course Phy 201 Would like to have feedback. 匁仮}榍嗫誽H篡槚飝 壥assignment #001
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20:04:36 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 --> ok
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20:05:36 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 --> Average speed is equal to distance divided by time. confidence assessment: 3
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20:06:47 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 --> v=40cm/5s v=8cm/s This is similar to the ball down the ramp experiment in which the ball's velocity is determined by lenth over time confidence assessment: 3
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20:09:08 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 --> v1=(40cm/2)/3s v1=20cm/3s = 6.7cm/s v2=20cm/2s=10cm/s The second half is slightly faster than the first confidence assessment: 3
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20:14:14 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's precision must be accurate to a hundredth of a second. b.The precision of the human triggering is as accurate to time measured minus reaction time. Since reaction time cannot be accurately measured then the precision is accurate to the time measured by human triggering. c. The actual differences in the time required for the object to travel the same distance are only present due to the changes in placement, time accuracy, and certain elements that might be present. d. if the object is placed further away or closer to the end which in turn changes the distance and will have an effect on time. e. Human error will be a result of reaction time which this error can not be eliminated because the reaction time cannot be measured. confidence assessment: 3
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20:14:54 06-16-2008 20:14:54 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 think that there is plenty of certainty with the precision of the timer program. b. I think that human reaction throws off the timing by milliseconds. C. If the same starting point was used every time then there would be very minimal error in accuracy D. if the object is placed further away or closer to the end which in turn changes the distance and will have an effect on time. E. Human error will be a result of reaction time which this error can not be eliminated because the reaction time cannot be measured.
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20:15:20 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. Use of a different timer for comparison would clear up uncertainty. B. Use of machine in place of human might take more accurate measurement. C. If all elements were exactly the same then there should be no differences in time measurment. D. If the same distance was used with the same elements then nothing should change the accuracy of measurement. E. Machine could possibly eliminate human error. confidence assessment: 3
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20:28:25 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 think that doubling the length would result in more than half of the initial frequancy. I determined that the frequency decreased along a logarithmic curve. confidence assessment: 3
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20:31:59 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 normal starting point or resting state is 0,0. confidence assessment: 2
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20:43:05 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 think two thoughts. 1 this is impossible and the measurements are in accurate. 2 perhaps the pendulum completes circle but that doesn't make sense either to cause an intersection at the vertical axis leading to a negative length. Perhaps the measurements are inaccurate. confidence assessment: 1
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20:44:39 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 --> Again i think that the measurements must be inaccurate resulting in a negative frequency. This is not possible. confidence assessment: 2
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20:45:27 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=30 cm The distance is 30cm. confidence assessment: 3
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20:46:08 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 the equation and concept and answered it correctly. self critique assessment: 3
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20:48:37 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 understand to a point but not fully the idea of significant figures. I remember the concept being used in grade school but don't quite remember the rules. I don't think that a question is necessary at this time because I will do th research i need to fully understand. confidence assessment: 3
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20:50:28 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 --> Again I feel I need to go back and relearn about significant figures and how to determine place value to be used. confidence assessment: 3
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牙観緱盅n蹆揖晚沐璥纮ニD酒筡緛� assignment #000 000. `Query 0 Physics I 06-16-2008