course Phy201 I believe this completes assignment 0 for me. gՈݕ|{yeԁassignment #000
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11:09:27 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 --> To check for my understanding about the basic ideas of motion and timing.
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11:10:59 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 on the incline is equal to the distance the object rolls divided by the time required to cover that distance. confidence assessment: 3
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11:13:26 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 --> 40cm/5seconds = 8 cm per second I took the width of the book that my cylinder was rolling on and divided it by the time it took the cylinder to roll across the book. That result gave me the average velocity of the cylinder on the book. confidence assessment: 3
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11:16:49 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 --> 20cm/3seconds = 6.667 cm per second for first half 20cm/ 2 secs = 10 cm per second for second half confidence assessment: 3
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11:23:55 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. the lack of precision of the timer could cause the times to vary when they in fact do not. b. the delay that it takes for your eyes to send the signal to your brain and from your brain to your finger could result in a slower time. c. Discrepancies could be maybe a slight draft, which would speed up or slow down the object. d. This could have casued the object to roll a longer or shorter distance each time. It was definitely possible that the object was not placed in exactly the same starting position each time. e. I believe there is a lot of discrepancy in this category. Using the naked eye to see that it crossed the end and timing that exactly. I thought it to be quite hard to observe when it crossed the ending. I believe the most discrepancies lie in categories b and e. confidence assessment: 3
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11:25:24 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. very little b. majority c. little d. little e. majority confidence assessment: 3
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11:29:11 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 know of nothing that could be done b.Have something so that when the object began to roll it would automatically trigger the timer to start c. ? d. set up some sort of guide so that the object would start in the same position each time e.Have something at the end of the incline so that we the object touched it, it would stop the timer. confidence assessment: 2
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11:36:41 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 --> I believe that would result in less than half of the frequency. When my pendulum length was approximately from my wrist to finger tip, the frequency was 60. After extending the length of the pendulum from my fingertip to my elbow, the frequency was in the 40's. Less than half. confidence assessment: 3
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11:39:01 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 --> This is so because when you are on the axis, you haven't gone in the other direction. If we go up on the y axis, we haven't gone left or right on the x-axis yet, therefore the x value is a distance of 0, the same principle applies in the opposite direction. confidence assessment: 3
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11:42:41 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 --> In this case where the graph intersects the vertical axis that would mean that the length of the pendulum is 0. Based on the graph you would assume the pendulum to have a very high frequency, but I do not believe this to be the case. I believe in this case the pendulum would not swing because of the length of the string. confidence assessment: 3
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11:44:32 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 --> This would tell me that no matter how long the pendulum is, there would be no frequency. That is the pendulum would not swing in any direction. confidence assessment: 3
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11:45:30 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 * 5 sec = 30 cm confidence assessment: 3
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11:45:49 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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15:17:07 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 --> Uncertainty, estimated uncertainty, and percent uncertainty. I do not understand how to apply it to problem 10 in ch 1 problems in the book. I am also having problems with number 11. On problem 11, I took the estimated uncertainty,.09, and divided it by 2.86, and then multiply by 100 which gives me 3%, the book says it is 9%. I'm struggling with #10 and #11 confidence assessment: 0
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15:18:22 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 think I answered this in the last question. confidence assessment: 0
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