Consider a very small hole in the bottom of a tank that is 17.0 cm in diameter and filled with water to a height of 90.0 cm. Find the speed at which the water exits the tank through the hole.
A) 48.3 m/s B) 44.1 m/s C) 17.64 m/s D) 4.20 m/s
To find the speed at which water exits the tank through the hole, we can use Torricelli's law, which states that the speed of fluid exiting a hole in a container is given by the equation:
v = √(2gh)
Where:
v is the speed of the fluid
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height of the fluid above the hole.
In this case, the height of the fluid above the hole is 90.0 cm, which is equal to 0.9 meters.
Now we can plug these values into the equation:
v = √(2 * 9.8 m/s^2 * 0.9 m)
v = √(17.64 m^2/s^2)
v ≈ 4.20 m/s
Therefore, the answer is D) 4.20 m/s.
To find the speed at which the water exits the tank through the hole, we can use Torricelli's Law, which states that the velocity of a liquid flowing out of an opening at the bottom of a container is given by the formula:
v = √(2gh)
Where:
v = velocity of the liquid exiting the tank
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height of the liquid above the opening
Given:
diameter of the tank (D) = 17.0 cm
radius of the tank (r) = D/2 = 17.0 cm ÷ 2 = 8.5 cm = 0.085 m
height of the water (h) = 90.0 cm = 0.9 m
Now, let's calculate the velocity using the formula:
v = √(2gh)
v = √(2 × 9.8 m/s² × 0.9 m)
v = √(17.64 m²/s²)
v ≈ 4.20 m/s
Therefore, the speed at which the water exits the tank through the hole is approximately 4.20 m/s.
The correct answer is D) 4.20 m/s.