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course Phy 201
7/14 4
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 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
First of all, the timer program only measures the time to the hundredths place, meaning that it estimates the time to the hundredths place. Also it seems that the timer does some more estimating as well. The times for each trip vary by several hundredths of a second. So the timer program could have intervals in which if the time follows in that interval then the time is estimated at a certain number. For example, the first trip was 2.42 seconds. The closest times to 2.42 seconds are 2.38 and 2.47. So it could be that any times in between like 2.40 and 2.44 are estimated to 2.42 seconds. This could help contribute to discrepancies. For example a time of 2.56 seconds could actually be more like 2.52, which is much closer to 2.47 seconds. And the 2.47 seconds could be more like 2.50 seconds. Now the difference between these two points is 0.02 instead of 0.09.
I think the discrepancies are explained by this factor but there are other factors that also cause discrepancies as well. An example of this is a person’s ability to judge when the steel ball reaches the end of the incline and then the amount of time it takes that person to hit the timer button. These factors along with the lack of precision with the program cause the discrepancies.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I think this factor plays a large role in the discrepancies. A person’s reaction time is not going to be exactly the same with every trial. Prematurely hitting the timer button or being slow to hit the timer button can cause pretty large time discrepancies. But I think this factor doesn’t affect the times as much as the lack of precision of the timer program.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I think the actual differences in the time required for the object to travel the same distance does not explain the discrepancies much at all. There can be differences in the time due to the ball hitting like a lump in the wood on one trip that it doesn’t hit on another trip but these differences are going to be extremely miniscule. I think that the precision of the timer program and the uncertainty of human triggering play a much larger role in the discrepancies.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The differences in positioning the object prior to release can have an effect on the time. One of the balls could have been set slightly lower down or higher up on the incline causing a discrepancy with time. I think this discrepancy does play a larger role than the actual differences in time required for the ball to travel the same distance. But I think this discrepancy is smaller than the human uncertainty of triggering and timer precision discrepancies.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Human uncertainty in observing exactly when the object reached the end of the incline definitely probably contributed to the discrepancies in time. There is a very slight chance that a human can estimate the point at which the object reached the end of the incline at the exact same point every time. And these few extra millimeters or few shorter millimeters can cause a difference of a few hundredths of a second. But once again, I think the timer precision probably contributes more to the discrepancies.
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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
This factor would fairly largely contribute to the uncertainty. I think this factor makes us able to only be certain up to the tenths place. If this program was more accurate then I believe we could potentially be certain up to the hundredths place.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
This factor would contribute to uncertainty as well but not a great deal. Human reaction times don’t differ but seconds or even tenths of a second. I feel like they differ more by hundredths or thousandths of a second. Thus this causes uncertainty to the thousandths place. We can be fairly certain of our times to the hundredths places in regards to the precision of human triggering.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
This factor does not contribute to the uncertainty much at all. I feel like this probably makes the thousandths or millionths place uncertain but nothing before. The times required for the object to travel the same distance should be extremely similar.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Differences in positioning the object prior to release probably causes an uncertainty in the hundredths place. This differences does add or subtract a little bit of time and can make the times a little more uncertain.
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� 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
Human uncertainty in observing exactly when the object reached the end of the incline will cause some uncertainty. I’d say that this would cause uncertainty in the hundredths place. Differences in where the ball is considered to be at the end of the incline can cause several hundredths of a second increase or decrease in the time measured.
