course phy 121 I am back from Europe. My converter did not have a three prong adaptor so I was unable to charge my laptop. My son played on it during the flight there and it was dead by the time we got to Europe. All in all that means I did very little work while I was gone. I read the first two chapters and did the assigned problems, but that was it. I now have to play catch up and I lugged that laptop all over Europe for nothing!
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06:11:47 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 determine my understanding of basic ideas related to motion and timing and to get me used to the procedures used int he course
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06:13:12 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 --> divide the change in distance by the change in time to get the average speed. confidence assessment: 2
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06:14:03 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 --> ds = 40cm dt = 5 s vAve = ds/dt vAve = 40cm/5s = 8 cm/s confidence assessment: 2
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06:15:43 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 --> ds1 = 20cm dt1 = 3s ds2 = 20 cm dt2 = 2s vAve1 = 20cm/3s = 6.6cm/s vAve2 = 20cm/2s = 10 cm/s confidence assessment: 2
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06:19:29 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) timer precision - very little relation b) human precision - the largest factor in differences c) actual differences - plays a small part d) object position - plays a small part e) human observation - plays a bigger part confidence assessment: 2
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06:20:10 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 --> The same as the last question confidence assessment: 2
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06:23:18 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) timer precision - could get a more precise timer b) human precision - could take the human out of the process and have an electronic sensor that starts and stops the timer. c) actual differences - nothing could be done, this is what is being tested d) object positioning - Have a set ""start gate"" that the object is placed at that ensures the same positioning each time. e) human uncertainty - the same as human precision, take the human out of the process and use an electronic sensor. confidence assessment: 2
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06:27:20 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 --> No, it would not result in half the frquency. doublind would result in more than half the frequency. The graph is decreasing at a decreasing rate, in order for doubling to result in half of the frequency the graph would have to be decreasing at a constant rate. confidence assessment: 2
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06:28:59 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 --> Because the line crosses the axis at the zero coordinate. When it crosses the x axis then the y coordinate is zero and when it crosses the y axis the x coordinate is zero. confidence assessment: 2
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06:31:23 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 graph interesected the vertical axis that would tell me that the length of the string was zero. I assume that if my length is zero then we have no pendulum and therefore no frequency either, there would be no movement. confidence assessment: 2
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06:33:10 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 interesected the horizontal axis it would tell me that the frequency was zero, that there was no movement. It is possible to have the frequency to be zero at any lenght, so I would need to know the x coordinate in order to determine anything about the length. confidence assessment: 2
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06:34:07 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 --> ds = vAve * dt ds = 6 cm/s * 5 s ds = 30 cm confidence assessment: 2
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06:34:24 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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06:52:17 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 significant figures up until the book started discussing percent error. You count how many digits are in the number to determine how many significant figures there are. Zeros are considered place holders, unless the numbes is stated as ""exactly"" that number. When numbers are multiplied then the answer should only have as many figures as the number with the smallest number of siginificatn figures. I do not understand then why this rule should be broken when it comes to uncertainty and how you know when to use more digits or less.
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06:57:50 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 --> In the very first problem I had no problem determining that 14 billion years was 1.4 x 10 ^ 10 years. In a chart I found that 1 year - 3 x 10^7 seconds. And I know I have to multiply the two to get how many seconds in 14 billion years however, I was unsure of what to do with my exponents. I came up with 4.2 x 10 ^ 17 seconds in 14 billion years. I added the exponents, was this correct? confidence assessment: 2
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