What is the rate at which the cowboy fills the horse trough?

[imnotsmart]

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?

 

[stunner5000pt]

$\mbox{mass rate flow = a constant =} \ R_{V}= Av$

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 $$R_{V} t = V$$

 

[imnotsmart]

How do I go about finding how long it takes.

 

[stunner5000pt]

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 $R_{V} = Av$
what are the UNITS of Rv?

by unit elimination you'll figure that
rate of flow x time = volume

 

[imnotsmart]

So 1.4 m/s * time= .387m
which =.28s?

 

[stunner5000pt]

READ EVERTYHING before you just jump to conclusion

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

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

 

[imnotsmart]

Don't think i totally sound stupid here but am i looking for the area of the hose?

 

[stunner5000pt]

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

 

[imnotsmart]

so 3.8e-4 is the area and the volume is .387
so the answer is 1018.1s?

 

[imnotsmart]

I figured it out...the answer is 12.1 min...thanks buddy