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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
You could find a timer program that is more precise. You could also do several trials for each time and take the average so you can hope to level out the discrepancies caused by the timer program. You can also not use all of the decimal places that the time gives. Using less decimal places lets others know that you are less certain of the certainty of your data.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
You could practice hitting the timer button at the same pace over and over again in order to decrease time discrepancies. Also you could adjust your hand and finger to be at the same place every time you hit the timer button. This causes the time between your reaction and your finger pressing the timer button to be the same each time. Also taking several time readings for each time measurement and averaging them would also help to diminish this discrepancy.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
You could try to make sure that the piece of wood is even, sloped the same way, etc. every single time so that the physical incline will not affect any or much discrepancies in time. You can also control this uncertainty by making sure to use the same materials every single time. Don’t switch the piece of wood or object rolling down the incline because these will cause discrepancies.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
You could try and make sure you are releasing the ball at the exact same point every time. You could do this by making a mark on the wood and making a mark on the ball. And then every time you do the experiment you could make sure that the marks on the wood and the ball match up so that you know your positioning of the ball each time is the same.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
You could pick a spot or something and always use that spot to determine the end of the incline. You could mark that spot or carefully watch it in order to decrease uncertainty. You could also not use your eyes but use your ears. You could decide to not try and judge the end of the incline but to press the timer button every time you hear the ball hit the table. This will cause some discrepancy in distance but it will be the same discrepancy every time and will diminish this human uncertainty that we are talking about.
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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:
An object’s average speed is the distance that object travels per unit of time. Thus you can take the distance that the object rolled and divide by the time it took the object to roll that distance and you will have the average speed in units of distance per units of time.
For example, if the ball traveled a distance of 20 cm and traveled that distance in 3 seconds, then the average speed would be 20cm/2 sec = 10 cm per second
Now this is the object’s average speed because we know that the speed changes as the ball goes down the incline. So we cannot find the exact speed because that changes constantly.
confidence rating #$&*: 3
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Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.
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Your solution:
Average velocity = 40cm / 5 seconds = 8 cm per sec
My answer means that the object traveled about 8 centimeters per second as it traveled down the slope. Now this is the average velocity. We know that the velocity changes as the ball goes down the slope. The speed is faster at the end and slower at the beginning compared to the average. All of what I have said and my answer definitely connect to my experience. My distance was shorter and thus there was a shorter amount of time as well but the general differences in the initial average and final speeds are about the same.
confidence rating #$&*: 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:
Average velocity of the first half = (40cm/2) / 3 sec = 6.67 cm per sec
Average velocity on the second half = (40cm/2) / (5 sec - 3 sec) = 10cm per sec
confidence rating #$&*: 3
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Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?
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Your solution:
I think that doubling the length of the pendulum length will result in half the frequency. Doubling the pendulum length will cause the pendulum to have to ravel a greater distance and thus the frequency will decrease. Since we doubled the length, I think the frequency will decrease by half.
confidence rating #$&*: 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:
Coordinates are (x,y) and the x position is how far left or right you go (up and down the x axis) and the y position is how far up or down you go on the y axis in the same column as the x position. So a point on the x axis has a y coordinate of zero because you would not want to move up or down any. If you moved up or down by any number on the y axis then you would no longer rest of the x axis, you would be above or below it. Thus the y coordinate must be 0 in order to remain on the x axis.
A point on the y axis has an x coordinate of 0 because if the x coordinate was anything but 0 then the coordinate would no longer be on the y axis, it would be either left or right of the y axis.
confidence rating #$&*: 3
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Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?
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Your solution:
In order for the graph to intersect the vertical axis the horizontal coordinate, or pendulum length, must be 0. This means that there is really no pendulum because it has no length. It is impossible for a nonexistent pendulum to have a frequency.
confidence rating #$&*: 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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
For the line or curve to intersect the horizontal axis then the vertical (y) coordinate, or frequency, must be 0. The x coordinate, or the point where the x axis is intersected, is the length of the pendulum. What this means is that there is a pendulum with a certain length but it is not moving thus it does not have a frequency.
confidence rating #$&*: 3
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Question: `qIf a ball rolls down between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far apart are the points?
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Your solution:
Average Velocity = distance traveled / time it took to travel the distance
( 6cm/sec = x / 5 sec ) 5 sec= 30 cm
The points are 30cm apart.
confidence rating #$&*: 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:
Solution given above. My solution was the dame as the given solution and thus no critique is needed.
confidence rating #$&*:
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Confidence assessment for problem is given above. My colution was correct therefore not critique or critique evaluation needs to be given.
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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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
I was a little confused when the text talked about significant figures and how the number of significant figures can depend on certainty. The book gave the example of 80 km. It said that if we say that something is ABOUT 80 km then there is only 1 sig fig. But if we say that something is EXACTLY 80 km then there is 2 sig figs.
