course Phy 121 ףҫRStudent Name:
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16:42:57 `q001. We obtain an estimate of the acceleration of gravity by determining the slope of an acceleration vs. ramp slope graph for an object gliding down an incline. Sample data for an object gliding down a 50-cm incline indicate that the object glides down the incline in 5 seconds when the raised end of the incline is .5 cm higher than the lower end; the time required from rest is 3 seconds when the raised end is 1 cm higher than the lower end; and the time from rest is 2 seconds when the raised end is 1.5 cm higher than the lower end. What is the acceleration for each trial?
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RESPONSE --> The average velocity can be found by dividing the length of the ramp by the time it takes to reach the end: 50cm/5s = 10cm/s. The final velocity must be twice the average velocity because the object started from rest, vo=0. So, final velocity is 20cm/s. Change in velocity is then 20cm/s. The acceleration is change in velocity over change in time: 20cm/s / 5s = 4cm/s^2. We follow the same process for the other cases and get these answers: For 3-second trial: vAve = 50cm/3s = 16.67cm/s vf = 2(16.67cm/s) = 33.33cm/s `dv = 33.33cm/s aAve = (33.33cm/s)/3s = 11.11 cm/s^2 For 2-second trial: vAve = 50cm/2s = 25cm/s vf = 2(25cm/s) = 50cm/s `dv = 50cm/s aAve = (50cm/s)/2s = 25cm/s^2
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16:47:11 We can find the accelerations either using equations or direct reasoning. To directly reason the acceleration for the five-second case, we note that the average velocity in this case must be 50 cm/(5 seconds) = 10 cm/s. Since the initial velocity was 0, assuming uniform acceleration we see that the final velocity must be 20 cm/second, since 0 cm/s and 20 cm/s average out to 10 cm/s. This implies a velocity change of 20 cm/second a time interval of 5 seconds, or a uniform acceleration of 20 cm/s / (5 s) = 4 cm/s^2. The acceleration in the 3-second case could also be directly reasoned, but instead we will note that in this case we have the initial velocity v0 = 0, the time interval `dt = 3 sec, and the displacement `ds = 50 cm. We can therefore find the acceleration from the equation `ds = v0 `dt + .5 a `dt^2. Noting first that since v0 = 0 the term v0 `dt must also be 0,we see that in this case the equation reduces to `ds = .5 a `dt^2. We easily solve for the acceleration, obtaining a = 2 `ds / `dt^2. In this case we have a = 2 * (50 cm) / (3 sec)^2 = 11 cm/s^2 (rounded to nearest cm/s^2). For the 2-second case we can use the same formula, obtaining a = 2 * (50 cm) / (2 sec)^2 = 25 cm/s^2.
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RESPONSE --> Ok.
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16:47:33 `q002. What are the ramp slopes associated with these accelerations?
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RESPONSE --> Slope = rise / run = height / length For 5-second trial: slope = 0.5cm / 50cm = 0.01cm For 3-second trial: slope = 1.0cm / 50cm = 0.02cm For 2-second trial: slope = 1.5cm / 50cm = 0.03cm
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16:50:39 For the 5-second trial, where acceleration was 4 cm/s^2, the 'rise' of the ramp was .5 cm and the 'run' was nearly equal to the 50-cm length of the ramp so the slope was very close to .5 cm / (50 cm) = .01. For the 3-second trial, where acceleration was 11 cm/s^2, the 'rise' of the ramp was 1 cm and the 'run' was very close to the 50-cm length, so the slope was very close to 1 cm / (50 cm) = .02. For the 2-second trial, where the acceleration was 25 cm/s^2, the slope is similarly found to be very close to 1.5 cm / (50 cm) = .03.
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RESPONSE --> Ok
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16:51:46 `q003. Sketch a reasonably accurate graph of acceleration vs. ramp slope and give a good description and interpretation of the graph. Be sure to include in your description how the graph points seem to lie with respect to the straight line that comes as close as possible, on the average, to the three points.
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RESPONSE --> When I sketched my graph of this experiment. The line I drew went almost directly through the first and last points which were (0.01, 4) and (0.03, 25). These two points were slightly above the line and the middle point (0.02, 11.11) was below it. The line would probably be a better average if it were a little lower because it would be between the points instead of almost exactly on the first and third.
