assignment0-1

course Phy 231

ٖ|~Hzض̾\assignment #000

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Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

000. `Query 0

Physics I

09-09-2007

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21:47:44

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 -->

It is a query about the labs and does not require self critique, it is meant to see how well I understand the concepts of the labs.

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21:48:17

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 would take the distance and divide it by the time.

confidence assessment:

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21:48:58

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 -->

40cm / 5s = 8cm/s.

You divide the distance by the time to get the average speed. This is how I calculated my average speed in my own expirement.

confidence assessment: 3

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21:50:52

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 -->

Half of the distance is 20m and the time is 3s so you divide them to get 6.6667cm/s for the first half.

Half of the distance is still 20cm and the time is now 2s, so divide them to get 10 cm/s for the average speed on the second half.

confidence assessment: 3

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21:52:54

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) This could explain it, but none of the times are really closely related so it seems that the discrepancy is somewhere else.

b) This would more accurately be the reason for the problems.

c) This wouldn't explain it, the times should be very similar.

d) This would also explain the time differences quite well because you are using the same measurements each time, and if it was actually a shorter distance you will get a faster average velocity.

e) This also very accurately explains the differences.

confidence assessment: 3

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21:53:56

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) This would add a moderate amount of uncertainty.

b) This would also add a moderate amount.

c) This would have a minimal effect on uncertainty.

d) This would have a moderate amount.

e) This would have a moderate amount.

confidence assessment: 2

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21:54:54

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 a different timer

b) Use a ruler to be certain

c) Nothing you can really do here

d) Once again use a ruler

e) Use a ruler and try to stop it at a certain mark.

confidence assessment: 2

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21:55:19

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 -->

It would almost halve the frequency. This seemed to be the results I got anyway.

confidence assessment: 2

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21:56:25

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 -->

These are the x and y intercepts, or the points where the line crosses the x and y axis. It means that in the function, there is a place where x will equal 0 and y will equal 0.

confidence assessment: 1

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21:59:06

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 give a negative length, but this might correspond to the length of the swing with the origin being vertical.

confidence assessment: 1

A point on the vertical axis corresponds to length 0. Length 0 isn't possible, but it might be that the frequency approaches some finite limiting value as length approaches 0. It is also possible that there is no limit to the frequency as length approaches 0.

Note that frequency is represented by the vertical coordinate, and there is nothing about a vertical intercept that implies a negative vertical coordinate.

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21:59:34

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 -->

This would mean negative frequencies. I don't understand really.

confidence assessment: 0

A horizontal intercept would mean 0 vertical coordinate, hence zero frequency. If the graph actually went through the horizontal axis, that would imply negative frequency past a certain length.

Either way a horizontal intercept wouldn't make sense.

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21:59:58

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 -->

30 cm. 6cm/s * 5s = 30cm. The seconds cancel.

confidence assessment: 3

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22:00:31

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 this.

self critique assessment: 3

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22:01:18

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 have had a problem getting my textbook so I haven't reviewed the problems yet,but I have Phy 201 and 202 credit, and I've attempted this class before so I understand the concepts.

confidence assessment: 3

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22:01:31

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 -->

Delay in my textbook.

confidence assessment: 0

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Good work.

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