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A dairy farmer notices that a circular water trough near the barn has become rusty and now has a hole near the base. The hole is 0.38 m below the level of the water that is in the tank. If the top of the trough is open to the atmosphere, what is the speed of the water as it leaves the hole

Respuesta :

okiki5

Answer: 2.73m/s

Explanation:

Potential energy of the water is turned into kinetic energy

mgd = mv^2/2

Where m = mass of water

g = acceleration due to gravity = 9.81m/s^2

d = depth of the water = 0.38m

v = velocity of leaking water

m × 9.81 × 0.38 = mv^2/

eliminate m on both sides and multiply through by 2

7.4456 = v^2

v = sqrt(7.4456)

v = 2.73m/s

The speed of the water as it leaves the hole should be considered as the 2.73m/s.

Calculation of the speed:

Since

The Potential energy of the water should be turned into kinetic energy

So,

[tex]mgd = mv^2/2[/tex]

here

m = mass of water

g = acceleration due to gravity = 9.81m/s^2

d = depth of the water = 0.38m

v = velocity of leaking water

So,

[tex]m \times 9.81 \times 0.38 = mv^2[/tex]

Now we eliminate m on both sides and multiply by 2

So,

[tex]7.4456 = v^2\\\\v = \sqrt(7.4456)[/tex]

v = 2.73m/s

hence, The speed of the water as it leaves the hole should be considered as the 2.73m/s.

Learn more about speed here: https://brainly.com/question/24759048

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