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4 years ago

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A large, cylindrical water tank with diameter 3.00 m is on a platform 2.00 m above the ground. The vertical tank is open to the

air and the depth of the water in the tank is 2.00 m. There is a hole with diameter 0.420 cm in the side of the tank just above the bottom of the tank. The hole is plugged with a cork. You remove the cork and collect in a bucket the water that flows out the hole.
A. When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank? Express your answer in millimeters.
B. How long does it take you to collect 1.00 gal of water in the bucket? Express your answer in seconds.

1 answer:

Viefleur [7K]4 years ago

4 0

Answer:

1999.46 mm

45.59 s

Explanation:

given that

cylindrical water tank with diameter, D = 3 m

Height of the tank above the ground, h = 2 m

Depth of the water in the tank, d = 2 m

Diameter of hole, d = 0.420 cm

We start by calculating the volume of water in the tank, which is given as

Volume = πr²h

V = (πD²)/4 * h

V = (3.142 * 3²)/4 * 2

V = 28.278/4 * 2

V = 7.07 * 2

V = 14.14 m³

If 1.0 gal of water is equal to 0.0038m³, then

1 gal is 0.0038 = A * h

the area of the tank is 7.07 m²

therefore, 0.0038 = 7.07 * h

h₁ =0.00054 m = 0.54 mm is the height of water that flow out

the change in height of water in the tank = h - h₁ = 2 - 0.00054 = 1.99946 m

b)

Like we stated earlier, 1.0 gal of water is 0.0038m³

to solve this we use the formula

Q = Cd * A * √2gH

where Cd is a discharge coefficient, and is given by 0.9 for water

A is the area of the small hole

A = (πD²)/4

A = (π * 0.0042²)/4

A = 5.54*10^-5 / 4

A = 1.39*10^-5 m²

H= height of the hole from the tank water level = 2m - 0.0042 = 1.9958 m

g = 9.8 m/s²

Q = 0.9 * 1.39*10^-5 m² * √2 * 9.8 * 1.9958

Q = 1.251*10^-5 * 6.25

Q = 7.82*10^-5 m³/s

Q = V/t

t = V/Q = 0.0038m³ / 7.82*10^-5 m³/s

t = 45.59 s

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