course Phy 201 assignment #000000. `Query 0
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12:16:22 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 --> to find the average speed of an object, all that needs to be done is : divide the distance traveled by the object by the time that has elapsed = average speed confidence assessment: 3
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12:18:44 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 cm / 5 s = 8 cm/s finding the average speed of an object was the same process used when doing the experiment with the can of veggies confidence assessment: 3
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12:21:49 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 halfway point would be 20 cm then : 20 cm / 3 s = 6.7 cm/s 20 cm / 2 s = 10 cm/s confidence assessment: 3
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12:24:15 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 feel that all of the answers given could be possible discrepancies to the accuracy of the experimental data: 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. confidence assessment: 2
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12:30:48 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 : possibly accurate up to .001 second b : +/- .25 second c : +/- .05 second d : 0 (the position of the object should be precise) e : +/- 1% confidence assessment: 2
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12:34:13 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 : nothing could be done unless I re-wrote the program b : practice and possibly take the average of several clicks to get a baseline time c : make sure the object is on a surface that is smooth and free from defects d : there shouldn't be any problems with positioning the object e : maybe film the experiment and replay to see when and where the object stopped confidence assessment: 2
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12:36:40 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 doubling the pendulum would give slightly more than half of the cycles than that of a pendulum at its initial length confidence assessment: 2
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12:39:26 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 --> I think that y would be the release point and x is the point at which the pendulum stops moving confidence assessment: 1
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12:44:38 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 --> Is it possible that if the graph intersects the y axis that the frequency of the cycles is greater the shorter the pendulum length? As for the x axis, if the graph intersects it I would think that the longer the pendulum the shorter the cycles become. confidence assessment: 1
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12:47:49 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 --> average velocity = displacement / elapsed time 6 cm/s = ? / 5 s 6 cm/s * 5 s = 30 cm confidence assessment: 3
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12:48:27 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 self critique assessment: 3
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12:51:12 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 think I understand most of chapter 1, however the section 1-6, converting units, may give me a little trouble until I am able to memorize the conversion rates, 1 in. = 2.54 cm., etc. confidence assessment: 2
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12:54:15 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 --> scientific notation, percent uncertanty, significant figures are pretty simple. I'm curious about something........ will we be allowed to take a formula sheet with us to the proctoring facility for the tests or should we memorize everything? confidence assessment: 3
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