I don't really get why there is a difference here. I always learned that a number like 80 or 500 or 10000 has 1 sig fig. I could just be remembering wrong but I thought that is what I learned. I get how there are 3 sig figs in a number like 508 or how there are 2 sig figs in a number like 0.000029. But the 80 and dealing with uncertainty confuses me.
Why does the “about” or “exactly” make the number of significant figures change?
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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
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:
Percent uncertainty is really hard for me to grasp. Question 5 asks for the percent uncertainty for the measurement 1.57m^2. My guess would be +/- .01, which is in the hundredths place and is thus 1%. 1% is the answer given in the back of the book but I feel that I merely had a lucky guess. Is the uncertainty always +/- 1 as the last point given in the number? Like is the uncertainty of 1.987 +/- .001 or 1/1000 or .1%?
And if the given number is 3.76 +/- 0.25 then is the percent of uncertainty 25%?
(just read the common questions and now I believe I get it but I would still like your input)
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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.
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:
"
Self-critique (if necessary):
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Self-critique rating:
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course Phy 201
7/14 4
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 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
First of all, the timer program only measures the time to the hundredths place, meaning that it estimates the time to the hundredths place. Also it seems that the timer does some more estimating as well. The times for each trip vary by several hundredths of a second. So the timer program could have intervals in which if the time follows in that interval then the time is estimated at a certain number. For example, the first trip was 2.42 seconds. The closest times to 2.42 seconds are 2.38 and 2.47. So it could be that any times in between like 2.40 and 2.44 are estimated to 2.42 seconds. This could help contribute to discrepancies. For example a time of 2.56 seconds could actually be more like 2.52, which is much closer to 2.47 seconds. And the 2.47 seconds could be more like 2.50 seconds. Now the difference between these two points is 0.02 instead of 0.09.
I think the discrepancies are explained by this factor but there are other factors that also cause discrepancies as well. An example of this is a person’s ability to judge when the steel ball reaches the end of the incline and then the amount of time it takes that person to hit the timer button. These factors along with the lack of precision with the program cause the discrepancies.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I think this factor plays a large role in the discrepancies. A person’s reaction time is not going to be exactly the same with every trial. Prematurely hitting the timer button or being slow to hit the timer button can cause pretty large time discrepancies. But I think this factor doesn’t affect the times as much as the lack of precision of the timer program.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I think the actual differences in the time required for the object to travel the same distance does not explain the discrepancies much at all. There can be differences in the time due to the ball hitting like a lump in the wood on one trip that it doesn’t hit on another trip but these differences are going to be extremely miniscule. I think that the precision of the timer program and the uncertainty of human triggering play a much larger role in the discrepancies.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The differences in positioning the object prior to release can have an effect on the time. One of the balls could have been set slightly lower down or higher up on the incline causing a discrepancy with time. I think this discrepancy does play a larger role than the actual differences in time required for the ball to travel the same distance. But I think this discrepancy is smaller than the human uncertainty of triggering and timer precision discrepancies.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Human uncertainty in observing exactly when the object reached the end of the incline definitely probably contributed to the discrepancies in time. There is a very slight chance that a human can estimate the point at which the object reached the end of the incline at the exact same point every time. And these few extra millimeters or few shorter millimeters can cause a difference of a few hundredths of a second. But once again, I think the timer precision probably contributes more to the discrepancies.
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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
This factor would fairly largely contribute to the uncertainty. I think this factor makes us able to only be certain up to the tenths place. If this program was more accurate then I believe we could potentially be certain up to the hundredths place.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
This factor would contribute to uncertainty as well but not a great deal. Human reaction times don’t differ but seconds or even tenths of a second. I feel like they differ more by hundredths or thousandths of a second. Thus this causes uncertainty to the thousandths place. We can be fairly certain of our times to the hundredths places in regards to the precision of human triggering.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
This factor does not contribute to the uncertainty much at all. I feel like this probably makes the thousandths or millionths place uncertain but nothing before. The times required for the object to travel the same distance should be extremely similar.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Differences in positioning the object prior to release probably causes an uncertainty in the hundredths place. This differences does add or subtract a little bit of time and can make the times a little more uncertain.
