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course Phy 201
2/1/13 at 3:46
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.
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:
Are the discrepancies in timing on the order of 0.1 second, 0.01 second, or 0.001 second?
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I would say that the discrepancies in the timing would be on the order of .01 second because all of the times that were taken are to two decimal places.
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To what extent do you think the discrepancies in the time intervals could be explained by each of the following:
· The lack of precision of the TIMER program. Base your answer on the precision of the TIMER program as you have experienced it. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?
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I think 5% of the discrepancies come from the lack of precision of the TIMER program. This is a good program, but it is 100% completely precise.
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· The uncertainty associated with human triggering (uncertainty associated with an actual human finger on a computer mouse). What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?
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I think 40% of the discrepancies come from the uncertainty associated with human triggering because to me human error seems like the biggest problem here that could cause the times to be different. It’s really hard to hit the timer at the exact right time on every trial.
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· Actual differences in the time required for the object to travel the same distance. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?
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I think 5% of the discrepancies is because of the time required for the object to travel the same distance because it should take the same time to travel the same distances unless there is something there changing the rate that the ball is rolling down the ramp. Since it is unlikely that anything is going to change the conditions of the ramp or anything that will affect its rate, but it is possibly that something could I gave it a 5%.
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· Differences in positioning the object prior to release. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?
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I think 30% of the discrepancy if due to the differences in positioning the object prior to release because once again here is where human error comes into play. It is likely that they will start the ball in about the same place, but probably not in the exact right place. I gave it 10% less than the timing discrepancy because they only place the ball once so that is one time they could cause an error and when timing, they have to start and stop the timer so there is two times they could cause an error.
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· Human uncertainty in observing exactly when the object reached the end of the incline. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?
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I think 20% of the discrepancy is due to the human uncertainty in observing exactly when the object reached the end of the incline because this is another time where human error comes into play. Sometimes it may be hard to tell the exact second that they ball reached the end. I gave it less than the other two with human error because I think it is more likely that someone could mess up when pushing the timer program or when placing the ball in the start position than observing when the ball reached the end.
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Question: If you had carefully timed the ball and obtained the results given above, how confident would you be that the mean of those five intervals was within 0.1 seconds of the actual mean? (Note that the mean of the given intervals is 2.43 seconds, as rounded to three significant figures)? Briefly explain your thinking.
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I would be about 99% confident that the 2.43 mean is within .1 seconds because 2.43 suggests that it has an uncertainty of + or - .001 and we are wanting to know if it’s within .1 seconds. I can calculate the percent uncertainty by doing .001 / .1 = .01 then doing .01 * 100 = 1% uncertainty which would mean that I am 99% confident because 100 - 1 = 99.
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How confident would you be that the 2.43 second mean is within .01 second? Briefly explain your thinking.
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I would be about 90% confident that the 2.43 mean is within .01 seconds because 2.43 suggests that it has an uncertainty of + or - .001 and we are wanting to know if it’s within .01 seconds. I can calculate the percent uncertainty by doing .001 / .01 = .1 then doing .1 * 100 = 10% uncertainty which would mean that I am 90% confident because 100 - 10 = 90.
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How confident would you be that the 2.43 second mean is within .03 second?
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I would be about 97% confident that the 2.43 mean is within .03 seconds because 2.43 suggests that it has an uncertainty of + or - .001 and we are wanting to know if it’s within .03 seconds. I can calculate the percent uncertainty by doing .001 / .03 = .03 then doing .03 * 100 = 3% uncertainty which would mean that I am 97% confident because 100 - 3 = 97.
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At what level do you think you can be confident of the various degrees of uncertainty?
Do you think you could be 90% confident that the 2.43 second mean is within 0.1 second of the actual mean?
Do you think you could be 90% confident that the 2.43 second mean is within 0.01 second of the actual mean?
Do you think you could be 90% confident that the 2.43 second mean is within 0.03 second of the actual mean?
Give your three answers and briefly explain your thinking:
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I think I could be at least 90% confident that the 2.43 second mean is within .1 second of the actual mean beacuse 2.43 suggests that it has an uncertainty of + or - .001 and we are wanting to know if it’s within .1 seconds. I can calculate the percent uncertainty by doing .001 / .1 = .01 then doing .01 * 100 = 1% uncertainty which would mean that I am 99% confident because 100 - 1 = 99.
I think I could be 90% confident that the 2.43 second mean is within .01 second of the actual mean because 2.43 suggests that it has an uncertainty of + or - .001 and we are wanting to know if it’s within .01 seconds. I can calculate the percent uncertainty by doing .001 / .01 = .1 then doing .1 * 100 = 10% uncertainty which would mean that I am 90% confident because 100 - 10 = 90.
