SOLUTION: two cylindrical water tanks stand side by side. one has a radius of 4 meters and contains water to a depth of 12.5 meter. the other has a radius of 3 meters and is empty. water is
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Question 868703: two cylindrical water tanks stand side by side. one has a radius of 4 meters and contains water to a depth of 12.5 meter. the other has a radius of 3 meters and is empty. water is pumped from the first tank to the second tank at a rate of 10 cubic meters per minute. how long, to the nearest tenth of minute, must the pump run before the depth of the water is the same in both tanks?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
two cylindrical water tanks stand side by side. one has a radius of 4 meters and contains water to a depth of 12.5 meter.
the other has a radius of 3 meters and is empty. water is pumped from the first tank to the second tank at a rate of 10 cubic meters per minute.
how long, to the nearest tenth of minute, must the pump run before the depth of the water is the same in both tanks?
:
Let t = time (in minutes) for the depth in both tanks is the same
:
Find the original volume of water in first tank
V =
V = 628.3 cu/m
The amt of water in the first tank as a function of time: (628.3-10t)
then d = depth
= (628.3-10t)
d =
:
The 2nd tank which is being filled, vol = 10t
= 10t
d =
:
the equation when the depths are equal
=
simplify,mult both by pi, we have:
=
cross multiply
16(10t) = 9(628.3-10t)
160t = 5654.7 - 90t
160t + 90t = 5654.7
250t = 5654.7
t = 5654.7/250
t = 22.6 minutes, the depths should be equal
:
:
See if that checks out by finding the actual depth of each tank
d = = 8.0 meters
and tank being filled
d = = 8.0 meters
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