A Water Tank That Needs Cleaning Mastering Physics
Clean water tank, physical problem.
Open cylindrical tanks need cleaning. The tank is filled with water up to a height of h0 = 1.00 m. Therefore, you decide to drain the water from time to time through a hole in the side of the tank near the bottom. The cross section of the tank is A1 = 0.785 square meters, the opening width is A2 = 0.002 square meters.
How long does it take t1 / 2 to empty half the tank? (Note: is a useful primitive Â Â "xÃ Â'1 / 2dx = 2x1 / 2.)
Express your answers digitally in seconds. g = 9.81 meters per second Consider the speed of free fall due to gravity per second.
The speed of emptying the tank.
Q = dV / s = v * A2
Where Q m / s flows. it is
And the rate of change in the volume of water in the dV / dt tank.
And v is the exit speed.
dV / s = A1 * dh / s
Since the sum is cylindrical and A1 is constant, then
A1 * dh / s = v * A2
It could be that v = ÂˆÂš (2gh), so adding it up gives us.
A1 * dh / s = ÂˆÂš (2gh) * A2
dh / ÂÂšh = (A2 / A1) * ÂˆÂš (2g) * s
Connect both sides:
ÂˆÂ "[, h] dh / ÂˆÂšh = (A2 / A1) ÂˆÂš (2g) ÂˆÂ" [0, t] s
2ÃÂšh | [, h] = (A2 / A1) (2g) t | [0, T]
2 (Ã Âšh ÂÂš) = (A2 / A1) Âš (2g) * t
t = 2 (1 Âš Âšh) (A1 / A2) / ÂˆÂš (2g) = ÂˆÂš2 * (g Âš (/ g) ÂˆÂš (h / g)) (A1 / A2)
t = Âš2 * (Âš Âš (1.00 m / 9.81 m / s²) Âº (0.50 m / 9.81 m / s²)) (0.785 m / 0.002 m)
t = 51.9 s Â - Â "
Bonus: How long does it take to empty the tank?
t = Â ± 2 * Â Â (1.00 m / 9.81 m / s²) * (0.785 m / 0.002 m)
t = 177 s
This means that it takes 125 seconds to empty the bottom of the tank.
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