course Phy 121
1/29/10 at 9:09 p.m.Hoping that I don't lose power so I can complete assignments. I have AEP so you know what that can mean...
ph1 query 0
Most queries in this course will ask you questions about class notes, readings, text problems and experiments. Since the first two assignments have been lab-related, the first
two queries are related to the those exercises. While the remaining queries in this course are in question-answer format, the first two will be in the form of open-ended
questions. Interpret these questions and answer them as best you can.
Different first-semester courses address the issues of experimental precision, experimental error, reporting of results and analysis in different ways and at different levels.
One purpose of these initial lab exercises is to familiarize your instructor with your work and you with the instructor 's expectations.
Comment on your experience with the three lab exercises you encountered in this assignment or in recent assignments.
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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
I think this does play a small role but feel that human response time is the biggest factor.
· 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
I definitely think this is a factor, especially as more people are involved the human error factor compounds.
· 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
I think this may play a small role in the discrepancy.
· Differences in positioning the object prior to release.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Since we are using a cylinder it is so difficult to be accurate with the starting point. My husband tried to place the Pepsi can
just barely off center each time so that when it was released it wouldn't go the opposite direction and fall off the book, but there
is no way that we can guarantee that we started the Pepsi can at the same exact spot each time.
· 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
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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
· 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
· 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
I think the difference would be minimal.
· 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 contribute somewhat because if an item is a cylinder, there is no guarantee that one is positioning it at the same spot 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 that this is a factor as well because of the human response time in noticing this and triggering the timer.
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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
I think the best thing to do is just realize that this exists and just factor it in.
· 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
Try to come up with a method to be as consistent as possible.
· 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 not much you can do besides take a few measurements and then determine the average.
· 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
Perhaps I could place a special mark on the object and book and try to line them up each time to be more consistent; it still wouldn't be perfect.
· 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
The placement of an item to create a sound or something to detecht the objects arrival so that it is not just the human eye should help with this.
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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:
Distance divided by time will tell us the speed. If we take multiple measurements of speed(for example three) and then add these up and divide by the number of measurements
(three), we will have the average speed.
confidence rating #$&*: 2
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Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your
experience.
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Your solution:
40 cm/5 sec = 8 cm/sec This would be considered an average because we know that it actually starts down the incline at a slower velocity and then speeds up toward the end.
confidence rating #$&*: 2
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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:
20 cm/3 sec = 6.67 cm/sec on the first half
20 cm/2 sec = 10 cm/sec on the second half
confidence rating #$&*: 2
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Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can
be thought of as the number of cycles per minute), more than half or less than half?
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Your solution:
When looking back at my data, it appears that when the length of the pendulum is doubled, the frequency is more than half.
confidence rating #$&*: 3
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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:
When graphing any point on the x axis, it would have a y coordinate of zero; the x could be any negative number, 0, or any positive number. When graphing any point on the y
axis, the x coordinate must be 0, and the y can be any number.
confidence rating #$&*: 2
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Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for the graph to intersect
the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the vertical axis)?
What would this tell you about the length and frequency of the pendulum?
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Your solution:
The length of the pendulum would be 0; therefore, the frequency would also be 0.
confidence rating #$&*:
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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:
The only place that it will intersect the horizontal axis is at 0,0.
confidence rating #$&*:
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Question: `qIf a ball rolls down between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far apart are the points?
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Your solution:
6cm/1s= x/5s x=30 cm.
confidence rating #$&*: 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 just set up a simple equation and solved for x. I do understand the above.
confidence rating #$&*: 2
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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 am starting to understand the significant figures. I've randomly done this in the past ususally rounding to the second decimal place consistently. The rationale does
make sense to me about how this can help identify the uncertainty. I understand now that the result of an equation should be only as many digits as the number with the least
number of significant figures used in the calculation. It's just a little confusing when we start adding all the zeros and doing scientific notation plus figuring out when 0 is
a significant number. For example, on p. 7 in the second paragraph it says that when you divide 2.0 by 3.0, the proper answer is .67. Why would the 0's be needed here(just to
show the uncertainty? that's exactly it--to indicate the accuracy of the information
), and why shouldn't the answer simply be .7 or .6 with a line over it(which I think is the way that I learned to do this long ago 2/3 = .6666...., but that assumes that 2 and 3 are exact figures; when the numbers are known only to with some uncertainty, the extra significant figures are meaningless and, worse, misleading. The number .6666.... implies infinite accuracy and presision
)? I also don't think I
fully undestand how to determine the level of uncertainty when conducting an experiment. For example, I was asked to determine the uncertainty of the timer that we're using,
and I honestly don't know for sure. you were asked to estimate the uncertainty; at this point we're just trying to get a handle on the idea of uncertainty, so any reasonable estimate would be OK
It varies how many decimal places are shown sometimes when you click it.
confidence rating #$&*: 2
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Question: `qTell your instructor 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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Your solution:
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:
&#Good responses. See my notes and let me know if you have questions. &#