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16:56:36 The graph will have acceleration in cm/s^2 on the vertical axis (the traditional y-axis) and ramp slope on the horizontal axis (the traditional x-axis). The graph points will be (.01, 4 cm/s^2), (.02, 11.1 cm/s^2), (.03, 25 cm/s^2). The second point lies somewhat lower than a line connecting the first and third points, so the best possible line will probably be lower than the first and third points but higher than the second. The graph indicates that acceleration increases with increasing slope, which should be no surprise. It is not clear from the graph whether a straight line is in fact the most appropriate model for the data. If timing wasn't particularly accurate, these lines could easily be interpreted as being scattered from the actual linear behavior due to experimental errors. Or the graph could indicate acceleration vs. ramp slope behavior that is increasing at an increasing rate.
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RESPONSE --> Ok
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16:57:06 `q004. Carefully done experiments show that for small slopes (up to a slope of about .1) the graph appears to be linear or very nearly so. This agrees with theoretical predictions of what should happen. Sketch a vertical line at x = .05. Then extend the straight line you sketched previously until it intersects the y axis and until it reaches past the vertical line at x = .05. What are the coordinates of the points where this line intersects the y-axis, and where it intersects the x =.05 line? What are the rise and the run between these points, and what therefore is the slope of your straight line?
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RESPONSE --> The straight line passes through the x = 0.05 line at about 43cm/s^2. So the point is (0.05, 43cm/s^2) The line also passes through the y-axis at about (0,-7cm/s^2) The slope of the line is therefore: slope = y2 -y1 / x2 - x1 = [43-(-7)cm/s^2] / .05 - 0 = 50cm/s^2 / 0.05 = 1000cm/s^2
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16:59:38 A pretty good straight line goes through the points (0, -6 cm/s^2) and (.05, 42 cm/s^2). Your y coordinates might differ by a few cm/s^2 either way. For the coordinates given here, the rise is from -6 cm/s^2 to 42 cm/s^2, a rise of 48 cm/s^2. The run is from 0 to .05, a run of .05. The slope of the straight line is approximately 48 cm/s^2 / .05 = 960 cm/s^2. Note that this is pretty close to the accepted value, 980 cm/second^2, of gravity. Carefully done, this experiment will give us a very good estimate of the acceleration of gravity.
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RESPONSE --> Ok
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17:01:42 `q005. The most accurate way to measure the acceleration of gravity is to use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) for the period of a pendulum. Use your washer pendulum and time 100 complete back-and-forth cycles of a pendulum of length 30 cm. Be sure to count carefully and don't let the pendulum swing out to a position more than 10 degrees from vertical. How long did it take, and how long did each cycle therefore last?
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RESPONSE --> The whole process took 110.6 seconds Therefore each cycle took 110.6/100 = about 1.1 seconds.
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17:03:07 100 cycles of a pendulum of this length should require approximately 108 seconds. This would be 108 seconds per 100 cycles, or 108 sec / (100 cycles) = 1.08 sec / cycle. If you didn't count very carefully or didn't time very accurately, you might differ significantly from this result; differences of up to a couple of cycles are to be expected.
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RESPONSE --> Ok
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17:06:25 `q006. You now have values for the period T and the length L, so you can use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) to find the acceleration g of gravity. Solve the equation for g and then use your values for T and L to determine the acceleration of gravity.
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RESPONSE --> T = 2 `pi / `sqrt(g) * `sqrt(L) Multiply both sides by `sqrt(g): T * `sqrt(g) = 2 `pi `sqrt(L) Divide both sides by T: `sqrt(g) = 2 `pi `sqrt(L) / T Sqaure both sides: g = 4 `pi^2 L / T^2 g = 4 `pi^2(30cm) / 1.1s^2 g = (1184)/(1.21) = about 979 cm/s^2
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17:11:38 Solving T = 2 `pi / `sqrt(g) * `sqrt(L) for g, we can first multiply both sides by `sqrt(g) to obtain T * `sqrt(g) = 2 `pi `sqrt(L). Then dividing both sides by T we obtain `sqrt(g) = 2 `pi `sqrt(L) / T. Squaring both sides we finally obtain {}g = 4 `pi^2 L / T^2. Plugging in the values given here, L = 30 cm and T = 1.08 sec, we obtain g = 4 `pi^2 * 30 cm / (1.08 sec)^2 = 1030 cm/s^2. You should check these calculations for accuracy, since they were mentally approximated.
