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course Phy 201
`q001. If the area of a 'graph rectangle' is 30 meters and its altitude is 5 meters / second, what is its width?Area = altitude * width
30 m = 5m/s * width
Width = 6 s
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If the area of a 'graph rectangle' is 60 Newton * meters and its width is 10 meters, what is its 'graph altitude'?
Area = altitude * width
60 N * m = altitude * 10 meters
Altitude = 6
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If the slope of a 'graph trapezoid' is 40 Newtons / meter and its width is 2 meters, what is its rise?
80 N/ 2 m = 40 N/m
80 Newtons
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If the rise of a 'graph trapezoid' is 50 meters / second and its slope is 200 meters / second^2, what is its width?
200 m/s^2 = 50m/s * run
0.25 s = width
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If the area of a 'graph trapezoid' is 30 meters, its average altitude is 5 meters / second and its slope is 10 meters / second^2, then what is its width and what are its altitudes?
30 m = 5m/s * width
Width = 6 s
Slope = rise / run
The altitudes are 7m and 3m
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Those altitudes would give you a slope of 4 m / (6 s) = .67 m/s, not 10 m/s^2.
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Interpret the preceding question in terms of the motion of an object, specifying its initial, average and final velocities, the change in velocity, the displacement, its acceleration and the time interval.
The object is traveling at the slope of 10 m/s^2. The initial v = 0 the final v= 10 and the Vave = 5. The change in Velocity is from 0 to 10 m/s. Displacement is 6 seconds. The acceleration is 5 m/s / 6 s = 0.8333 m/s^2. The time interval was from 0 seconds to 6 seconds
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10 m/s^2 would be the rate at which velocity changes. You don't travel at 10 m/s^2, velocity changes at 10 m/s^2.
Displacement is change in position, which isn't measured in seconds.
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The five quantities associated with a 'graph trapezoid' are its two altitudes, its area, its slope and its width. If we know any three of these quantities we can find the other two. List all possible combinations of these five quantities.
7m/s, 3m/s, 10m/s^2
7m/s, 3m/s, 30m
7m/s, 3m/s, 6s
7m/s, 10m/s^2, 6s
7m/s, 30m, 6s
3m/s, 30m, 6s
3m/s, 30m, 10m/s^2
30m, 6s, 10m/s^2
30m, 10m/s^2, 7m/s
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Explain how you would find the area and slope of a 'graph trapezoid', given its two altitudes and its width.
Area = altitude * width
Slope is the rise / run or change in y / change in x
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Your statements are OK, but neither is given in terms of the two altitudes and the width.
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Explain how you would find the area and width of a 'graph trapezoid', given its two altitudes and its slope.
Slope is rise / run
The run can be used along with drawing a graph to figure out the width.
The altitudes need to be averaged and multiplied by the width to get the area.
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Your third statement is specific and correct.
Your second statement doesn't quite specify the calculation in terms of the two altitudes and the slope. You mention a graph but don't say how it might be used.
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Explain how you would find the area and 'right altitude' of a 'graph trapezoid', given its 'left altitude', its slope and its width.
Area is altitude * width. The slope tells you the rise / run. The given width * the ave altitude = area. Take the ave altitude and subtract the left altitude and whatever you get add that back to the left altitude to get the right altitude.
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'The given width * the ave altitude = area' is a correct statement. You don't say how to get the average altitude.
Your last sentence would be correct if you made one change, so it would read:
'Take the ave altitude and subtract the left altitude and whatever you get add that back to the average altitude to get the right altitude.'
Note the replacement of 'left' with 'average'.
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Explain how it's more complicated than in previous situations to find the 'right altitude' and width of a 'graph trapezoid', given its left altitude, and its slope and its area.
Its not that it is more complicated it is that there are a lot more calculations that have to be conducted to get the final answer. Like taking slope and knowing the run but having one rise value and trying to find the other value.
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For each of the preceding four questions, interpret in terms of uniformly accelerated motion, assuming that the graph trapezoid represents velocity vs. clock time.
Given the values that are in the problems you can assume uniformly accelerated motion. The graphs being described in the above are assumed to have an initial V of 0 and that time started at 0 also.
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For each of the preceding four questions, interpret in terms of forces, work and displacement, assuming that the graph trapezoid represents rubber band force vs. length.
Since the slopes are positive then you would conclude that the force is increasing as the length of the rubber band increases. As the force increases the length of the rubber band is stretched so the change from point a to point be changes in a positive direction. As the rubber band is stretched it requires an increase in work to stretch the rubber band.
