Water drives a waterwheel (or turbine) of radius 3.5 m. The water enters at a speed v_1 = 7.5 m/s and exits from the waterwheel at a speed v_2 = 4.0 m/s.

a) If 84 kg of water passes through per second, what is the net torque?
b) If the water causes the waterwheel to make one revolution every 5.3 s, how much power is delivered to the wheel?

Force = change in momentum/ change in time

Force = F= 84 kg/s *(7.5 -4)m/s = 294 Newtons
Torque = force*radius =F R = 294*3.5 = 1029 Newton meters

Power = force *distance/time =F*2piR/time = torque *2 pi/time = 2 pi * 1029/5.3 = 1212 watts

To solve this problem, we need to use the principles of torque and power.

a) To find the net torque, we need to calculate the torque applied by the incoming water and subtract the torque exerted by the outgoing water.

The torque applied by the incoming water is given by:
τ1 = Fr1, where F is the force exerted by the water on the waterwheel and r1 is the radius of the waterwheel.

To find the force F, we can use Newton's second law: F = m * a, where m is the mass of water passing through per second and a is its acceleration.

The acceleration a can be calculated using the change in velocity (v2 - v1) and the time taken for the water to pass through the wheel:
a = (v2 - v1) / t, where t is the time taken for one kilogram of water to pass through the wheel.

Since the mass of water passing through per second is given as 84 kg/s, the time taken for one kilogram of water to pass through is 1 / 84 s.

Substituting the values, we can find the acceleration:
a = (4.0 m/s - 7.5 m/s) / (1 / 84 s) = -288 m/s^2 (negative sign indicates acceleration opposite to the motion)

Now we can calculate the force exerted by the water:
F = m * a = 84 kg * (-288 m/s^2) = -24,192 N (negative sign indicates force in the opposite direction)

Finally, we can calculate the torque applied by the incoming water:
τ1 = F * r1 = -24,192 N * 3.5 m = -84,672 Nm (negative sign indicates torque opposite to the direction of rotation)

Similarly, we can calculate the torque exerted by the outgoing water:
τ2 = F * r2, where r2 is the radius at which the water exits the waterwheel.

Since the waterwheel is a closed system and the water enters and exits at different radii, the outgoing torque will be opposite in direction to the incoming torque.

Therefore, the net torque is given by:
Net Torque = τ1 - τ2 = -84,672 Nm - (-24,192 Nm) = -60,480 Nm

So, the net torque exerted on the waterwheel is -60,480 Nm.

b) To find the power delivered to the waterwheel, we use the formula:
Power = Torque * Angular Speed

First, let's calculate the angular speed of the waterwheel.

The waterwheel makes one revolution in 5.3 seconds, which means it completes one full circle in that time.

So, the angular speed (ω) is given by:
ω = 2π / t, where t is the time taken for one revolution.

Substituting the value, we get:
ω = 2π / 5.3 s ≈ 1.186 rad/s

Now we can calculate the power:
Power = Net Torque * Angular Speed
Power = -60,480 Nm * 1.186 rad/s ≈ -71,741.28 W

The negative sign indicates that power is being extracted from the waterwheel.

Therefore, the power delivered to the waterwheel is approximately -71,741.28 Watts.