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� 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
Human uncertainty in observing exactly when the object reached the end of the incline will cause some uncertainty. I’d say that this would cause uncertainty in the hundredths place. Differences in where the ball is considered to be at the end of the incline can cause several hundredths of a second increase or decrease in the time measured.
#$&*
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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
You could find a timer program that is more precise. You could also do several trials for each time and take the average so you can hope to level out the discrepancies caused by the timer program. You can also not use all of the decimal places that the time gives. Using less decimal places lets others know that you are less certain of the certainty of your data.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
You could practice hitting the timer button at the same pace over and over again in order to decrease time discrepancies. Also you could adjust your hand and finger to be at the same place every time you hit the timer button. This causes the time between your reaction and your finger pressing the timer button to be the same each time. Also taking several time readings for each time measurement and averaging them would also help to diminish this discrepancy.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
You could try to make sure that the piece of wood is even, sloped the same way, etc. every single time so that the physical incline will not affect any or much discrepancies in time. You can also control this uncertainty by making sure to use the same materials every single time. Don’t switch the piece of wood or object rolling down the incline because these will cause discrepancies.
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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?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
You could try and make sure you are releasing the ball at the exact same point every time. You could do this by making a mark on the wood and making a mark on the ball. And then every time you do the experiment you could make sure that the marks on the wood and the ball match up so that you know your positioning of the ball each time is the same.
#$&*
� 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 pick a spot or something and always use that spot to determine the end of the incline. You could mark that spot or carefully watch it in order to decrease uncertainty. You could also not use your eyes but use your ears. You could decide to not try and judge the end of the incline but to press the timer button every time you hear the ball hit the table. This will cause some discrepancy in distance but it will be the same discrepancy every time and will diminish this human uncertainty that we are talking about.
#$&*
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Question: If, as in the object-down-an-incline experiment, you know the distance an object rolls down an incline and the time required, explain how you will use this information to find the object 's average speed on the incline.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
An object’s average speed is the distance that object travels per unit of time. Thus you can take the distance that the object rolled and divide by the time it took the object to roll that distance and you will have the average speed in units of distance per units of time.
For example, if the ball traveled a distance of 20 cm and traveled that distance in 3 seconds, then the average speed would be 20cm/2 sec = 10 cm per second
Now this is the object’s average speed because we know that the speed changes as the ball goes down the incline. So we cannot find the exact speed because that changes constantly.
confidence rating #$&*: 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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Average velocity = 40cm / 5 seconds = 8 cm per sec
My answer means that the object traveled about 8 centimeters per second as it traveled down the slope. Now this is the average velocity. We know that the velocity changes as the ball goes down the slope. The speed is faster at the end and slower at the beginning compared to the average. All of what I have said and my answer definitely connect to my experience. My distance was shorter and thus there was a shorter amount of time as well but the general differences in the initial average and final speeds are about the same.
confidence rating #$&*: 3
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Question: If the same object requires 3 second to reach the halfway point, what is its average velocity on the first half of the incline and what is its average velocity on the second half?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Average velocity of the first half = (40cm/2) / 3 sec = 6.67 cm per sec
Average velocity on the second half = (40cm/2) / (5 sec - 3 sec) = 10cm per sec
confidence rating #$&*: 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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
I think that doubling the length of the pendulum length will result in half the frequency. Doubling the pendulum length will cause the pendulum to have to ravel a greater distance and thus the frequency will decrease. Since we doubled the length, I think the frequency will decrease by half.
confidence rating #$&*: 3
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Question: `qNote that for a graph of y vs. x, a point on the x axis has y coordinate zero and a point on the y axis has x coordinate zero. In your own words explain why this is so.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Coordinates are (x,y) and the x position is how far left or right you go (up and down the x axis) and the y position is how far up or down you go on the y axis in the same column as the x position. So a point on the x axis has a y coordinate of zero because you would not want to move up or down any. If you moved up or down by any number on the y axis then you would no longer rest of the x axis, you would be above or below it. Thus the y coordinate must be 0 in order to remain on the x axis.