I think I could be at least 90% confident that the 2.43 second mean is within .03 second of the actual mean because 2.43 suggests that it has an uncertainty of + or - .001 and we are wanting to know if it’s within .03 seconds. I can calculate the percent uncertainty by doing .001 / .03 = .03 then doing .03 * 100 = 3% uncertainty which would mean that I am 97% confident because 100 - 3 = 97.
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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.
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You could use a more precise TIMER program to do the experiment to cut down on the uncertainty.
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· The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)
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I think you could get a computer program to observe the experiment happening and have it start timing when the ball starts moving and stop timing when the ball reaches the end or you could have a recording of each trail. Then go back and time the trial 5 times from watching the same recording and take the average of those recordings to try to get a more precise time that that trial actually took. Another option would be have something holding the ball at the top of the ramp. When you push a button to start the ball rolling down the ramp have it automatically start a timer program. Then have a sensor at the bottom of the ramp so when the ball gets to the bottom it automatically stops the timer program.
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· Actual differences in the time required for the object to travel the same distance.
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I think you could just make sure that the environment that you are doing the experiment doesn’t change while you are doing the experiment, for example, use the same objects in the experiment, keep the same temperature, ect. If nothing changes, then nothing should affect the rate in which the same ball goes down the ramp.
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· Differences in positioning the object prior to release.
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You could make something that holds the ball at the top of the ramp that is just big enough for the ball to fit in. Then you could have a release button to push to let the ball start rolling down the ramp.
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· Human uncertainty in observing exactly when the object reached the end of the incline.
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You could have a sensor at the bottom of the ramp that automatically stops the timer program so you would not have any human error here because they wouldn’t have to observe when the ball reaches the bottom or has to stop the time when it gets to the bottom.
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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:
Speed = distance/time so to fine the object’s average speed on the incline I would take how far the object had to go divided by how long it took for that object to do so.
confidence rating #$&*: 3
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Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.
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Your solution:
If an object travels 40 cm in 5 seconds then I can use this to find velocity by doing displacement/seconds. So 40 cm/ 5 seconds = 8 cm/s
This answer is related to my experience because I can use this type of thinking every day in figuring out things like how long it will take me to get somewhere while I’m driving.
confidence rating #$&*: 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:
For the first half we could find the average velocity for it by doing 20 cm (since that is half the distance) / 3 seconds (since that is how long it took for it to get there so 20 cm / 3 seconds = 6.67 cm/s
For the second half we could find the average velocity for it by doing 20 cm (since that is half the distance) / 2 seconds (which is the time left to travel the rest of the way down the incline) so 20 cm / 2 seconds = 11.5 cm/s
confidence rating #$&*: 3
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Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?
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Your solution:
Based on the results from my introductory pendulum experiment, I think that doubling the length of the pendulum will result in more than half of the frequency. Some of my data was:
Length Frequency What half the frequency would be
6 103 51.5
12 75 37.5
24 54 27
48 38 19
96 27 13.5
By looking at this data you can see that when the length is doubled, the frequency is not cut in half. It is more than half.
confidence rating #$&*: 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 a point is sitting on the x axis the y coordinate has to be zero in order for that point to be sitting there. If the y coordinate was anything but zero then the point would have to be moved vertically and the point could no longer be sitting on the x axis. The same goes for when a point is sitting on the y axis, the x coordinate has to be zero because if it wasn’t then that point would have to be moved horizontally and could no longer be on the y axis.
confidence rating #$&*: 2
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Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?
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Your solution:
If the line would have crossed the vertical axis which would be for the frequency, then that would mean that the length had changed. It could have already crossed in if the length was getting bigger or smaller depending on which side of the vertical axis the line started on. However, I am unsure of how you could have negative length because of it to cross the vertical axis, then it would have either had to start on the left side of the vertical axis meaning that the x coordinate was negative meaning negative length or it would have started on the right and decreased length, but to get to the left side of the axis the x coordinate has to be less than zero.
confidence rating #$&*: 2
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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 line intersected the horizontal axis, then that would mean that the frequency had changed. It could start above the horizontal axis and then once the frequency reaches zero, then it would intersect the horizontal axis.
confidence rating #$&*: 2
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Question: `qIf a ball rolls between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far apart are the points?
