course Phy 201 Ԧwꏰxassignment #000
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16:16:53 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 --> The purpose of this first Query is for the instructor to see my understanding of basic ideas of motion and timing.
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16:20:16 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 --> To find the average speed take the distance traveled and divide it by the time required. confidence assessment: 2
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16:23:32 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 --> Average velocity = 40 cm / 5 sec = 8 cm/sec. I don't understand what you mean about explaining how my answer is connected to my experience..... confidence assessment: 2
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16:26:57 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 --> The first halfway point's average velocity is: 1/2 of 40 cm is 20 cm, thus 20 cm / 3 sec = 7 m/s The second halfway point's average velocity is: 20 cm / 5 sec = 4 m/s confidence assessment: 2
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16:39:53 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. Lack of precision of the TIMER program: The use of significant figures. B. Human triggering probably has alot to do with the discrepancies. C. Actual differences in the time required could be due to the 'shape' of the incline in different places. Examples are: a dip/hump or wavey-ness of the incline. D. Differences in positioning of the object before release also can cause alot of the discrepancies. It is difficult to get an object in the same place. E. When observing the exact time the end of the incline is reached can also cause discrepancies. Example: Noting when the object reaches the end at the end of the book or when it rolls off the end of the book. confidence assessment: 1
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16:43:09 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 --> A. 5% B. 20% C. 10% D. 20% E. 45% confidence assessment: 2
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16:52:45 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. I don't think anything can be done about the lack of precision of the timer. B. Possibly help from another person could help with the uncertainty of the human triggering because it is somewhat difficult to concentrate on the click of mouse and release the object at the same time. C. One way to try to get more precise times could possibly be to make sure the surface of the incline is flat in all places. D. For more accurate positioning one could put a piece of tape just behind the start area and placing the object's back part flush against the tape. E. I think it would be difficult to do anything about the uncertainty of observation of the human eye. confidence assessment: 2
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18:54:50 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 length of the pendulum will result in half the frequency. confidence assessment: 2
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18:58:12 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 --> For a point on the x-axis there is no rise only the run. For a point on the y-axis there is only a rise, no run. confidence assessment: 1
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19:01:46 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 --> For the graph to intersect the y-axis the length would have to be less than zero and the frequency would have to be between 0 and infinity. confidence assessment: 1
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19:04:34 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 --> For the line/curve to go through the x-axis there would have to be a negative number for a frequency and the length would be greater than zero. confidence assessment: 2
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19:07:14 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 ball rolls 6 cm in 1 second. To know how far it rolls in 5 seconds multiply 6 cm by 5 seconds. The result would be 30 centimeters between the two points. confidence assessment: 3
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19:09:24 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 --> I understand. I should have went into more detail on the formal calculation. self critique assessment: 2
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19:16:08 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 --> What I have the most difficulty with in Chap. 1 is significant figures. I don't understand why zeros can't be significant figures instead of place holders because not having zeros before a decimal or after signficant figures can really mess up a problem. I'm also still not sure on how to get the correct significant figures. Can you explain how to get correct significant figures in adding, subtracting, multiplying and dividing? confidence assessment: 3
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19:20:58 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 --> I had trouble with problem # 11, but you answered it for me. The only other thing I was not sure I got the right answer was for # 12. #12 has more to do with lack of confidence in myself about getting the right answer. I don't have any problem understanding it. confidence assessment: 3
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