course phy201 ????????H??b?assignment #000
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11:47:23 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 is to figure out how much we know about motion and timing
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11:48:16 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 --> by taking the distance over the time you will be able to determine the avrage speed on the incline confidence assessment: 3
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11:51:25 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 --> by taking 40cm/5sec=8cm/s is the average velocity of the object traveling down an incline this is connect to my experiance because if you know the distance of an incline and how long it takes to move down the incline you can figure out the average speed confidence assessment: 3
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11:54:12 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 --> by taking 40cm/2 to get the half way distance to be 20 cm and then i took 20cm/3sec to get 6.67 m/s for the average velocity for the first half of the incline then i took the remaining 20cm/2sec because if it took 5 sec to do the whole incline then 5-3 = 2 and i got the the average velocity for the second half to be 10m/s confidence assessment: 3
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11:59:54 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. this could explain somethings due to the anount of sig figs the program will show b. this is most likely the biggest player because where ever there is human intervention then there is human error so there is a reaction time to the object moving and when the human can click the mouse c. the differences in the time would be a small contributer but still could contribute the wing could be blowing one time or the object could hit a rough patch on the ramp d. this could have a small influence on the timing results because if the object has to travel farther then it will take more time e. again this is probably a big player in the time differences because this is human error and there is a reaction time between when the ball crosses and when the person clicks or even there can be a difference as to where the end is based 0on the angle of the person to the end of the ramp confidence assessment: 3
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12:05:34 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 --> when timing a number of trials this will help narrow down the amount of uncertanty because each of the things that could be a discrepance would be negated by doing multilple trials for example if one trial the person clicked late then chances are on another trial the person will click early confidence assessment: 3
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12:08:25 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.go into the program and list more sig figs for the program to read to b.have a computer automated release function so the computer starts timing as soon as the ball is released c. have the ball travel the same path under the same conditions d.have a gate that the ball is in so it will always start at the same spot e.have a laser reader at the bottom that is attached to the computer so when the ball crosses and the laser is broken then the timer will stop confidence assessment: 3
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12:10:52 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 when you double the length then it will be exactly half of the frequency confidence assessment: 3
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12:12:50 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 because that is y repersents frequency and x represents length there is a length that the frequency is zero and a frequency when the length is zero confidence assessment: 2
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12:15:10 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 --> iif the line went through the vertical axis that means that the frequency is zero confidence assessment: 3
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12:16:48 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 --> when it crosses the horrizontal it means the frequency is zero and when it crosses the vertical it means the length is zero so when it crosses the horizontal it will tell you the length that would make the frequency zero confidence assessment: 3
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12:17:22 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/s*5s=30cm confidence assessment: 3
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12:17:40 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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12:21:05 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 --> this was things that i understood pretty well and really have no questions about the material or any questions when i was reading it confidence assessment: 3
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12:22:24 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 fully understood all of the problems they were straight forward confidence assessment: 3
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