Asked by Eriko
A rectangular trough is 8ft long, 2ft across the top, and 4ft deep. if water flows in at the rate 2 ft^3/min, how fast is the surface rising when the water is 1 ft deep?
CAN ANYONE PLEASE GIVE ME SOME IDEAS TO DO IT???????THANKS A LOT!!!!!
CAN ANYONE PLEASE GIVE ME SOME IDEAS TO DO IT???????THANKS A LOT!!!!!
Answers
Answered by
bobpursley
what is the shape of the trough at the bottom?
Answered by
Damon
change in volume of water * area of surface of water * dH
where dH is the change in depth
therefore
flow rate =dV/dt = area of surface of water * dH/dt
or
dH/dt = dV/dt / area of water surface
the depth has noting to do with it. Only the surface area matters.
where dH is the change in depth
therefore
flow rate =dV/dt = area of surface of water * dH/dt
or
dH/dt = dV/dt / area of water surface
the depth has noting to do with it. Only the surface area matters.
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