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 0.29 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

2.385 m/s

Explanation:

Here there is a transformation of potential energy of the water at the top to Kinectic energy at the hole

This can be expressed as

Potential energy= Kinect energy

mgh= 1/2m v²

Where m= mass of water

h= Lenght below the level of the water that is in the tank.

v= the speed of the water as it leaves the hole

g= acceleration due to gravity= 9.81m/s

Here there is 'm' at both sides which can cancelled beach other out.

Then we have

gh= ¹/₂v²

making v² subject of the formular

v²=gh/(¹/₂)

v=√(2gh)

V=√2g h

V= √(2×9.81×0.29)

V=2.385 m/s

Therefore, the speed of the water as it leaves the hole is 2.385 m/s

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this