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2 years ago

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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.

1 answer:

NNADVOKAT [17]2 years ago

6 0

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, $\dfrac{m}{t}=30\ kg/s$

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 :

$F=\dfrac{\Delta P}{\Delta t}$

$F=\dfrac{m(v-u)}{\Delta t}$

$F=30\ kg/s\times (-16-16)\ m/s$

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.

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