course Phy 201 ¯öäz†aèýí¢þ˜k„f˜ŽiÃe™Ïή˜ÛF–îassignment #000
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22:27:44 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 to let me understand basic ideas about motion and timing observed throughout the labs that I did. If I have problems, my instructor will help me understand the problem fully.
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22:29:54 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 --> I can find the average speed on the incline by using the distance and the time required, and doing multiple trials to see the specific time and then determine the average by the times that I have calculated. confidence assessment: 2
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22:35:17 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 --> 40 centimeters / 5 seconds = 8 centimeters per second. The average velocity is 8 centimeters per second. In my experience I could find the velocity by dividing the distance travled by the time it took to travel. confidence assessment: 2
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22:37:51 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 --> Given that the distance traveled is 40 centimeters, the half way point would be 20 centimeters. So the average velocity on the first half of the incline would be 20 centimeters / 3 seconds= 6.6 centimeters per second. The average velocity on the second half would be 5 seconds (which is how long it took to complete the whole distance - 3 from the first half of the incline= 2 seconds) therefore, the velocity of the second half would be 20 centimeters / 2 seconds = 10 centimeters per second. confidence assessment: 2
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22:44:24 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 timer times by intervals and not exact precision of decimal places such as .0001 b- each person reacts differently to when they think the object reaches the end of the incline. What one's eye may see, and there reaction may be quicker or slower depending upon their reaction to what was taking place. c- different times could make one think that their results are misleading, but there could be different times based on how the object was placed when released, how smooth the surface the object was travling on was, whether or not their was some slight wind, any factor could cause a difference in the time. d- I could have placed the object in a different location than previous, causing it to be more off balance, more on balance, causing it to go slower or faster depending on where I let go of the object going down the incline. e- when someone is looking directly at the object their reaction rate may be slower or faster than the actual time of when the object was at the end of the incline. Sometimes people guess when the object is near the end causing errors in the results. confidence assessment: 2
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22:48:06 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- i don't think that the TIMER program would cause much uncertainty because that is where you count on all of your data for, for accuracy, which is very important. b- I think that a lot of uncertanity comes from the human triggering, their reflexes may be off. c-i don't think that the actual time differences cause that much of an uncertainty because you are looking for different times and coming up with an average based upon those times. d- i think that positioning causes some what of uncertainty because if you misplace where you release it may cause a time difference and may cause an error in your results. Always try to keep your release in the same positon. e- i think this causes a lot of uncertainty, because your eyes can be very misleading when trying to decide when the object actually reached the end of the incline. confidence assessment: 2
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22:51:06 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- you really can't control the timer program, so I would just trust that it is correct!! b- i would try to let my finger be as precise as possible, trying to hold it really still, and be precise on clicking trying not to click to much, or to soft. c- if I were to get an answer that really didn't fit my skeme of numbers, I would then re-due that time trial and see if there was a difference, but as long as the numbers some what correspond, then I wouldn't worry too much about it. d- i would always try to release in the same position every single time to try and not get errors in my work. e- i would just try and watch as closely as possible and use my best judgment when looking for the end result. confidence assessment: 2
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22:52:46 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 --> when doubling the length of the pendulum it will result in less than half the frequency. confidence assessment: 2
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22:55:48 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 --> The center of orgin concludes wether or not a point is postive or negative and provides conveinence for directions of the graph. confidence assessment: 2
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22:58:28 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 --> this means that the curve or line would cross through the vertical axis, meaning the line runs completely through the vertical axis. This would tell us that the frequency depends on the length and that the higher the length the lower the frequency since the curve crosses through the vertical axis. confidence assessment: 1
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23:00:46 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 crossed the horizontal axis it would mean that the curve is crossed through the horizontal axis meaing that frequency depends on length, but that this case it would be positive since it crosses through the horizontal axis confidence assessment: 1
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23:02:41 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 --> 6 cm /sec * 5 sec = 30 cm So the points are 30 cm apart from one another. confidence assessment: 2
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23:03:04 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 and I got the same answer. self critique assessment: 2
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23:09:33 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 --> after reading over dimensional analysis a few times, I understand that you only add/subtract quantites with the same dimensions, but I don't fully understand how you can check to see if you are doing equations correctly by using dimensional analysis. confidence assessment: 1
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23:15:42 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 a hard time finding the approximate uncertanity of the radius of a circle. I know how to calculate the area of a circle when the radius is given, but then how do you go from there to finding the approximate uncertanity? confidence assessment: 1
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