course phy 201
2/5/2010 2:08 pm
ph1 query 0Most 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.
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Comment on your experience with the three lab exercises you encountered in this assignment or in recent assignments.
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I felt that these assignments were good to introduce us to the the world of physics. The assignments all had simple caluclations that allowed
us to get an idea of the role that using mathmatical formulas combined with physical laws of our enivironment plays in helping us understand what is
happening during the exercises we have been working on.
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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?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The timer program is accurate since the number are very close in value, however the timer program is not precise because the number are never repeated
even though the conditions of the experiment have not changed.
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· 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?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The uncertainty associated with an actual human finger on a computer mouse plays a significant role in explaining the discrepancies of the times
indicated by the timer. If we were using a timer that ran by use of a computer program that had been preset to calculate time our results would be more
precise because those devices are using a numerical system to count down actual time. But since we are manually clicking on a mouse we are less likely
to get repetitive results because the rate in which we click varies by fraction of seconds, a computer program can be set to calculate fractions of
seconds sequentially, something we cannot achieve by clicking the mouse manually.
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· 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?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I dont think that this factor would play as much of a role as clicking the mouse to operate the timer has, but it may play a small part, but very little
I would think, since the conditions of the object rolling down the ramp have not changed, it would seem very unlikely that this would change the rate
since it is the same ball rolling down the same ramp with no change to the incline of the ramp.
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· Differences in positioning the object prior to release.
To what extent to you think the discrepancies are explained by this factor?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The differences in the positioning of the object prior to release could be a factor in the discrepancies in the timer. This may be due to the way the
ball rolls down the ramp, it goes at an angle instead of a straight line. The surface of the board may not be consistently level, perhaps more bumpy or
more rigid, which play a role in the discrepancies in time.
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· 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?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Humane uncertainty could be a factor because you may not be stopping, or stop clicking the timer, at the same exact point to which you think the end is.
If the ball were to stop on its own at the end then the timing may be a little more precise, because the ball simply cannot roll any longer. But since
we must visually estimate when the ball reaches the end of the ramp and this may be hard to judge, the timing may have variances.
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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?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I think that this factor would contribute a great amount to the uncertainty in timing a nmber of trials. If the timer is not precise, there is no way
to be sure exactly which trial is accurate
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· 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?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
This factor, I think, has the most uncertainty. Since there is not humanly way possible to click the mouse with any precision, and it is in fact done
almost iradically, being that one may click faster or slower at certain points. This would be the, I think, the most important factor concerning any
discrepancies or uncertainties in timing.
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· 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?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
This may contribute to uncertainties because, for one we are judging the starting points and stopping point of the ball as it rolls down the ramp. It
would be almost impossible to precisly jusdge at waht point the ball starts to roll and reaches the end of the ramp with each trial. So the distances
are not going to be exact each trial.
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· Differences in positioning the object prior to release.
To what extent to you think this factor would contribute to the uncertainty?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The Differneces in positioning contribute because you may have a starting point that is closer or farther away to the end of the ramp in each trial, if
there are any discrepencies in the condition of the surface of the ramp, like bumps or ridges it may make the ball roll slower, and if the ball is
rolling at more of an angle then just straight down, there will be some uncertainty in each trial.
· 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?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
This would account for much uncertainty, because it is nearly impossible to stop the timer at the exact point that the ball reaches the end the same
for each trial.
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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?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
You could run several trials, the trials that have repeated occurances in timing that occur most often, average those trails together to get as close to
precise as you could.
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· 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?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Perhaps try to develop a systematic approach, maybe like beats per second, when clicking the mouse, so that you are doing it at the same rate each time.
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· 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?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
maybe get a percent of increase or decrease in each trial, and then average those number together, so that you could be as accurate as possible.
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· Differences in positioning the object prior to release.
What do you think you could do about the uncertainty due to this factor?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Try to mark the positioning and release the object from that point only.
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· 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?
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your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
You could place a shield at the end of the ramp so that the ball stops at that point only, each and every time.
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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:
D=r * t If you have the distance (D) and the time (t), then speed or rate r=D/t
confidence rating #$&*:3
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Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer
is connected to your experience.
