# A Water Tank That Needs Cleaning Mastering Physics

## 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

I rearranged

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)

Connection value:

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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