course phy 201
11/6 8:30 pm
`q001. If the area of a 'graph rectangle' is 30 meters and its altitude is 5 meters / second, what is its width?****
6 seconds.
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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'?
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6 Newtons/meter
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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?
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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?
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0.25 seconds
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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?
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width: 6 seconds; Starting altitude: -25 m/s; final altitude: 35 m/s
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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.
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The area of the trapezoid is the distance covered, ds; The average altitude is the average velocity, vAve. The width is the change in time: dt. The slope is a measure of acceleration: a. The change in altitude is the change in velocity: dv.
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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.
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starting altitude, final altitude, width; starting altitude, area, slope; starting altitude, slope, width; final altitude, slope, width; final altitude, slope, area; slope, width, area;
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You have most of them. There are actually 10 possible combinations of three.
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Explain how you would find the area and slope of a 'graph trapezoid', given its two altitudes and its width.
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The difference between the two altitudes over the width will give the slope of the trapezoid. The point directly between the two altitudes, multiplied by the time, will give the area.
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Explain how you would find the area and width of a 'graph trapezoid', given its two altitudes and its slope.
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Take the change between the two altitudes, and divide it by the slope, to get the width of the trapezoid. Once you have that, find the average between the two altitudes and multiply it by the width to find the area.
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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.
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Multiply the slope by the width, and you will have the result for the “right altitude”.
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only after adding to the 'left altitude'
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With that found, take the average between the two altitudes, and multiply it by the width to find the area.
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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.
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It is more complicated due to the fact that you need to find more values. The first problem you could take the variables given and find the two others with just two equations. The second problem requires you to find a value, then determine another value, THEN find the final 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.
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simply take the answers from the previous questions, but substitute the following words: Left altitude = initial velocity, Right altitude = final velocity, average altitude = average velocity, width = seconds, area = displacement.
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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.
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the requested interpretations haven't been included
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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?
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mass is 2/3 as great as before, force is the same; acceleration = 4/3 as great; 2.5 * 4/3 = 3.33 m/s^2;
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There are 4 dominoes instead of 10, so the ratio of masses is 2/5. Given equal forces the acceleration would be 5/2 as great.
There are 5/3 as many paperclips, which will exert 5/3 the force, which will increase the acceleration by another factor of 5/3.
You'll get 5/2 * 5/3 = 25/6 the acceleration.
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If you double the number of dominoes and the number of paperclips, what happens to the acceleration?
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force doubles, mass doubles, acceleration remains the same.
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If you double the number of dominoes and halve the number of paperclips, what happens to the acceleration?
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mass is doubled, force is halved, accleration is ¼ as great as before.
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If you double the number of paperclips and halve the number of dominoes, what happens to the acceleration?
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force is doubled, mass is halved; acceleration quadruples;
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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).
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10 dominos and 6 paperclips = 2.5 cm/s^2 acceleration; 40 is 16 times greater than 2.5; cut mass in half (5 dominoes) gives multiple of 2; increase paperclips by 8 times (48 paperclips) gives multiple of 8. 8*2 = 16. 5 dominos and 48 paperclips will give desired acceleration.
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Good.
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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?
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1 kg is 62.5 times as great as 16 grams; 1 newton = 250 times as great as 4 millinewtons; force: 250 times as great, mass, 62.5 times as great. 250 / 62.5 = 4; acceleration is 4 times as great; 4 cm/s^2 *4 = 16 cm/s^2;
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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?
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my weight: roughly 59 kg; force: 20 newtons; force is 20 times as great, mass is 59 times as great; 20 /59 = 0.34; acceleration is 0.34 times as great; 16 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?
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0.34 m/s, or 34 cm/s (force/mass);
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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.
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v0: 30 cm/s; ds: 60 cm; dt: 1.5 seconds;
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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.
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ds: 60 cm; a: 40 cm/s^2; v0: 0 m/s;
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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).
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v0: 5 m/s; a: -10 m/s^2; ds: 6 meters;
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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.
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omega_0: 400 degrees/second; omega_F: 100 degrees per second; dTheta: 1800 degrees
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A rotating wheel rotates through 1200 degrees in 15 seconds, starting with angular velocity 100 degrees / second.
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omega_0: 100 degrees/s; d_theta: 1200 degrees; dt: 15 seconds;
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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.
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d_theta: 40 pi radians; dt: 20 seconds; omega_0: 0 pi radians/second
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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.
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1) first equation of motion
2) fourth equation of motion
3) fourth equation of motion
4) second equation of motion
5) first equation of motion
6) first equation of motion
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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.
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You haven't included your solutions. I assume that you know how to do this, as equations seem to be your preferred mode of solving problems.
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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.
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`q006. Give the data you obtained in class today, including an explanation of what was measured and how.
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measured the time it took for a rotating plank to come to rest, as well as how far it rotated. The time was measured in oscillations of a pendulum. The pendulum has been clocked at 41 oscillations per minute
amount of rotation(in degrees) time (in oscillations)
1440 13
2520 18.5
2498 5
922 16.5
585 6.25
540 6
810 11.75
90 3.75
1440 17.5
2025 23
702 11
2135 23
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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.
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The acceleration would be the slope of the graphs. Another way to find it, is to find the average angular velocity of each trial. Since this is the average, and the final velocity was 0 (at rest), doubling the given average will give you the initial angular velocity, which is also the change in velocity. Taking that over the time will give the acceleration..
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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?
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it is somewhat consistent. There is an obvious spread amongst the data, as well as a few outliers, but there is a trend. There is evidence that the velocity has an effect on acceleration.
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