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RESPONSE --> I understand this problem.
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ĕQ[zζoGᣱ assignment #004 ۶mSۥ} Physics I Vid Clips 06-12-2006
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20:26:40 Physics video clip 09 displacement for linear v vs. t graph: common sense, formula, area If we know the initial and final velocities over some time interval, and if the rate which velocity changes is constant, then how do we calculate the displacement over a the time interval?
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RESPONSE --> The displacement is the average velocity times the time interval. Since velocity increases at a constant rate, the average of the initial and final velocities is the average velocity. `ds = vAve x `dt = `ds =(vf + v0)/2 x `dt
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20:29:50 ** Displacement is the product of average velocity and time interval. Since acceleration is uniform average velocity is average of initial and final velocities. Displacement could therefore be calculated from the final and initial velocities => `ds =[(vf + v0)/2] * `dt. **
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RESPONSE --> Ok
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20:31:24 How do we use a graph of v vs. t to depict the calculation of the displacement over a time interval?
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RESPONSE --> The area under the graph represents the displacement of the object over the time interval. The midpoint of the graph represents the average velocity because it is between the initial and final velocities and change in velocity is constant. The time interval is obviously, the time duration from initial to final velocity. `ds = (midpoint)(time interval) `ds = (vAve)(`dt)
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20:34:34 ** Looking at the graph, we notice a trapezoid created by the slope line across the top, and an imaginary line drawn from the y value down to and perpendicular to the x axis. The area of this trapezoid represents the displacement or 'signed distance' the object travels. The displacement for any time interval can be found finding the average of the two 'altitudes' of the trapezoid, which represent initial and final velocities. Multiplying the average 'altitude' by the width is therefore equivalent to multiplying the approximate average velocity by the time interval, giving us the area of the trapezoid, which represents the approximate displacement. In the case where the graph is linear (which corresponds to uniform acceleration) the average of the two altitudes in fact represents the average velocity, and the result is the displacement, not the approximate displacement. **
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RESPONSE --> OK.
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20:35:27 What aspect of the graph gives the displacement during the time interval?
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RESPONSE --> The area under the graph represents change in position over the time interval.The average altitude which is the midpoint of the graph multiplied by the width which is the time interval will give us the displacement.
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20:38:09 ** STUDENT ANSWER: It is the average of the two sides that are as high as 'y' in each case, multiplied by the width-units of 'x'. **
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RESPONSE --> Ok
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[XxЖqˋޏ assignment #007 ۶mSۥ} Physics I 06-12-2006
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21:17:29 Describe the flow diagram you would use for the uniform acceleration situation in which you are given v0, vf, and `dt.
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RESPONSE --> First level: Given v0, vf, and `dt Second level: Subtract vo from vf to find `dv Add vf and vo together and divide by 2 to get the average velocity, vAve. Third level: Divide `dv by `dt to get acceleration, a. Multiply vAve by `dt to get displacement, `ds.
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21:19:35 ** We start with v0, vf and `dt on the first line of the diagram. We use vO and vf to find Vave, indicated by lines from v0 and vf to vAve. Use Vave and 'dt to find 'ds, indicated by lines from vAve and `dt to `ds. Then use `dv and 'dt to find acceleration, indicated by lines from vAve and `dt to a. **
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RESPONSE --> Ok
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21:21:56 Describe the flow diagram you would use for the uniform acceleration situation in which you are given `dt, a, v0
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RESPONSE --> First level: Given vo, a, `dt Second level: Multiply a and `dt to get change in velocity, `dv Third level: Add vo to `dv to get final velocity, vf Fourth level: Add vo and vf, then divide by 2 to get average velocity, vAve Fifth level: Multiply vAve by `dt to get displacement, `ds
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21:25:03 ** Student Solution: Using 'dt and a, find 'dv. Using 'dv and v0, find vf. Using vf and vO, find vave. Using 'dt and Vave, find 'ds. **
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RESPONSE --> Ok
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21:28:08 Explain in detail how the flow diagram for the situation in which v0, vf and `dt are known gives us the two most fundamental equations of motion.
