course Phy 201 ??^??????????assignment #009
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21:37:51 `questionNumber 90000 Do the data seem to indicate, within the limits of the errors inherent in the experiment, that acceleration is the same on a constant incline regardless of where the cart is or how fast it is going?
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RESPONSE --> Despite fluctuations with velocities, within the limits of errors inherent to the experiemet, there appears to be consistency in acceleration.
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21:37:55 `questionNumber 90000 ** The usual results do show that the acceleration is the same, within about 10%, for any section of the ramp. Since different sections can have different lengths, which are associated with different average and final velocities, this indicates that within these limits the acceleration does appear to be constant. **
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RESPONSE --> Ok
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21:39:51 `questionNumber 90000 Where does the force that accelerates the cart come from?
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RESPONSE --> The force that accelerates the cart comes from gravitational force of Earth's pull. Increase in slope increased the gravitational force
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21:39:55 `questionNumber 90000 ** The force is the result of the gravitational force exerted on the cart by the Earth. On an incline a component of this force acts in the direction down the incline. The greater the incline the greater the proportion of the total gravitational force that acts down the incline. **
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RESPONSE --> Correct
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21:41:18 `questionNumber 90000 List all the forces that act on the cart, and discuss how they affect its motion.
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RESPONSE -->
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21:44:34 `questionNumber 90000 ** gravity pulls straight down, but the ramp isn't able to push straight up. That leaves a component of the gravitational force parallel to the ramp; the greater the slope the greater this component. Friction opposes motion. If there is a force holding the car back, then it needs to be included in the list, as it is in your list. **
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RESPONSE --> Ok, I was not sure where I could find my sources for this question
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21:48:04 `questionNumber 90000 Explain why a `ds is equal to the change in 1/2 v^2.
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RESPONSE --> 'ds is equal to the change in 1/2^2 as 'ds = (vf^2 - v0^2)/2 and so vf^2 - v0^2 is the change in v^2. Divide 'ds by 2 is equal to v^2.
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21:48:12 `questionNumber 90000 ** By the fourth equation of motion vf^2 = v0^2 + 2 a `ds. Thus a `ds = (vf^2 - v0^2) / 2. Since vf^2 - v0^2 is the change in v^2, it follows that a `ds is half the change in v^2. So a `ds is proportional to v^2. **
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RESPONSE --> Correct!
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21:51:11 `questionNumber 90000 Explain in detail why we expect Fnet `ds to be proportional to the change in v^2.
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RESPONSE --> Fnet * mass = a 'Fnet * 'ds = ' m* a * 'ds = ' v^2
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21:51:15 `questionNumber 90000 ** a `ds is proportional to the change in v^2 (see the preceding question), and Fnet * m = a. So the change in Fnet * `ds is the change in m a * `ds, which for a given object is proportional to the change in v^2. **
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RESPONSE --> Ok
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