass m, moving at ‘high speeds’ through a fluid of density ρ (kg m^3), is given by
V(terminal)=√((2mg)/(DρA))

where A is the cross-sectional area of the object (m2) and D is a dimensionless “drag coefficient”.
Show that equation is dimensionally correct.

Now, not really being certain what the question is asking for regards 'dimensions' hasn't helped but! I did make an attempt by substituting each variable with it's corresponding units.
e.g.

2mg= 2((m/s^2)x(kg))=((m x kg)/ s^2)and ρA=((Kg/m^3)x(m^2))=Kg x m^(-1)

which yields V(ter)=√((mKg)/ s^2)/mKg
=√(s^2) x D
=s x D

This seems more or less nonsensical.
I'm sure it's probably mathematical error or just a failure to grasp the concept of proving an equations dimensions.

Am I wrong?
what is going on?

 

[vela]

Hey guys, this is my er...first post.
It's a first year university physics assignment that I'm having a bit of trouble with, any help will be rewarded with kind words!(bit of an empty gift, but it's all I have)

Problems from introductory courses don't belong in the advanced physics homework forum. I moved your thread.

Ok, here's the problem.

The terminal velocity of a mass m, moving at ‘high speeds’ through a fluid of density ρ (kg m^3), is given by
V(terminal)=√((2mg)/(DρA))

where A is the cross-sectional area of the object (m2) and D is a dimensionless “drag coefficient”.
Show that equation is dimensionally correct.

Now, not really being certain what the question is asking for regards 'dimensions' hasn't helped but! I did make an attempt by substituting each variable with it's corresponding units.
e.g.

2mg= 2((m/s^2)x(kg))=((m x kg)/ s^2)and ρA=((Kg/m^3)x(m^2))=Kg x m^(-1)

Fine up to here. Your dimensions for ρA are therefore kg/m, right?

which yields V(ter)=√((mKg)/ s^2)/mKg

You used kg m instead of kg/m for the dimensions of ρA.

=√(s^2) x D
=s x D

This seems more or less nonsensical.
I'm sure it's probably mathematical error or just a failure to grasp the concept of proving an equations dimensions.

Am I wrong?
what is going on?