course Phy 231 Ÿ¶ù½¢}¾±rô¢sÍcÍФâ±Wò¢Ÿ˜assignment #000
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11:13:33 The Query program normally asks you questions about assigned problems and class notes, in question-answer-self-critique format. Since Assignments 0 and 1 consist mostly of lab-related activities, most of the questions on these queries will be related to your labs and will be in open-ended in form, without given solutions, and will not require self-critique. The purpose of this Query is to gauge your understanding of some basic ideas about motion and timing, and some procedures to be used throughout the course in analyzing our observations. Answer these questions to the best of your ability. If you encounter difficulties, the instructor's response to this first Query will be designed to help you clarify anything you don't understand. {}{}Respond by stating the purpose of this first Query, as you currently understand it.
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RESPONSE --> This query will gauge my understanding of the lab material that has been covered thus far - essentially ideas about motion and timing. Most of the questions in this query will be open-ended with no self-critique required.
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11:14:55 If, as in the object-down-an-incline experiment, you know the distance an object rolls down an incline and the time required, explain how you will use this information to find the object 's average speed on the incline.
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RESPONSE --> The average speed of an object moving down an incline can be found by taking the distance traveled and dividing that by the time required to traverse that distance. Example: if the distance is 10 cm and the time needed is 5 s, then the average speed is = 10 cm / 5 s = 2 cm/s. confidence assessment: 3
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11:22:13 If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.
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RESPONSE --> If an object travels 40 cm in 5 s, then the average velocity is = 40 cm / 5 s = 8 cm/s This answer was found in the same manner as the average velocity of my own experiment. I found an average time required to travel a particular distance and from that was able to determine the average velocity. My particular experiment had an average velocity that was greater then 8 cm/s though. confidence assessment: 3
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11:25:24 If the same object requires 3 second to reach the halfway point, what is its average velocity on the first half of the incline and what is its average velocity on the second half?
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RESPONSE --> We first need to determine the distance at the halfway point. 40 cm / 2 = 20 cm = halfway distance of incline. We then solve the problem in the same manner, by dividing 20 cm by the time interval (3 s and 2 s). 20 cm / 3 s = 6.67 cm/s = avg velocity of 1st half 20 cm / 2 s = 10 cm/s = avg velocity of 2nd half confidence assessment: 3
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11:29:54 Using the same type of setup you used for the first object-down-an-incline lab, if the computer timer indicates that on five trials the times of an object down an incline are 2.42 sec, 2.56 sec, 2.38 sec, 2.47 sec and 2.31 sec, then to what extent do you think the discrepancies could be explained by each of thefollowing: {}{}a. The lack of precision of the TIMER program{}{}b. The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse){}{}c. Actual differences in the time required for the object to travel the same distance.{}{}d. Differences in positioningthe object prior to release.{}{}e. Human uncertainty in observing exactly when the object reached the end of the incline.
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RESPONSE --> a. I can't 100% attest to the accuracy of the TIMER program, but I would say it's probably very accurate and doesn't account for the differences in time. b. The uncertain precision in triggering the timer is most likely the cause of the differences in time. c. All things being equal, I don't believe the object would take different amounts of time to reach the same endpoint on the same incline. d. This is also a good possibility for differences in timing. Coupling this with letter b is very likely. e. Again, this is a definitely possibility. The human factor in starting, stoping the timer, in placing the object at the same spot, and in determining exactly when the item has reached its end point are all hard to maintain as constant. confidence assessment: 3
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11:32:06 How much uncertainty do you think each of the following would actually contribute to the uncertainty in timing a number of trials for the object-down-an-incline lab? {}{}a. The lack of precision of the TIMER program{}{}b. The uncertain precision of human triggering (uncertainty associated bLine$(lineCount) =with an actual human finger on a computer mouse){}{}c. Actual differences in the time required for the object to travel the same distance.{}{}d. Differences in positioning the object prior to release.{}{}e. Human uncertainty in observing exactly when the object reached the end of the incline.
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RESPONSE --> I believe this was just asked, though perhaps I misunderstood the previous question. To restate, I would say letters b, d, and e are all contributing factors. Starting and stopping the timer, errors in original placement of the item, and knowing exactly when the item reaches its end point are all very human dependent and therefore somewhat subjective or unreliable. There will be variances in the measurements. confidence assessment: 3
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11:42:00 What, if anything, could you do about the uncertainty due to each of the following? Address each specifically. {}{}a. The lack of precision of the TIMER program{}{}b. The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse){}{}c. Actualdifferences in the time required for the object to travel the same distance.{}{}d. Differences in positioning the object prior to release.{}{}e. Human uncertainty in observing exactly when the object reached the end of the incline.
