course Phys201 sorry once again, for not sending this all at once...if I do any more of these tonight I will continue to use this format, however I will try not to make this same mistake again. I am just afriad you will not receive some of my material.
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16:17:01 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
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RESPONSE --> Force = mass * acc Fnet is the total force exerted. confidence assessment: 3
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16:17:14 ** Net force is the sum of all forces acting on an object. If you're pushing your car you are exerting a force, friction is opposing you, and the let force is the sum of the two (noting that one is positive, the other negative so you end up with net force less than the force you are exerting). Your heart rate responds to the force you are exerting and the speed with which the car is moving; the accel of the car depends on the net force. **
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RESPONSE --> self critique assessment: 3
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16:18:47 In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds?
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RESPONSE --> acceleration * change in distance is proportional to the change in velocity ^2, because to obtain acceleration you need change in velocity over distance. the change in v^2 will be proportional to the net force and change in time because Fnet `ds = k if there is a change in v^2, for the appropriate k (specifically for k = mass / 2.) Force = mass of the object * its acceleration. confidence assessment: 2 It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for k = 1/2. Now since Fnet = m a we see that Fnet is proportional to a for a given mass m, and it follows that Fnet `ds = k * change in v^2, for the appropriate k (specifically for k = mass / 2. ** confidence assessment: 3
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16:19:26 ** It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for k = 1/2. Now since Fnet = m a we see that Fnet is proportional to a for a given mass m, and it follows that Fnet `ds = k * change in v^2, for the appropriate k (specifically for k = mass / 2. **
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RESPONSE --> self critique assessment: 2
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16:20:14 How do our experimental results confirm or refute this hypothesis?
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RESPONSE --> They confirm the hypothesis, because I knew my dominant hand (hand I threw with, shoot with, write with, point with) is used alot more, and I felt that it was quicker. The results proved me correct because it was .1 seconds faster in the trials. confidence assessment: 3
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16:21:19 ** We didn't actually do this part of the experiment, but on a ramp with fixed slope a `ds is simply proportional to `ds. When we measured `dt for different distances from rest down the same ramp, we were then able to determine the average and final velocities. The change in v^2 for each timing would be from 0 to vf^2. The change would therefore be just vf^2. If a `ds is proportional to the change in vf^2 then a graph of vf^2 vs. a `ds should be linear. Since a is constant we don't even need to determine it--a graph of vf^2 vs. `ds would be linear. This would confirm the hypothesis. **
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RESPONSE --> self critique assessment: 3
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course Phys201 sorry once again, for not sending this all at once...if I do any more of these tonight I will continue to use this format, however I will try not to make this same mistake again. I am just afriad you will not receive some of my material.
......!!!!!!!!...................................
16:17:01 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
......!!!!!!!!...................................
RESPONSE --> Force = mass * acc Fnet is the total force exerted. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:17:14 ** Net force is the sum of all forces acting on an object. If you're pushing your car you are exerting a force, friction is opposing you, and the let force is the sum of the two (noting that one is positive, the other negative so you end up with net force less than the force you are exerting). Your heart rate responds to the force you are exerting and the speed with which the car is moving; the accel of the car depends on the net force. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
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16:18:47 In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds?
......!!!!!!!!...................................
RESPONSE --> acceleration * change in distance is proportional to the change in velocity ^2, because to obtain acceleration you need change in velocity over distance. the change in v^2 will be proportional to the net force and change in time because Fnet `ds = k if there is a change in v^2, for the appropriate k (specifically for k = mass / 2.) Force = mass of the object * its acceleration. confidence assessment: 2 It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for k = 1/2. Now since Fnet = m a we see that Fnet is proportional to a for a given mass m, and it follows that Fnet `ds = k * change in v^2, for the appropriate k (specifically for k = mass / 2. ** confidence assessment: 3
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16:19:26 ** It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for k = 1/2. Now since Fnet = m a we see that Fnet is proportional to a for a given mass m, and it follows that Fnet `ds = k * change in v^2, for the appropriate k (specifically for k = mass / 2. **
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RESPONSE --> self critique assessment: 2
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16:20:14 How do our experimental results confirm or refute this hypothesis?
......!!!!!!!!...................................
RESPONSE --> They confirm the hypothesis, because I knew my dominant hand (hand I threw with, shoot with, write with, point with) is used alot more, and I felt that it was quicker. The results proved me correct because it was .1 seconds faster in the trials. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:21:19 ** We didn't actually do this part of the experiment, but on a ramp with fixed slope a `ds is simply proportional to `ds. When we measured `dt for different distances from rest down the same ramp, we were then able to determine the average and final velocities. The change in v^2 for each timing would be from 0 to vf^2. The change would therefore be just vf^2. If a `ds is proportional to the change in vf^2 then a graph of vf^2 vs. a `ds should be linear. Since a is constant we don't even need to determine it--a graph of vf^2 vs. `ds would be linear. This would confirm the hypothesis. **
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RESPONSE --> self critique assessment: 3
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