course Phy 231
9/04/09xxxx
3:45 am" "Question: This question, related to the use of the TIMER program in an experimental situation, is posed in terms of a familiar first-semester system.
Suppose you use a computer timer to time a steel ball 1 inch in diameter rolling down a straight wooden incline about 50 cm long. If the computer timer indicates that on five trials the times of an object down an incline are 2.42sec, 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 the following:
· The lack of precision of the TIMER program.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Not to a large extent since the lack of precision of the timer program mainly only affects very small times.
· The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Probably not that much if we're just talking about the mouse click and not human uncertainty in when to click (3 questions below), although I suppose it depends on the reaction time of the person.
· Actual differences in the time required for the object to travel the same distance.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Not by much as long as the variables (mass, distance) are the same.
· Differences in positioning the object prior to release.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv That could be a large (or large in terms of the .1 or .2 sec) discrepancy, depending on how much difference there is in the positioning.
· Human uncertainty in observing exactly when the object reached the end of the incline.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv That's probably the major discrepancy, or at least it has been in my experience since .1 or .2 sec out of a 2-3 window is a normal variation in measuring when something crosses a line (at least for me).
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Question: How much uncertainty do you think each of the following would actually contribute to the uncertainty in timing a number of trials for the ball-down-an-incline lab?
· The lack of precision of the TIMER program.
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Not very much (since it rounds after .0001, it would be minimal compared to the numbers we have).
· The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Depending on the discrepancy of human reaction time, it could possibly introduce a small amount of uncertainty.
· Actual differences in the time required for the object to travel the same distance.
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv As long as it's the same distance/mass, and all other variables are controlled, it wouldn't really contribute to the uncertainty.
· Differences in positioning the object prior to release.
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv This could also introduce a small or large amount of uncertainty (depending on the person setting up the experiment---the person could vary wildly in positioning the object or could be very close to the same positioning each time).
· Human uncertainty in observing exactly when the object reached the end of the incline.
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv I think this would typically be the largest factor in contributing to uncertainty. It's very likely that the greatest difference in the numbers came from human observation of the end of the incline, which would give it the greatest uncertainty.
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Question: What, if anything, could you do about the uncertainty due to each of the following? Address each specifically.
· The lack of precision of the TIMER program.
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Very little, since it is preprogrammed.
· The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv I suppose you could try your best to be ready for the click and not to hesitate.
· Actual differences in the time required for the object to travel the same distance.
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv There is nothing to be done (and the uncertainty is minimal anyway).
· Differences in positioning the object prior to release.
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv You could attach a strip or mark an x for the spot where you position the object the first time, and be sure to replicate this positioning.
· Human uncertainty in observing exactly when the object reached the end of the incline.
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv You could put up a barrier at the end of the incline to make the ending point more observable or you could put a strip of neon color at the end to highlight the point it needs to reach, and just be observant when it gets there.
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Question: 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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Your solution:
You will divide the distance required by the time required to get the average speed.
confidence rating: 3
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Question: 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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Your solution:
40 centimeters/5 seconds = 8 cm/sec
Since velocity has always been the change in distance/the change in time in my experience, this makes sense.
confidence rating: 3
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Question: 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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Your solution:
The average velocity on the first half is 20 cm/3 sec = 6.67 cm/sec.
The average velocity on the second half is 20 cm/ (5 sec- 3 sec) = 10 cm/sec
confidence rating: 3
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Question: `qAccording 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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Your solution:
According to my experiment, it was more than half. My graph was decreasing at a decreasing rate, so every time the pendulum doubled, frequency lessened by by a smaller rate.
confidence rating: 2
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Question: `qNote 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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Your solution:
These are the x and y intercepts. If you have an equation, you can plug in 0 for y to see where it crosses the y axis and 0 for x to see where it crosses the x axis.
confidence rating: 2
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Question: `qOn 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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Your solution:
confidence rating:
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Question: `qOn 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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Your solution:
Well, I don't see how it could intersect since the x axis is the length and intersecting/crossing the horizontal axis would imply that the length of the pendulum is 0 (intersect) or negative (crossing). Since we can't have a 0 or negative length for the pendulum, it can't cross.
confidence rating: 2
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Question: `qIf 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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Your solution:
(6 cm/sec) * 5 sec = 30 cm apart
If it takes 5 sec at an average velocity of 6 cm/sec, it has traveled 30 cm.
confidence rating: 3
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Given Solution:
`aOn 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 as follows:
* 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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Your solution: I probably need to fully write out each step and not skip steps.