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RESPONSE --> a. This could be verified to ensure accuracy. If the program was deemed to be inaccurate, it could be modified or a new program could be used. b. This is difficult because it is hard to accurate trigger the timer and release the object at the same time. I'm not sure much could be done unless the trigger was somehow automated so that the person wouldn't have to manually trigger the timer. c. I don't believe there's much to do here. Unless you were using an object that didn't have a ""true"" roll to it, there shouldn't be much change. d. To rectify this, we could draw a starting line or somehow annotate the exact starting point, so that it's easy to know exactly where to place the object each and every time. e. Again, unless this is automated, it's going to be hard to observe the exact mount that the object reaches the end point. We could perhaps position the incline (or ourselves) such that we have a good vantage point to see the object move past the end point. confidence assessment: 2
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12:03:57 According to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?
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RESPONSE --> Doubling the lengths of the pendulums resulting in fewer cycles (or a smaller frequency), but this was still more the half. For example, at 16"" length, the frequency was 71, and at 32"", the frequency was 53. confidence assessment: 3
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12:08:56 Note that for a graph of y vs. x, a point on the x axis has y coordinate zero and a point on the y axis has x coordinate zero. In your own words explain why this is so.
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RESPONSE --> A point on the x axis has a y coordinate of zero because it is not moving up or down on the graph at all. It is only moving left or right. A point on the y axis has an x coordinate of zero because it is not moving right or left - it is moving up and down. confidence assessment: 2
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12:14:31 On a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?
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RESPONSE --> If the graph intersected the vertical axis, that would mean that the length of the pendulum is zero. As the length of the pendulum is decreased, the frequency increases and vice versa. confidence assessment: 2
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12:17:21 On a graph of frequency vs. pendulum length, what would it mean for the graph to intersect the horizontal axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the horizontal axis)? What would this tell you about the length and frequency of the pendulum?
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RESPONSE --> If the curve went through the horizontal axis, this would mean that there were zero cycles in a minute - essentially the pendulum is stationary. I would estimate that with the graph of frequency vs. length, the frequency approaches zero as the length approaches infinity, and the frequency approaches infinity as the length approaches zero. confidence assessment: 3
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12:19:07 If a ball rolls down between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far apart are the points?
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RESPONSE --> The points are 30 cm apart. 6 cm/sec * 5 sec = 30 (cm sec / sec) = 30 cm We know that distance = average velocity * change in time. confidence assessment: 3
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12:19:27 On the average the ball moves 6 centimeters every second, so in 5 seconds it will move 30 cm. {}{}The formal calculation goes like this: {}{}We know that vAve = `ds / `dt, where vAve is ave velocity, `ds is displacement and `dt is the time interval. {}It follows by algebraic rearrangement that `ds = vAve * `dt.{}We are told that vAve = 6 cm / sec and `dt = 5 sec. It therefore follows that{}{}`ds = 6 cm / sec * 5 sec = 30 (cm / sec) * sec = 30 cm.{}{}The details of the algebraic rearrangement are asfollows:{}{}vAve = `ds / `dt. We multiply both sides of the equation by `dt:{}vAve * `dt = `ds / `dt * `dt. We simplify to obtain{}vAve * `dt = `ds, which we then write as{}`ds = vAve *`dt.{}{}Be sure to address anything you do not fully understand in your self-critique.
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RESPONSE --> OK self critique assessment: 3
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12:23:14 You were asked to read the text and some of the problems at the end of the section. Tell me about something in the text you understood up to a point but didn't understand fully. Explain what you did understand, and ask the best question you can about what you didn't understand.
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RESPONSE --> On exercise 1.16A, it asks the reader to find the area of a rectangle and the ""uncertainty"" in the area. The length is 5.10 + or - .01cm and 1.90 + or - .01 cm. I read section 1.3 which briefly explains uncertainty, but I'm unsure how best to calculate this. I can easily calculate the area as 5.10 * 1.90, but I don't understand how best to calculate the uncertainty of the area.
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12:25:57 Tell me about something in the problems you understand up to a point but don't fully understand. Explain what you did understand, and ask the best question you can about what you didn't understand.
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RESPONSE --> There is a similar question (#17) which I had the same difficulty with. I can find the approx. volume of the bag, but not the uncertainty of the bag. Diameter of cylinder = 8.50 + or - 0.02 cm and thickness of 0.050 + or - 0.005 cm
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