course phy 201 ?P????[???????w?assignment #015
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22:25:26 `q001. Note that this assignment contains 5 questions. . Suppose that a net force of 10 Newtons acts on a 2 kg mass for 3 seconds. By how much will the velocity of the mass change during these three seconds?
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RESPONSE --> Fnet = 10N m = 2kg dt = 3sec p=mv Fnet *dt = mvf - mv0 10N * 3sec = mvf-mv0 30N/s = mvf - mv0 30n/2kg 15 confidence assessment: 2
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22:27:07 The acceleration of the object will be accel = net force / mass = 10 Newtons / (2 kg) = 5 m/s^2. In 3 seconds this implies a change of velocity `dv = 5 m/s^2 * 3 s = 15 meters/second.
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RESPONSE --> this way makes more sense I was not sure of my answer but figured I could divide by m of one since both were multiplied by it. self critique assessment: 2
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22:28:48 `q002. By how much did the quantity m * v change during these three seconds? What is the product Fnet * `dt of the net force and the time interval during which it acted? How do these two quantities compare?
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RESPONSE --> fnet *dt = mvf - mv0 15m/s * 2kg 30 Fnet = 5 * 3 = 15 they are the same confidence assessment:
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22:31:55 The impulse of the force will be Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. Note that the 1200 kg mass has nothing to do with the magnitude of the impulse.STUDENT COMMENT: That's a little confusing. Would it work to take the answer I got of 3234 N and add back in the weight of the person at 647 N to get 3881? INSTRUCTOR RESPONSE: Not a good idea, though it works in this case. Net force = mass * acceleration. That's where you need to start with problems of this nature.Then write an expression for the net force, which will typically include but not be limited to the force you are looking for. *&*&
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RESPONSE --> impulse = Fnet * dt momentum = mvf-mv0 do not include mass in finding the impulse it is only includ . in momentum I accident. skipped and did not see this question self critique assessment: 2
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22:34:37 `q004. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be change in the velocity of the vehicle?
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RESPONSE --> F = 2000N m=1200kg dt = 1.5s F=ma a = f/m * dt a = 2000N /12000kg * 1.5s dv = 2.5m/s confidence assessment: 3
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22:36:43 The impulse of the 2000 Newton force is equal to the change in the momentum of the vehicle. The impulse is impulse = Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. The change in momentum is m * `dv = 1200 kg * `dv. Thus 1200 kg * `dv = 3000 kg m/s, so `dv = 3000 kg m/s / (1200 kg) = 2.5 m/s. In symbols we have Fnet * `dt = m `dv so that `dv = Fnet * `dt / m.
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RESPONSE --> i got the same answer i assume this is a different approach self critique assessment: 3
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22:42:19 `q005. Use the Impulse-Momentum Theorem to determine the average force required to change the velocity of a 1600 kg vehicle from 20 m/s to 25 m/s in 2 seconds.
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RESPONSE --> Fnet * dt = mvf-mv0 fnet = 1600*25 - 1600*20 / 2sec fnet= 4000 f=ma * dt a = dv/t a = 5m/s / 2sec a = 2.5m/s/s f = 1600kg * 2.5m/s * dt sec f = 4000 confidence assessment: 3
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22:42:49 The vehicle changes velocity by 5 meters/second so the change in its momentum is m * `dv = 1600 kg * 5 meters/second = 8000 kg meters/second. This change in momentum is equal to the impulse Fnet * `dt, so Fnet * 2 sec = 8000 kg meters/second and so {} Fnet = 8000 kg meters/second / (2 seconds) = 4000 kg meters/second^2 = 4000 Newtons. In symbols we have Fnet * `dt = m * `dv so that Fnet = m * `dv / `dt = 1600 kg * 5 m/s / ( 2 s) = 4000 kg m/s^2 = 4000 N.
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RESPONSE --> self critique assessment: 3
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