It's good to get into that habit with your submitted work. You'll find that this causes you to think more deeply about the meanings and the relationships involved, which will be helpful when you later encounter the more challenging problems of this course.
confidence rating: 3
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Question: `qYou were asked to read the text and some of the problems at the end of the section. Tell your instructor 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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Your solution:
I think I understand the section pretty well, although I will admit that I had to look up a conversion factor or two I should have known. Also, I remember there were one or two things I wanted to ask about vectors, but since those problems weren't assigned, I'll save them for after I attempt that section.
Oh, also, I'm a bit worried that I might not be doing the right questions since I have the 12th edition of the book, but the concepts are the same (conversions, uncertainty, etc.), so I at least get the concepts.
The 12th edition will be fine.
If by the end of the weekend you haven't gotten more information about the 12th edition vs. the 11th, please let me know.
Also, you're welcome to ask questions about the vectors.
confidence rating: 3
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SOME COMMON QUESTIONS:
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QUESTION: I didn’t understand how to calculate uncertainty for a number such as 1.34. When given examples we had problems such as 1.34 ±0.5 and with that we had a formula (0.5/1.34)*100. So I do not understand how to compute uncertainty when no estimated uncertainty is given.
INSTRUCTOR RESPONSE:
The +- number is the uncertainty in the measurement.
The percent uncertainty is the uncertainty, expressed as a percent of the number being observed.
So the question in this case is simply, 'what percent of 1.34 is 0.5?'.
*
0.5 / 1.34 = .037, approximately. So 0.5 is .037 of 1.34.
*
.037 is the same as 3.7%.
I recommend understanding the principles of ratio, proportion and percent as opposed to using a formula. These principles are part of the standard school curriculum, though it does not appear that these concepts have been well mastered by the majority of students who have completed the curriculum. However most students who have the prerequisites for this course do fine with these ideas, after a little review. It will in the long run save you time to do so.
There are numerous Web resources available for understanding these concepts. You should check out these resources and let me know if you have questions.
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QUESTION: I understood the main points of changing the different units, but I’m not sure when in the problem I should change the number to 10 raised to a certain power. In example 1-8 I did not understand why they changed 70 beats/min to 2 x 10^9 s.
2 * 10^9 is about the number of seconds in 70 years.
70 beats / min were not changed to 2 * 10^9 seconds; in changing the beats / minute to beats in a lifetime, there was a step where it was necessary to multiply by 2 * 10^9 seconds.
The example actually used 80 beats / min as a basis for the solution. This was converted to beats / second by the calculation
80 beats / min * 1 minute / (60 seconds), which would yield about 1.33 beats / second.
This was then multiplied by 2 * 10^9 seconds to get the number of beats in a lifetime:
2 * 10^9 seconds * 1.33 beats / second = 3 * 10^9 beats.
In the given solution 80 beats / min * 1 minute / (60 seconds) was not actually calculated; instead 80 beats / min * 1 minute / (60 seconds) was multiplied by 2 * 10^9 seconds in one step
80 beats / min * 1 minute / (60 seconds) * 2 * 10^9 seconds = 3 * 10^9 beats.
In your instructor's opinion the unit 'beats' should have been left in the result; the text expressed the result simply as 3 * 10^9, apparently ignoring the fact that the unit 'beats' was included in the quantities on the left-hand side.
Also the text identified this number as 3 trillion. In the British terminology this would be correct; in American terminology this number would be 3 billion, not 3 trillion.
COMMENT:
I thought that these problems were pretty basic and felt that I understood them well. However, when I got to questions 14 (determine your own mass in kg) and 15 (determining how many meters away the Sun is from the Earth), I did not understand how to complete these. I know my weight in pounds, but how can that be converted to mass in kilograms? I can look up how to convert miles to meters, but is this something I should already know?
INSTRUCTOR RESPONSE:
Both of these questions could be answered knowing that an object with a mass of 1 kg has a weight of 2.2 lb, and that an inch is 2.54 centimeters. This assumes that you know how many feet in a mile, and that the Sun is 93 million miles away. All these things should be common knowledge, but it doesn't appear to be so.
For my own weight I would reason as follows:
I weigh 170 lb and every kg of my mass weighs 2.2 lb. I'll have fewer kg of mass than I will pounds of weight, so it's reasonable to conclude that my mass is 170 / 2.2 kg, or about 78 kg.
More formally 170 lb * (1 kg / (2.2 lb) ) = 170 / 2.2 kg = 78 kg, approx.. (technical point: this isn't really right because pounds and kilograms don't measure the same thing--pounds measure force and kg measure mass--but we'll worry about that later in the course).
Converting 93 million miles to kilometers:
93 million miles * (5280 feet / mile) * (12 inches / foot) * (2.54 cm / inch) * (1 meter / (100 cm) ) = 160 billion meters (approx.) or 160 million kilometers.
Please feel free to include additional comments or questions: :)
&#This looks good. See my notes. Let me know if you have any questions. &#