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RESPONSE --> We can find change in velocity from the initial and final velocities. Then multiply change in velocity by time interval to get acceleration. When we have a, `dt, vo, and vf we can use the fundamental equation: vf = vo + a `dt We can also use vo and vf to find average velocity. Then we multiply `dt by average velocity to get displacement. After we have `ds, vo, vf, `dt, and vAve we can use this fundamental equation: `ds = (vf + vo) / 2 x `dt
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21:31:16 **Student Solution: v0 and vf give you `dv = vf - v0 and vAve = (vf + v0) / 2. `dv is divided by `dt to give accel. So we have a = (vf - v0) / `dt. Rearranging this we have a `dt = vf - v0, which rearranges again to give vf = v0 + a `dt. This is the second equation of motion. vAve is multiplied by `dt to give `ds. So we have `ds = (vf + v0) / 2 * `dt. This is the first equation of motion Acceleration is found by dividing the change in velocity by the change in time. v0 is the starting velocity, if it is from rest it is 0. Change in time is the ending beginning time subtracted by the ending time. **
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RESPONSE --> Ok
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21:36:14 Explain in detail how the flow diagram for the situation in which v0, a and `dt are known gives us the third fundamental equations of motion.
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RESPONSE --> For the third fundamental equations of motion all we need to know is the initial velocity, acceleration, and time interval. Then we can find displacement. `ds = (vo)(`dt) + 1/2(a)(`dt^2)
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21:36:46 ** a and `dt give you `dv. `dv and v0 give you vf. v0 and vf give you vAve. vAve and `dt give you `ds. In symbols, `dv = a `dt. Then vf = v0 + `dv = v0 + a `dt. Then vAve = (vf + v0)/2 = (v0 + (v0 + a `dt)) / 2) = v0 + 1/2 a `dt. Then `ds = vAve * `dt = [ v0 `dt + 1/2 a `dt ] * `dt = v0 `dt + 1/2 a `dt^2. **
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RESPONSE --> Ok
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21:37:02 Why do we think in terms of seven fundamental quantities while we model uniformly accelerated motion in terms of five?
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RESPONSE --> The five fundamental quanities are a, `dt, `ds, vo, and vf. Average velocity and change in velocity can be found from the other five quantities. These two quantities make up the rest of the seven. Average velocity and change in velocity are vital in solving problems for other quantities.
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21:37:48 ** ONE WAY OF PUTTING IT: The four equations are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. However to think in terms of meanings we have to be able to think not only in terms of these quantities but also in terms of average velocity vAve and change in velocity `dv, which aren't among these five quantities. Without the ideas of average velocity and change in velocity we might be able to use the equations and get some correct answers but we'll never understand motion. ANOTHER WAY: The four equations of unif accelerated motion are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. The idea here is that to intuitively understand uniformly accelerated motion, we must often think in terms of average velocity vAve and change in velocity `dv as well as the five quantities involved in the four fundamental equations. one important point is that we can use the five quantities without any real conceptual understanding; to reason things out rather than plugging just numbers into equations we need the concepts of average velocity and change in velocity, which also help us make sense of the equations. **
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RESPONSE --> Ok
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21:38:01 Accelerating down an incline through a given distance vs. accelerating for a given time Why does a given change in initial velocity result in the same change in final velocity when we accelerated down a constant incline for the same time, but not when we accelerated down the same incline for a constant distance?
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RESPONSE --> The change in velocity does not change, it is constant. Change in velocity can change if the time interval changes. So if the time interval does not change, the change in velocity is constant. If distance is the same but initial velocity changes, acceleration will change and this will result in variation in change in velocity.
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21:38:32 ** If we accelerate down a constant incline our rate of change of velocity is the same whatever our initial velocity. So the change in velocity is determined only by how long we spend coasting on the incline. Greater `dt, greater `dv. If you travel the same distance but start with a greater speed there is less time for the acceleration to have its effect and therefore the change in velocity will be less. You might also think back to that introductory problem set about the car on the incline and the lamppost. Greater initial velocity results in greater average velocity and hence less time on the incline, which gives less time for the car to accelerate. **
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RESPONSE --> OK
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