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mmh A stream of water strikes a stationary turbine blade horizon- tally, as the drawing illustrates. The incident water stream has a velocity of 16.0 m/s, while the exiting water stream has a velocity of 16.0 m/s. The mass of water per second that strikes the blade is 30.0 kg/s. Find the magnitude of the average force exerted on the water by the blade.

960 $\mathrm{N}$

Physics 101 Mechanics

Chapter 7

Impulse and Momentum

Moment, Impulse, and Collisions

Rutgers, The State University of New Jersey

University of Washington

Hope College

McMaster University

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all right in question Number 10 we have a stream of water that strikes a stationary term mind blade horizontally, and there's a picture in the text. And then what we're asked to find is the magnitude of the average force exerted on the water by the blade. And one key piece of of learning from this problem or one trick to this problem is that it tells us the mass of water per second that strikes the latest 30 kilograms per second. So we're going to assume for our problems sake that this is all happening in it one second interval. So that's gonna help us to be able to solve this. Okay, let's get started. So we're given in this problem that the water has an initial velocity. So the I of 16 meters per second and when it strikes this blade, it comes back with a final velocity of negative 16 meters per second. So it's moving exactly in the opposite direction once it's done. And like you mentioned, we're gonna talk about this happening in a time interval of one second. So if we 1.0 seconds that allows us to identify the mass of the water that's moving in that one second to be 30 kilograms. The reason that we want to do this is because we can. We now can use the impulse momentum zero to solve this problem so we don't impose momentum. Theorem is f Delta t equals change in momentum. Delta P and that force is what we're looking for. So that is our unknown in this problem. Okay, so we have force times Delta t equals Delta P. We're and divide delta T over so that our force equals P final minus p initial, which is our Delta p divided by Delta t Our mass is going to be the 30 kilograms because we define this as a one second interval. I'm going right out this portion, though, so remember that momentum equals mass times velocity. So our momentum final would be mass times velocity final our momentum initial mass times velocity initial divided by Delta T because mass is the same both we can factor that out, which gives us force equals mass times The final My ass could be initial all over the time interval Delta T. Okay, So we have solved this as far as we can go out to break Lee so now we're ready to substitute in numbers. So he said the mass was 30 kilograms that travels by in one second. The V final is a negative 16 meters per second minus the initial velocity which is a positive 16 meters per second all in our time interval of one second and we get an average velocity exerted of 960 Newtons.

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