C) electrical energy is transformed into heat energy
Answer:
The magnitude of the average force exerted on the water by the blade is 960 N.
Explanation:
Given that,
The mass of water per second that strikes the blade is, 
Initial speed of the oncoming stream, u = 16 m/s
Final speed of the outgoing water stream, v = -16 m/s
We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :



F = -960 N
So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.
We need a and we have m and F . Now a = f÷m so therefore a = 4,9 ÷ 0,5 which is 0,98 metres per cubic second
Change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right
Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s
Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right