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Cowboy filling trough problem

  1. Apr 11, 2005 #1
    A cowboy at a dude ranch fills a horse trough that is 1.7 m long, 65 cm wide, and 35 cm deep. He uses a 2.2 cm diameter hose from which water emerges at 1.4 m/s. How long does it take him to fill the trough?

    I know the trough's volume is .387m^3.
    How do I find the rate at which the water fills?
     
  2. jcsd
  3. Apr 11, 2005 #2
    [itex] \mbox{mass rate flow = a constant =} \ R_{V}= Av [/itex]

    A is the corss sectional area, p is the density of the fluid (water = 1), and v is the velocity at the flow is going
    Rv is going to come in m^3/s
    now you have the Volume in m^3, and the flow is m^3/s
    then obviously [tex] R_{V} t = V [/tex]
     
  4. Apr 11, 2005 #3
    How do I go about finding how long it takes.
     
  5. Apr 11, 2005 #4
    :uhh: didnt u read??
    perhaos i was being UNCLEAR

    the RATE OF FLOW is given by the Cross section area from which the fluid comes out from times the VELOCITY at whiuch the fluid flows out

    thus [itex] R_{V} = Av [/itex]
    what are teh UNITS of Rv???

    by unit elimination you'll figure that
    rate of flow x time = volume
     
  6. Apr 11, 2005 #5
    So 1.4 m/s * time= .387m
    which =.28s?
     
  7. Apr 11, 2005 #6
    READ EVERTYHING before you just jump to conclusion

    what is the rate of flow(given as Rv)?? [itex] R_{V} = \mbox{AREA x VELOCITY} [/itex]

    and then the RATE OF FLOW times the time gives you the volume

    that is Rv (the rate of flow) x t (time) = Volume
    isolate for time (t) and solve
     
  8. Apr 11, 2005 #7
    Don't think i totally sound stupid here but am i looking for the area of the hose?
     
  9. Apr 11, 2005 #8
    area of the cross section i.e. if you cut the hose like it was a sausage that was being cut horizontally then find the area of that circular face
     
  10. Apr 11, 2005 #9
    so 3.8e-4 is the area and the volume is .387
    so the answer is 1018.1s?
     
    Last edited: Apr 11, 2005
  11. Apr 11, 2005 #10
    I figured it out...the answer is 12.1 min...thanks buddy
     
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