Question

7. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The oncoming water stream has a velocity of 16.0 m/sec, while the outgoing water stream has a velocity of 16.0 m/sec in the opposite direction. The mass of water per second that strikes the blade is 30.0 kg/sec. Find the magnitude of the average force exerted on the water by the blade.

Answer 1

The magnitude of the average force exerted on the water by the blade is 960 N.

Step-by-step explanation:

Given that,

The mass of water per second that strikes the blade is,
`(m)/(t)=30\ kg/s`

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=(\Delta P)/(\Delta t)`

`F=(m(v-u))/(\Delta t)`

`F=30\ kg/s* (-16-16)\ m/s`

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.