course Phys 201 Xpⵦ쫩ԡT assignment #000 000. `Query 0 Physics I 01-19-2009
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12:08:03 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 program is supposed to give basic ideas about motion, timing, and other proceedures.
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12:12:40 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 --> It is an example of average velocity. you the formula vAve= ds/ dt confidence assessment: 3
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12:25:02 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 --> ds=40cm dt=5sec vAve= ds/dt = 40 cm/5sec = 8m/s confidence assessment: 3
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12:59:13 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 --> First Half vAve=ds/dt = 20cm/3s vAve= 6.67cm/s Second Half vAve=ds/ dt = 20cm/2s = 10m/s confidence assessment: 2
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13:15:16 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 --> B confidence assessment: 2
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13:20:58 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 --> D confidence assessment: 1
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13:21:51 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 --> E confidence assessment: 2
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13:22:51 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 --> No. I think it will extend the frequency because it has a longer length and therefore, further away from the center confidence assessment: 3
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13:25:03 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 --> It is stationary. confidence assessment: 3
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13:31:55 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 --> i don't fully understand this question. confidence assessment: 0
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13:34:20 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 --> i don't understand this one either. confidence assessment: 0
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13:49:37 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 --> ds=vAve*dt = 6cm/s *5sec ds= 30cm confidence assessment: 3
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13:49:50 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 --> self critique assessment: 3
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13:51:23 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 --> i understood questions with regards to average velocity formulas, displacement, and time. I did not , however, understand the questions that were about graphing. I think it had to do with the way they were worded. confidence assessment: 3
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13:51:32 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 --> confidence assessment: 3
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dĤϵwLھR assignment #000 000. `Query 0 Physics I 01-19-2009"