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Your solution:
You could find the answer by making an equation with the information we already know and solve for x. So velocity = distance / seconds we already know it took 5 seconds to go between two points and that the average velocity is 6 cm / s. We are trying to find distance. We could find that by doing this:
X / 5 (seconds) = 6 (cm/sec) here we need to get x alone so we multiply 5 to both sides
X = 6 * 5
X = 30 cm
So the ball would be rolling 30 cm.
confidence rating #$&*: 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:
You could find the answer by making an equation with the information we already know and solve for x. So velocity = distance / seconds we already know it took 5 seconds to go between two points and that the average velocity is 6 cm / s. We are trying to find distance. We could find that by doing this:
X / 5 (seconds) = 6 (cm/sec) here we need to get x alone so we multiply 5 to both sides
X = 6 * 5
X = 30 cm
So the ball would be rolling 30 cm.
confidence rating #$&*: 3
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Self-critique: OK
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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:
When I was reading over the percent uncertainty in the book, it seemed like it was pretty easy. I thought that to get it you take the uncertainty / the answer and then multiply that by 100 to make it a percent. However when I was doing the questions at the end of the chapter, I couldn’t seem to come up with the right answer (which is in the back of the book). For example I couldn’t get number 5 right. Since it’s asking for the percent uncertainty of 1.57 m^2 I think the uncertainty of that number should be + or - .01. So I took the uncertainty and divided it by the original number .01 / 1.57 = .006 then I multiplied that by 100 so .006 * 100 = .6% uncertainty. However in the back of the book it said that answer was 1%. I’m not sure what I am doing wrong.
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I expect that the text rounded the .6% up to 1%.
I would be more likely to round it to .5%, if I was going to round it.
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I had the same problem with question 11. It wanted to know the percent uncertainty in volume of a spherical ball with a radius of 2.86 + or - .09 m.
So first I had to find the volume of the sphere by using the formula V = 4/3 * pie * r^3
V = 4/3 * 3.14159* 2.86 ^ 3
V = 4/3 * 3.14159 * 23.393656
V = 97.99
V = 98 or 9.8 * 10^1
So now I need to find the percent uncertainty so:
.09 / 98 = .000918
.0009 * 100 = .09% uncertainty
However the back of the book said that the answer was 9%. I’m not sure how they came up with that
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There are two ways to look at this.
One way:
.09 m is about 3% of 2.86 m.
So each time you multiply by 2.86 you get an additional 3% uncertainty.
When you cube 2.86 you are multiplying by 2.86 three times.
Since you multiply by 2.86 three times you get the 3% uncertainty, three times, giving you a 9% uncertainty.
Another way:
If you calculate the volume based on 2.86, then calculate the volume based on 2.86 + .09 = 2.95, you can calculate the percent difference between the results. You will find that the difference between the two calculated volume is about 9% of the volume.
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STUDENT QUESTION
I understand that we cannot measure to exact precision, but when we are dealing with estimated uncertainty, do we always
just increment our lowest unit by one and that is our uncertainty? Is there a standard that is used to figure out this?
INSTRUCTOR RESPONSE
The standard answer is that we assume an uncertainty of +- 1 of our smallest unit of precision. However, depending on how well we can ‘see’ that smallest unit, we can get pretty close to +- 1/2 of a unit.
A more sophisticated answer can be given in terms of the statistics of the normal distribution, but in this course we’re not going to go into a whole lot of depth with that. A calculus background would be just about required to understand the analysis well enough to apply it meaningfully.
STUDENT QUESTION
I fully understand how to calculate uncertainty, but what if the uncertainty isn’t given? For example,
problem 6 asks us for the uncertainty of 1.67. Do we just use .01 as the uncertainty?
INSTRUCTOR RESPONSE
Depending on the nature of the instrument and the observation, +- .01 might be necessary, but we could go to +-.005 if can regard 1.67 as an accurate roundoff.
Without very good reason, though, +-.01 would be the safer assumption.
STUDENT QUESTION
I had trouble grasping the uncertainty. I understand the bit about significant figures, but I’m not sure how that applies
to the uncertainty. Is it just the last digit of the significant figure that could be wrong?
INSTRUCTOR RESPONSE
Any measurement is uncertain to some degree.
On some of the initial videos, despite the fact that the ruler was marked in inches and subdivided to eighths of an inch, the resolution of the image was poor and it wasn’t possible to observe its position within eighths of an inch. Had the videos been very sharp (and taken from a distance sufficient to remove the effects of parallax), it might have been possible to make a good estimate of position to within a sixteenth of an inch or better.
So for the videos, the uncertainty in position was probably at least +- 1/4 inch, very possibly +- 1/2 inch. But had we used a better camera, we might well have been able to observe positions to within +-1/16 inch.
The video camera is one instrument, and each camera (and each setup) introduces its own unique uncertainties into the process of observation.
