Question

A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air. Find the speed at which the water shoots out of the tank

Answer 1

The speed at which water shoots out of the hole in the tank is approximately 16.57 meters per second, calculated using Torricelli's theorem and the given height of 14.0 meters.

Step-by-step explanation:

The speed at which the water shoots out of the tank can be determined using Torricelli's theorem, which states that the speed of efflux under the action of gravity is the same as the speed that a body (in this case, a droplet of water) would acquire in free falling from a height equal to the depth of the orifice below the free surface of the liquid. The formula for the speed of efflux (v) is given by:

v = √(2gh), where g is the acceleration due to gravity (9.81 m/s²) and h is the height of the water level above the hole (14.0 m).

To find the speed, plug in the known values:

v = √(2 * 9.81 m/s² * 14.0 m) = √(274.44 m²/s²) = 16.57 m/s

Therefore, the speed at which the water shoots out of the hole in the tank is approximately 16.57 meters per second.

Answer 2

Answer: 16.6 m/s

Explanation: To get the speed of the efflux we will use the Torricelli‘s equation which states that the speed of the efflux through a small circular hole is given as
v = Square Root 2gh

After inserting the given variables one gets the following efflux velocity
v= square root 2 x 9.81 x 14
v= Square Root 276.54 = 16.6 m/s