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Your solution:
D=r *t
40cm=r * 5s
r= 8cm/s
so the velocity is 8 cm per second
confidence rating #$&*: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:
First Half
20cm=r * 3s
r= 6.67 cm/s
Second Half
20cm=r * 2s
10 cm/s
confidence rating #$&*:2
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I am not sure if If am calculating this right???????
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Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the
frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?
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Your solution:
I think that by doubling the lenght you would get less than half the frequency. This may be becasue the distance from the start of the cycle to the
end of the cylce and back is longer due to the excess length of the pendulum. Therefore the frequency decreases. The shorter the length the more
frequent the cycles are.
confidence rating #$&*: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:
I think that this may be where the points of y vs x imtercept, But I am really not sure???????
If x = 0 then the point is on the Y axis. For example (0, 1), (0, -4) and (0, 1005) are all on the y axis.
Similar if y = 0 then the point is on the x axis. For example (5, 0), (-12, 0) and (-357, 0) are all on the x axis.
confidence rating #$&*:1
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Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for
the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing
frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?
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Your solution:
I know that as the length decreases the frequency increases...so for the graph to intersect on the vertical axis the length would have to be getting
smaller and smaller. So the pendulums cycles have increased due to the shorter length of the pendulum length.
????????But I still cannot seem to wrap my brain around the graphs increasing as it decreases, or decreases as it increases, or if it is decreasing as
it decreases concept????
A graph can't increase as it decreases, but it can increase while its slope decreases. Such a graph is increasing at a decreasing rate.
Or it can decrease as while getting steeper and steeper, so that it decreases at an increasing rate.
confidence rating #$&*:1
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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:
If the graph intersect on the horizontal axis the lenghts of the pendulum must be getting longer and longer. The graph moves closer to intersecting on
the vertical axis as the length decreases, but as the length increases it seems to move toward the horizontal axis.
confidence rating #$&*:1
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Question: `qIf a ball rolls down between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far
apart are the points?
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Your solution:
V= 6cm
t= 5s
D=6cm*5s
D=30 cm/s
Right idea, but the calculation would be
6 cm / s * 5 s = 30 cm,
not
6 cm * 5 s,
which would give you 30 cm * s.
confidence rating #$&*: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:
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:
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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.
I did not fully understand these questions
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Your solution:
1)In the text question five asks for the percent uncertainty of a measurement given 1.57 m^2
I think that we figure this by an uncertainty of .01/1.57m^2 = .6369 or approximately one. ??????Am I correct in how I calculate this??????? Can I
asuume that if the number given was 1.579 then we would calculate it by .001/1.57 = .1 % approximately or am I incorrect?????
You're on the right track.
There are two ways to look at this.
1.57 m^2 represents a quantity which rounds off to 1.57, so presumably lies between 1.565 and 1.575.
This means that the quantity is within .005 of 1.57.
.005 / 1.57 = .003, approx., so the uncertainty is .003 of 1.57, which is the same as 0.3%, of 1.57.
Another way to look at it:
1.57 could be interpreted to mean a number between 1.56 and 1.58. The uncertainty would then be .01, which is .01 / 1.57 = .006, or 6%, of 1.57.
2)In the text question number 11 the book asks what is the percent uncertainty in the volume of a sphere whose radius is r=2.86 plus or minus .09
I know that the Volume of a spher is 4/3 pi r^3, so I calculated the volume to be 4/3 pi (2.86)^3 = 97.99 and to get the percent uncertainty I tried
to divide 0.09/97.99 * 100 =.091846, but the book answer is 9% ??????I am not sure what i am doing wrong here?????????????????
Again there are two ways to approach this.
I believe the book tells you that the uncertainty in the square of a number is double the uncertainty in the number, and the uncertainty in the cube of the number is trip the uncertainty in the number.
An uncertainty of .09 in a measurement of 2.86 is .09 / 2.86 = .03, approx., or about 3%. As you state, you cube the radius to find the volume. When 2.86 is cubed, the resulting number has three times the uncertainty, or about 9%.
Another approach:
Calculate the volume for r = 2.86.
Then calculate the volume for r = 2.86 - .09 = 2.77.
You will find that the resulting volumes differ by about 9%.
You could just as well have calculated the volume for r = 2.86 + .09 = 2.95. Again you would find that the volume differs from the r = 2.86 volume by about 9%.
????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????"
&#Your work looks good. See my notes. Let me know if you have any questions. &#