The same can be said of any setup and any instrument or combination of instruments.
STUDENT QUESTION: I understood the portion discussing the nature of science and felt familiar with much of the measurement. What I did not fully understand was how do you know when to write an answer using the powers of 10 or to leave it alone? Several of the tables had values in powers of 10 for metric prefixes such as centi and mili.
INSTRUCTOR RESPONSE
Whether you use scientific notation or not depends a lot on the context of the situation.
As a rule of thumb, I would recommend going to scientific notation for numbers greater than a million (10^6) and less than a millionth (10^-6). When numbers outside this range are involved in an analysis it’s a good idea to put everything into scientific notation.
And when you know that scientific notation is or is not expected by your audience, write your numbers accordingly.
QUESTION RELATED TO UNIVERSITY PHYSICS (relevant only to University Physics students)
I don’t fully understand the dot product rule
INSTRUCTOR RESPONSE
The dot product of vectors A = a_1 i + a_2 j + a_3 k and B = b_1 i + b_2 j + b_3 k is a_1 * b_1 + a_2 * b_2 + a_3 * b_3. The dot product is simply a number.
The magnitude of A is | A | = sqrt( a_1 ^ 2 + a^2 ^ 2 + a_3 ^ 2); the magnitude of B is found in a similar manner.
The dot product is equal to | A | * | B | * cos(theta), where theta is the angle between the two vectors.
If you have the coefficients of the i, j and k vectors, it is easy to calculate the dot product, and it’s easy to calculate the magnitudes of the two vectors. Setting the two expressions for the dot product equal to one another, we can easily solve for cos(theta), which we can then use to find theta.
More importantly for physics, we can find the projection of one vector on another. The projection of A on B is just the component of A in the direction of B, equal to | A | cos(theta). The projection of one vector on another is important in a number of situations (e.g., the projection of the force vector on the displacement, multiplied by the displacement, is the work done by the force on the interval corresponding to the displacement).
Dot products are a standard precalculus concept. Check the documents at the links below for an introduction to vectors and dot products. You are welcome to complete these documents, in whole or in part, and submit your work. If you aren’t familiar with dot products, it is recommended you do so.
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/pc2/pc2_qa_09.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/pc2/pc2_qa_10.htm
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SOME ADDITIONAL 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.
QUESTION
What proved to be most tricky in the problems portion was the scientific notation. I am somewhat familiar with this from
past math classes, but had trouble when dealing with using the powers of 10. I had trouble dealing with which way to move my decimal according to the problems that were written as 10^-3 versus 10^3. Which way do you move the decimal when dealing with negative or positive powers of 10?
INSTRUCTOR RESPONSE
Using your numbers, 10^3 means 10 * 10 * 10 = 1000.
When you multiply a number by 1000 you move the decimal accordingly. For example 3.5 * 1000 = 3500.
10^-3 means 1 / 10^3 = 1 / (10 * 10 * 10) = 1 / 1000.
When you multiply by 10^-3 you are therefore multiplying by 1 / 1000, which is the same as dividing by 1000, or multiplying by .001.
For example 3.5 * 10^-3 = 3.5 * .001 = .0035.
As another example 5 700 000 * 10^-3 would be 5 700 000 * (1 / 1000) = 5 700.
From these examples you should be able to infer how the decimal point moves.
You can also search the Web under ‘laws of exponents’, ‘arithmetic in scientific notation’, and other keywords.
There isn’t a single site I can recommend, and if I did find a good one its URL might change by the time you try to locate it. In any case it’s best to let you judge the available materials yourself.
When searching under ‘arithmetic in scientific notation’ using Google, the following appear as additional suggested search phrases:
scientific notation
exponents
scientific notation metric prefixes
significant digits
multiply with scientific notation
scientific notation decimal
scientific notation lessons
addition and subtraction with scientific notation
scientific notation metric system
‘scientific notation lessons’ might be a good place to look.
QUESTIONS AND RESPONSES
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 sphere 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%.
STUDENT QUESTION
When reading the section about the scientific notation some of the answers were written in powers of 10 and some were just
written regularly. How do I know when to turn my answer into a power of 10 or to leave my answer as is?
INSTRUCTOR RESPONSE
Good question.
Convenience and readability are the main factors. It’s a lot less typing or writing to use 438 000 000 000 000 000 000 than 4.38 * 10^20, and it’s easier for the reader to understand what 10^20 means than to count up all the zeros.
For readability any number greater than 100 000 or less than .001 should probably be written in scientific notation.
When scientific notation is first used in a calculation or result, it should be used with all numbers in that step, and in every subsequent step of the solution.
