# Homework Help: Mass falls in unit time

1. Dec 11, 2008

### physiker99

Assume there is an area S. Then we have dust particles falling over it having velocity v and density rho.
Is is correct to state mass falling over the S in unit time equals (rho)*(S)*v?

I guess it should be correct, because mass shouldn't have velocity component if none other cancels it.

2. Dec 12, 2008

### tiny-tim

Welcome to PF!

Hi physiker99! Welcome to PF!

Yes.
If you mean that, dimensionally, the 1/L3 in the M/L3 in rho has to be converted to 1/T, then that's correct.

3. Dec 12, 2008

### physiker99

I did a typo there. I don't understand how velocity is used to calculate mass.

4. Dec 12, 2008

### D.R.

It's not purely to calculate the mass. It also determines the rate per unit time through an area S. So you can rewrite S and v as:
$$Sv = S\frac{dr}{dt} = \frac{dV}{dt}$$
, with V a volume

The physical interpretation is that a volume V moves through an area S in dt time. See it as moving a cube through a square hole in dt time. The density is defined as mass per volume. So you can rewrite:
$$\rho \frac{dV}{dt} = \frac{d\rho V}{dt} = \frac{dM}{dt}$$
, with M the mass

Last edited: Dec 12, 2008
5. Dec 12, 2008

### tiny-tim

Welcome to PF!

I'll just add this to what D.R. says:

dimensionally, rho is M/L3, S is L2, and V is L/T

so multiply them all together, and you get M/L3 x L2 x L/T, = M/T, which is mass per time.

oh, and welcome to PF, D.R.!

6. Dec 12, 2008

### physiker99

It's understood. Thanks a lot.