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`q002. On an Atwood machine, a net force equal to the weight of 6 paperclips accelerates a system of 10 dominoes at 2.5 cm / s^2.
What would be the acceleration of a system consisting of 4 dominoes subject to a net force equal to the weight of 10 paperclips?
10 dom / 2.5 cm/s = 4 units of force / domino
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10 dom / 2.5 cm/s = 4 dom / (cm/s^2), not 4 units of force / domino.
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6/2.5 = 2.4 units of force / clip
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6 clips / (2.5 cm/s^2) = 2.4 clips / (cm/s^2), which could be expressed as 2.4 clips / unit of acceleration.
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4*4 = 16
2.4 * 10 = 24
The acceleration would be 8 cm/s^2
I am not sure if any of my calculations are right on this
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All other things being equal, what would be the effect on the acceleration of the system of increasing the number of clips from 6 to 10?
What would be the effect on acceleration of decreasing the number of dominoes from 10 to 4?
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If you double the number of dominoes and the number of paperclips, what happens to the acceleration?
Acceleration will stay constant
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If you double the number of dominoes and halve the number of paperclips, what happens to the acceleration?
Acceleration will decrease
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Right. It would be appropriate to speculate on the factor by which it would decrease.
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If you double the number of paperclips and halve the number of dominoes, what happens to the acceleration?
Acceleration will increase
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Right. Again would be appropriate to speculate on the factor by which it would change.
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How many paperclips and how many dominoes would result in an acceleration of 40 cm/s^2? There are many possible answers to this question. An answer which involves a fractional number of dominoes and/or paperclips is acceptable for General College Physics, though an answer involving a whole number of paperclips and dominoes is preferable (and is expected for University Physics).
40 cm/s = 5 paper clips* 2.4 + 7 dominos * 4
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You get more acceleration by adding clips, but you get less acceleration by adding dominoes. So your two terms would have opposite effects.
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`q003. A small paperclip has a weight of about 4 milliNewtons. A domino has a weight of about 16 grams. The acceleration of 2 dominoes, when subject to a net force equal to the weight of a small paperclip, is observed to be 4 cm/s^2. All results are to be regarded as accurate to within +-5%.
What therefore would be the acceleration of a 1-kilogram mass if accelerated by a net force of 1 Newton?
1000g = 1N
16 *2 = 32
32g = 3200 N
0.000004N = 4 mN
Acceleration is the sum of all forces
A = 2 cm/s^2
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A 1-kilogram mass would be about 30 times the 32-gram mass.
A 1-Newton force would be 250 times as great as a 4 milliNewton force.
What would be the effect on acceleration if mass was increased by a factor of 30 and force by a factor of 250?
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Based on your previous answer, what would be the acceleration of your mass if accelerated by a net force equal, in Newtons, to your age in years?
19 cm/s^2
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If based on our results we define a Newton and the force required to accelerate a kilogram at 1 m/s^2,what is the change in your preceding result?
The change is 18 cm/s^2
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`q004. For each of the following identify each given quantity as v0, vf, a, `dt, or `ds for the interval of uniform acceleration; if the motion is rotational identify instead the corresponding angular quantities omega_0, omega_f, alpha, `dt or `dTheta.
A ball is given a velocity of 30 cm/s at one end of a 60 cm ramp, and accelerates uniformly to the other end in 1.5 seconds.
Vo = 30 cm/s
Dt = 1.5 s
Ds = 60 cm
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A ball accelerates, starting from rest, down a 60 cm ramp. Its velocity changes with respect to clock time at 40 cm/s^2.
Vo = 0 cm/s
Ds = 60 cm
a = 40 cm/s^2
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A steel washer is given an upward velocity of 5 m/s at a height of 6 meters above the floor. It ends up on the floor. The acceleration of gravity is approximately 10 m/s^2 (more precisely it is 9.8 m/s^2, but we'll take a 2% 'hit' on our accuracy and use the more convenient number 10).
Vo = 5 m/s
Ds= 6m
A = 10 m/s^2
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A rotating strap slows from an angular velocity of 400 degrees / second to 100 degrees/second as it rotates through 1800 degrees.
Omega o = 400d/s
Omega = 100 d/s
D theta = 1800 d
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A rotating wheel rotates through 1200 degrees in 15 seconds, starting with angular velocity 100 degrees / second.
D theta = 1200 d
Dt = 15 s
Omega o = 100d/s
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A rotating sphere coasts to rest as it rotates through 40 pi radians in 20 seconds. You don't have to know what a radian is to answer this.