QUESTION
In my problems (I am working from the University Physics text- exercise 1.14) they are asking for the ratio of length to
width of a rectangle based on the fact that both of the measurements have uncertainty. ?????Is there anything special you
have to do when adding or multiplying numbers with uncertainty?????? I know that there are rules with significant figures,
but I don’t understand if the same is true for uncertain measurements.
INSTRUCTOR RESPONSE:
For example:
If there is a 5% uncertainty in length and no significant uncertainty in width, then area will be uncertain by 5%.
If there is a 5% uncertainty in length and a 3% uncertainty in width, then it is possible for the area result to be as much as 1.05 * 1.03 = 1.08 times the actual area, or as little as .95 * .97 = .92 times the actual area. Thus the area is uncertain by about 8%.
This generalizes. The percent uncertainty in the product or quotient of two quantities is equal to the sum of the percent uncertainties in the individual quantities (assuming the uncertainties are small compared to the quantities themselves).
(optional addition for University Physics students): The argument is a little abstract for this level, but the proof that it must be so, and the degree to which it actually is so, can be understood in terms of the product rule (fg) ‘ = f ‘ g + g ‘ f. However we won’t go into those details at this point.
QUESTIONs RELATED TO UNIVERSITY PHYSICS (relevant only to University Physics students)
I understand everything but the part on measuring the individual i j k vectors by using cosine.
INSTRUCTOR RESPONSE
It’s not completely clear what you are asking, but I suspect it has to do with direction cosines.
The vector A = a_1 i + a_2 j + a_3 k makes angles with the directions of the x axis, the y axis and the z axis.
Let’s consider first the x axis.
The direction of the x axis is the same as the direction of the unit vector i.
The projection of A on the x direction is just a_1. This is obvious, but it can also be found by projecting the A vector on the i vector.
This projection is just | A | cos(alpha), where alpha is the angle between A and the x direction.
Now A dot i = A = (a_1 i + a_2 j + a_3 k) dot i = A = a_1 i dot i + a_2 j dot i + a_3 k dot i = a_1 * 1 + a_2 * 0 + a_3 * 0 = a_1.
It’s also the case that A dot i = | A | | i | cos(alpha). Since | i | = 1, it follows that A dot i = | A | cos(alpha), so that
cos(alpha) = A dot i / | A | = a_1 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ).
Making the convention that alpha is the angle made by the vector with the x direction, we say that cos(alpha) is the direction cosine of the vector with the x axis.
If beta and gamma are, respectively, the angles with the y and z axes, reasoning similar to the above tells us that
cos(beta) = a_2 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ) and
cos(gamma) = a_3 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ).
cos(alpha), cos(beta) and cos(gamma) are called the ‘direction cosines of the vector A’ with respect to the three coordinate axes.
Recall that alpha, beta and gamma are the angles made the the vector with the three respective coordinate axes.
If we know the direction cosines and the magnitude of the vector, we can among other things find its projection on any of the coordinate axes.
STUDENT QUESTION (University Physics)
Chapter 1 wasn’t bad of course I had to read in detail the vector section there is little confusion on what is meant by
antiparallel. Does that mean that you wouldn’t displace anything if the magnitude was equal only the direction was different?
Also when handwritten vectors are written above the say A the arrow is only in one direction (to the right) not the direction
traveled?
INSTRUCTOR RESPONSE
I don’t have that reference handy, but my understanding of the word ‘antiparallel’ is two vectors, one of which is in the direction exactly opposite the other.
If two vectors are antiparallel, then their dot product would equal negative of the product of their magnitudes:
The angle theta between antiparallel vectors v and w would be 180 degrees, so v dot w = | v | * | w | * cos(180 deg) = - | v | * | w | .
STUDENT QUESTION
I do not understand the answer to problem 13b. I do not understand why it is not correct to write the total distance covered
by the train as 890,010 meters. I do not understand this because 890 km equals 890,000 meters and if you add the 10 meters
the train overshot the end of the track by, it seems to me the answer should be 890,010 meters. I think the answer has
something to do with uncertainty, but I cannot figure out how to apply it to this problem.
INSTRUCTOR RESPONSE
If the given distance was 890. kilometers instead of 890 km, then the 0 would be significant and it would be appropriate to consider additional distances as small as 1 km.
Had the given distance been 890 000. meters then all the zeros would be significant and additional distances as small as 1 meter would be considered.
As it is only the 8 and the 9 are significant, so that distances less than 10 km would not be considered significant.
Please feel free to include additional comments or questions:
#$&*
This looks good. See my notes. Let me know if you have any questions.