course PHY201 RՠkѰassignment #000
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20:42:42 `questionNumber 0 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 is designed to see how much I understand about basic ideas of motion, timing, and procedures used throughout this course.
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20:43:19 `questionNumber 0 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 speed you should divide the distance the object traveled by the time the object took to complete this distance. confidence assessment: 3
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20:44:13 `questionNumber 0 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 --> The velocity is calculated by dividing the distance by speed; therefore, 40cm/5s = 8cm/s. confidence assessment: 3
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20:45:21 `questionNumber 0 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 --> If the object reaches 20cm/3s, then the average velocity = 6.67cm/s. The second half therefore goes 20cm/2s = 20cm/s. confidence assessment: 3
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20:48:49 `questionNumber 0 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 --> The timer can only calculate to the 0.01 percentage of a second; therefore, the measurements can not be more accurate than this. However, the results do not come within 0.01seconds of each other, so this probably does not account for the majority of the discrepancies of the experiment. The uncertain precision of human triggering is probably the responsible factor for these discrepancies. Due to the short amount of time, it is hard to accurately start and stop the timer at the exact moment necessary. confidence assessment: 3
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20:51:31 `questionNumber 0 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 - This would probably contribute the least to the accuracy, according to the precision of the timers measuring capabilities. b. This probably accounts for much of the accuracy, as human reaction time is unpredictable and varies from person to person. c. This should not cause much inaccuracy, becuase the same object and surface should produce consistant results. d. A difference in positioning would have a minor influence on the accuracy of the experiment. e.HUman uncertainty would have a large influence on the accuracy of teh experiment, as perception can differer from trial-to-trial and between different people. confidence assessment: 3
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20:53:12 `questionNumber 0 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. Nothing could be done about the precision of the timer, except get a more accurate timer. b. The precision of a human triggering could be improved through practice, or facilitating the clicking with a device such as a hand held remote. c. Actual differences in the time required to travel the distance could not be improved, except to resolve any discrepancies of where the object is placed at the beginning. confidence assessment: 3
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20:56:04 `questionNumber 0 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 --> Doubling the length of the pendulum would result in more than half the frequency of the first pendulum. The reason the x axis has a y coordinate of 0 and vice-versa is becuase these are the points at which each axis intercepts the other. confidence assessment: 3
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20:57:07 `questionNumber 0 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 line went through the vertical axis, it would mean that the pendulum was 0cm long, and the frequency would be represented by wher eon the vertical axis the line intercepted. confidence assessment: 3
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20:58:24 `questionNumber 0 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 graph intercepted the horizontal length it would mean that there was a frequency of 0, and the length of the pendulum would be equal to the point where is line crossed the horizontal axis. confidence assessment: 3
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20:59:03 `questionNumber 0 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 --> 6cm / sec x 5 sec = 30 cm apart confidence assessment: 3
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20:59:16 `questionNumber 0 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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21:01:07 `questionNumber 0 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 all the problems, but had the most difficulty in figuring out the approximate accuracy of a number such as 38000. I came to the conclusion that this owuld be 1000, as that this is the smallest value of a significant number. However, how do you calculate the approximate accuracy of a large number, such as 38000. confidence assessment: 3
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21:01:19 `questionNumber 0 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 --> Already answered confidence assessment: 3
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