"The two equations below express conservation of energy and conservation of mass for water flowing from a circular hole of radiu
s 3 centimeters at the bottom of a cylindrical tank of radius 20 centimeters. In these equations, delta m is the mass that leaves the tank in time delta t, v is the velocity of the water flowing through the hole, and h is the height of the water in the tank at time t. g is the accelertion of gravity, which you should approximate as 1000 cm/s2 The first equation says that the gain in kinetic energy of the water leaving the tank equals the loss in potential energy of the water in the tank.
1/2 delta m v2 = delta m gh
The second equation says that the rate at which water leaves the tank equals the rate of decrease in the volume of water in the tank (which is conservation of mass because water has constant density)
pi 102^2 dh/dt = pi 3^2 v
Derive a differential equation for the height of water in the tank.
If the initial height of the water is 30 centimeters, find a formula or the solution.
According to the model, how long does it take to empty the tank?
Another way to solve this differential equation is to make the substitution w = underroot h. What is the differential equation that w satisfies?
The two equations below express conservation of energy and conservation of mass for water flowing from a circular hole of radius 3 centimeters at the bottom of a cylindrical tank of radius 10 centimeters. In these equations, delta m is the mass that leaves the tank in time delta t, v is the velocity of the water flowing through the hole, and h is the height of the water in the tank at time t. g is the acceleration of gravity, which you should approximate as 1000 cm/s2.
The process releases energy because the total mass of the resulting single nucleus is less than the mass of the two original nuclei. The leftover mass becomes energy.
