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Mass falls in unit time

  1. Dec 11, 2008 #1
    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. jcsd
  3. Dec 12, 2008 #2


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    Welcome to PF!

    Hi physiker99! Welcome to PF! :smile:

    Yes. :smile:
    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.
  4. Dec 12, 2008 #3
    I did a typo there. I don't understand how velocity is used to calculate mass.
  5. Dec 12, 2008 #4
    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:
    [tex] Sv = S\frac{dr}{dt} = \frac{dV}{dt}[/tex]
    , 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:
    [tex]\rho \frac{dV}{dt} = \frac{d\rho V}{dt} = \frac{dM}{dt}[/tex]
    , with M the mass
    Last edited: Dec 12, 2008
  6. Dec 12, 2008 #5


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    Welcome to PF!

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

    oh, and welcome to PF, D.R.! :smile:
  7. Dec 12, 2008 #6
    It's understood. Thanks a lot.
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