Dt = 20s
Omega f = 0 pi radians / s
Omega o = 40 pi radians /s
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`q005. For each of the situations in the preceding problem, identify which equation(s) of the four equations of uniformly accelerated motion will yield new information.
1) Ds= Vo * dt +1/2 a * dt^2
2) Vf^2 = Vo ^2 + 2a * ds
3) Vf^2 = Vo ^2 + 2a * ds
4) Omega f ^2 = omega o^2 + 2 alpha * dtheta
5) D theta = omega o * dt + ½ alpha * dt^2
6) Omega f = omega o + alpha * dt
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For each equation you identified, algebraically solve the equation for the unknown variable. Don't plug in any numbers, do the algebra with the symbols.
1) a = ½ ((ds-Vo*dt)/dt^2)
2) Vf^2 = Vo ^2 + 2a * ds
3) Vf^2 = Vo ^2 + 2a * ds
4) alpha = ((omega f^2 - omega o ^2)/ d theta)/2
5) alpha = ½ ((d theta-omega o*dt)/dt^2
6) alpha = (omega f - omega o)/dt
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For each symbolic solution obtained in the preceding question, plug in the values of the given quantities to find the value of the unknown variable.
1) a = ½ ((60 cm-30 cm/s*1.5s)/1.5s^2) = 10 cm/s^2
2) Vf^2 = (0cm/s) ^2 + 2(40 cm/s^2 * 60cm = 4800 = vf ^2
Sqrt 4800 = sqrt vf^2
Vf = 69.28 cm/s
3) Vf^2 = (5 m/s)^2 + 2(10 m/s^2) * 6m = 145 m/s = Vf^2
Sqrt 145 = sqrt Vf^2
Vf = 12.04
4) alpha = (((100d/s)^2 - (400d/s) ^2)/ 1800d)/2 = 41.7 d/s^2
5) alpha = ½ ((1200 d-100d/s*15s)/(15s)^2 = 73.3
6) alpha = (0 pi radians - 40 pi radians)/20s = -2 pi radians / s^2
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`q006. Give the data you obtained in class today, including an explanation of what was measured and how.
Degrees #’s correspond with time interval #’s
Degrees : first 6 are counter clockwise and last six are clockwise
1) 1260
2) 775
3) 1260
4) 1260
5) 630
6) 540
7) 1260
8) 1255
9) 1200
10) 1920
11) 1120
12) 2430
Time intervals
1) 4.1
2) 3.4
3) 5.4
4) 5.4
5) 3.8
6) 3.2
7) 6.3
8) 6.9
9) 5.9
10) 8.9
11) 5.1
12) 10.0
This experiment was measuring how many degrees the rotating strap rotated before coming to rest and the amount of time it took to come to rest.
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For each different system, construct a graph of average angular acceleration vs. average angular velocity. You may at this point use a spreadsheet to analyze your data. Explain how you obtained your accelerations, and describe your graphs.
Accelerations were obtained by (Vf - Vo) / dt
When the data was put into excel and a line of best fit was constructed the graph was linear. The graph of the clockwise measurements seemed to have a much more consistent trend than that of the graph that was counterclockwise.
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Does the strap rotating on the threaded rod appear to have reasonably consistent angular acceleration? Is there any evidence that angular velocity has an influence on angular acceleration?
Yes, with human error and different consistencies in rotation and making incorrect estimates about the degrees rotated by the strap. The threaded rod seems to have a more consistent angular acceleration. Depending on the clock time angular V influences angular acceleration. I would say that angular V is dependent on and acceleration with respect to clock time.
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`q006. University Physics
Give your data for today's experiment, along with a brief description of what you did.
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Estimate the uncertainties in your data.
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Graph rubber band force vs. separation of magnets and compare the shape of this graph to the shape of the 'slope graph' obtained from your graph of coasting angular displacement vs. separation of magnets.
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Give your best estimate of rubber band force vs. magnet proximity.
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Give your best estimate of magnet force vs. proximity. If you measured the distances at which the rubber band and the magnet acted on the system, use those distances. If not, you can assume for now that the ratio of the latter to the former is either 9 or 5, whichever you think is closer to the ratio for the system as you set it up.
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You're doing exceptionally well with the equations and the experiment.
You did pretty well with the graphs, but many of your explanations weren't complete.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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You're doing exceptionally well with the equations and the experiment. Your ability to apply equations will certainly be one of your biggest strengths in this course.
You did pretty well with the graphs, but many of your explanations weren't complete.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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