Find the depth of water in a leaking tank which ejects water from the bottom of a small hole near the buttom Of the tank at a speed of 7.0 m/s
The depth of water in the tank can be found using the principle of conservation of energy.
Assuming the tank is open to the atmosphere, the potential energy of the water at the surface of the tank is converted into kinetic energy as it exits the small hole at the bottom. The formula for conservation of energy can be written as:
ΔPE + ΔKE = 0
where ΔPE is the change in potential energy and ΔKE is the change in kinetic energy.
The potential energy change can be given by:
ΔPE = mgh
where m is the mass of water ejected, g is the acceleration due to gravity, and h is the depth of water in the tank.
The kinetic energy change is given by:
ΔKE = 0.5mv^2
where m is the mass of water ejected and v is the velocity of water ejecting from the hole.
Since the changes in potential and kinetic energy must sum up to zero:
mgh + 0.5mv^2 = 0
Solving for h:
h = -0.5v^2/(g)
Given v = 7.0 m/s and g = 9.81 m/s^2, we can calculate the depth of water in the tank:
h = -0.5(7.0)^2 / (9.81)
h ≈ -2.5 m
Therefore, the depth of water in the tank is approximately 2.5 meters.