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Math Help - Momentum of water from a hose

  1. #1
    Member alexgeek's Avatar
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    Momentum of water from a hose

    (think this is the right section).

    I'm stuck on a question in a physics paper:

    Water of density 1000kg\,m^{-3} flows out of a garden hose of cross-sectional area 7.2\times 10^{-4} m^2 at a rate of 2.0 \times 10^{-4} m^3 per second. How much momentum is carried by the water leaving the hose per second?

    I think it may have something to with F=ma and F \Delta t = \Delta mv. Maybe somehow using the rate of flow to work out acceleration and density to find mass but I'm clutching at straws there.

    Thanks

    edit, oh and probably \rho = \frac{m}{V}
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  2. #2
    A Plied Mathematician
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    I don't think you need Newton here. You just need your definition of momentum, as well as a little dimensional analysis to solve this problem. Think this way:

    \dfrac{p}{\text{sec}}=\dfrac{m}{\text{sec}}\,v=\le  ft(\dfrac{\text{mass of water}}{\text{sec}}\right)(\text{velocity of water})=\left(\rho\,\dfrac{\text{volume of water}}{\text{sec}}\right)\left(\dfrac{\text{volum  e of water}}{\text{sec}}\,\dfrac{1}{\text{area}}\right)  .

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  3. #3
    Member alexgeek's Avatar
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    I'm with you on:
     \frac{mv}{\Delta t} = \frac{\rho V}{\Delta t}v \because m = \rho V

    I'm not really sure what you're doing after that.

    Are you saying this in the last bit?  v = \frac{V}{\Delta t} \frac{1}{A}
    Thank you
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  4. #4
    Member alexgeek's Avatar
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    Oh wait, looking at the ratios of the units that does make sense, I just would of never have thought of it.
    Think I get it now.
    Thanks!
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  5. #5
    A Plied Mathematician
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    You're welcome. Have a good one!
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