water flows smoothly from the left-hand pipe section (radius 2.00R), through the middle section (radius R), and into the right-hand section (radius 3.00R). The speed of the water in the middle section is 0.460 m/s. What is the net work done on 0.430 m3 of the water as it moves from the left-hand section to the right-hand section?
To find the net work done on the water as it moves from the left-hand section to the right-hand section, we need to consider the change in the kinetic energy of the water.
The equation for the change in kinetic energy is given by:
ΔKE = (1/2) * m * (v^2)
Where:
ΔKE is the change in kinetic energy
m is the mass of the water
v is the velocity of the water
First, we need to find the mass of the water. We can do this by multiplying the volume of water by its density:
m = ρ * V
Where:
m is the mass of the water
ρ is the density of the water
V is the volume of the water
Given that the volume of the water is 0.430 m^3, we need to determine the density of water. The density of water is approximately 1000 kg/m^3.
m = 1000 kg/m^3 * 0.430 m^3
m ≈ 430 kg
Now, we can calculate the change in kinetic energy. The velocity of the water in the middle section is given as 0.460 m/s. We will use this value to determine the change in kinetic energy.
ΔKE = (1/2) * 430 kg * (0.460 m/s)^2
ΔKE ≈ 45.37 J
Therefore, the net work done on 0.430 m^3 of water as it moves from the left-hand section to the right-hand section is approximately 45.37 Joules.