Assignment 0

course phy 201

¡Ä¢ÜäT ð÷„¯ÔÑ×ï¿o¥—z¥L¥„GøÒassignment #000

000. `Query 0

Physics I

02-03-2009

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15:14:16

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 first Query is to gauge out understanding of some of the basic ideas about motion and timing and to get us ready for the proscedures to be used throughout the course.

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15:16:46

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 -->

An easy way to understand is to say that speed is given in distance per time (d/t). Therefore, you would take the overall distance traveled (say 100mm) and ratio that to the amount of time it took to travel that distance (4 seconds). The computation would look like 100mm/4s giving you 25mm per second (25mm/s).

confidence assessment: 3

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15:19:54

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 -->

From the previous response, 40cm/5s is the same as 8cm/s. This is obviously the average, but if you want to figure out how much it was speeding up or slowing down throughout its descent, you would have to measure the time traveled for the first half of the distance and again for the second half of the distance. There, you would be able to find if the speed changes, and by how much.

confidence assessment: 3

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15:22:37

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'm not 100 percent but I believe the distance was 40cm and the total time was 5s, so that is the data i'll use here. For the first half, 20cm/3s = 6.67cm/s and for the second half, 20cm/(5s-3s) = 10cm/s.

confidence assessment: 3

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15:25:00

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 don't understand what you're looking for here, but the differences in data could be directly related to any of these choices. There is no exact way to measure the time or the distance so the more trials you attempt could get a more precise average.

confidence assessment: 3

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15:28:14

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. the timer program would result in at least a tolerance of .2s.

b. only way to get a close answer is to anticipate when to hit the timer, so there is no delay in hand-eye coordination.

c. plus or minus .3s

d. plus or minus 4mm

e. plus or minus .2s

confidence assessment: 2

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15:30:02

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 -->

a. multiple runs

b. anticipate when to press the button

c. multiple runs

d. use a fixed lock and unlock to start the experiment

e. put the incline on a table so when the object touches the table, it's done.

confidence assessment: 3

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15:30:55

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 -->

would result in more than half the frequency.

confidence assessment: 3

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15:32:18

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 point on the x axis is not above or below the axis, therefore would not have a positive or negative number, resulting is 0. The pointo n the y axis is neither to the left or right of the axis line so would not be considered positive or negative, resulting in a 0 coordinate.

confidence assessment: 3

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15:35:03

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 -->

This would mean the object is fixed and has no length on the pendulum line.

confidence assessment: 3

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15:35:52

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 result in the object not swinging at all, which could only happen if the object was not set in motion, or if the time measured was 0.

confidence assessment: 3

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15:36:18

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 -->

6cm/s * 5s = 30cm

confidence assessment: 3

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15:36:35

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 -->

ok

self critique assessment: 3

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15:38:29

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 -->

the main thing i had trouble understanding was the approximate uncertainty when just given a length. I used the last digit and gave it a tolerance of plus or minus value of 1 (4.5cm +/-0.1cm). after practice, it started making more sense.

Good. Note that 4.5 is the 2-figure rounding of any number between 4.45 and 4.55, so 4.5 really means 4.50 +- .05.

confidence assessment: 3

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15:38:40

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 -->

previous...

confidence assessment: 3

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&#Good work. See my notes and let me know if you have questions. &#