course PHY 231 ŠmòŸÎÑÏxÎ[ë”ó ì¯çÙ¸ÁäÒüôaƒb–Êôassignment #000
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21:32:09 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 query is to assess our knowledge of some basic concepts of motion and timing because of the use of these concepts further in the course.
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21:33:45 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 --> You will use the distance traveled divided by the time interval and that will give you the average rate of change. confidence assessment: 2
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21:39:13 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 --> I think my last answer didn't get entered so again the object is moving at 8cm/s because 40 divided by 5 is 8. If it took the object 3 seconds to make it halfway then it would be moving at an average of 13.333333333cm/s because this is 40 divided by 3. Therefore on the backstretch the object is moving at a whopping 20cm/s indicating that the object is speeding up. This figure was obtained by dividing 40 by 2. confidence assessment: 2
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21:41:30 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 --> I believe that the lack of precision on the timer's part is the least of the three. Of course there is going to be human error (this is probably the most consequential) between clicks. There is also going to be actual differences in time due to the surface maybe leaning one way a little or it might not be that stable or the object might not be perfectly round. Again the most consequential being human error. confidence assessment: 2
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21:45:18 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 --> I believe that the timer program is the least of our worries although it is one. The uncertain precision of human triggering and human uncertainty in observing when the object actual reaches its' marks are probably the two biggest factors of uncertainty in this experiment. Actual differences in time required for the object to travel the same difference and difference in positioning are both problems, but they are ones that can be solved by getting everything level and marking the positions. confidence assessment: 2
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21:49:02 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 --> In order of keeping the timer honest we could repeat the experiment a thousand times. The same for uncertain human triggering, by doing the experiment over and over we reduce the percent-error. Do the experiment over and over so that the differences become smaller and smaller. Mark a position for the ball to be released every time. Again, correct human uncertainty by performing the experiment numerous times. confidence assessment: 2
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22:09:49 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 --> Doubling the length of the pendulum will result in less frequency but I do not believe that it will cut the frequency in half. It will however cut the frequency by a quarter or so. confidence assessment: 2
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22:13:23 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 these are considered inputs and when there is a point on the x and y axes are considered boundaries and when there is no y input then the point is simply on the x-axis if there is no x-input then the point is on the y-axis. confidence assessment: 0
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22:15:11 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 --> It would mean that the length and frequency would be a negative. confidence assessment: 0
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22:15:59 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 --> It would tell us that the numbers are negative. confidence assessment: 0
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22:17:18 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 --> The points are 30cm apart because the distance is equal to the average velocity multiplied by the change in time. confidence assessment: 2
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22:17:57 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 full understand. self critique assessment: 3
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22:30:05 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 --> Vectors gave me a little trouble. I pretty much fully understand the concept except for the products. I am not really sure what question to ask so maybe a little more practice will help. confidence assessment: 0
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22:30:40 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 --> confidence assessment: 0
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