A point on the y axis has an x coordinate of 0 because if the x coordinate was anything but 0 then the coordinate would no longer be on the y axis, it would be either left or right of the y axis.
confidence rating #$&*: 3
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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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
In order for the graph to intersect the vertical axis the horizontal coordinate, or pendulum length, must be 0. This means that there is really no pendulum because it has no length. It is impossible for a nonexistent pendulum to have a frequency.
confidence rating #$&*: 2
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Question: `qOn a graph of frequency vs. pendulum length, what would it mean for the graph to intersect the horizontal axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the horizontal axis)? What would this tell you about the length and frequency of the pendulum?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
For the line or curve to intersect the horizontal axis then the vertical (y) coordinate, or frequency, must be 0. The x coordinate, or the point where the x axis is intersected, is the length of the pendulum. What this means is that there is a pendulum with a certain length but it is not moving thus it does not have a frequency.
confidence rating #$&*: 3
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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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Average Velocity = distance traveled / time it took to travel the distance
( 6cm/sec = x / 5 sec ) 5 sec= 30 cm
The points are 30cm apart.
confidence rating #$&*: 3
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.............................................
Given Solution:
`aOn the average the ball moves 6 centimeters every second, so in 5 seconds it will move 30 cm.
The formal calculation goes like this:
• We know that vAve = `ds / `dt, where vAve is ave velocity, `ds is displacement and `dt is the time interval.
• It follows by algebraic rearrangement that `ds = vAve * `dt.
• We are told that vAve = 6 cm / sec and `dt = 5 sec. It therefore follows that
• `ds = 6 cm / sec * 5 sec = 30 (cm / sec) * sec = 30 cm.
The details of the algebraic rearrangement are as follows:
• vAve = `ds / `dt. We multiply both sides of the equation by `dt:
• vAve * `dt = `ds / `dt * `dt. We simplify to obtain
• vAve * `dt = `ds, which we then write as{}`ds = vAve *`dt
Be sure to address anything you do not fully understand in your self-critique.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Solution given above. My solution was the dame as the given solution and thus no critique is needed.
confidence rating #$&*:
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Confidence assessment for problem is given above. My colution was correct therefore not critique or critique evaluation needs to be given.
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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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
I was a little confused when the text talked about significant figures and how the number of significant figures can depend on certainty. The book gave the example of 80 km. It said that if we say that something is ABOUT 80 km then there is only 1 sig fig. But if we say that something is EXACTLY 80 km then there is 2 sig figs.
I don't really get why there is a difference here. I always learned that a number like 80 or 500 or 10000 has 1 sig fig. I could just be remembering wrong but I thought that is what I learned. I get how there are 3 sig figs in a number like 508 or how there are 2 sig figs in a number like 0.000029. But the 80 and dealing with uncertainty confuses me.
Why does the “about” or “exactly” make the number of significant figures change?
#$&*
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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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:
Percent uncertainty is really hard for me to grasp. Question 5 asks for the percent uncertainty for the measurement 1.57m^2. My guess would be +/- .01, which is in the hundredths place and is thus 1%. 1% is the answer given in the back of the book but I feel that I merely had a lucky guess. Is the uncertainty always +/- 1 as the last point given in the number? Like is the uncertainty of 1.987 +/- .001 or 1/1000 or .1%?
And if the given number is 3.76 +/- 0.25 then is the percent of uncertainty 25%?
(just read the common questions and now I believe I get it but I would still like your input)
@& Gladly.
The percent uncertainty is the uncertainty, divided by the value of the quantity.
1.57 would be regarded as being uncertain by +- .01, as you say.
.01 is about 0.6 % of 1.57 (the calculation is .01 / 1.57 = .006, which is 0.6 %). So the percent uncertainty is +-0.6%, which we would likely round up to +- 1%.
0.25 is about .0006 of 3.76, or 0.06%. We would probably round this up to 0.1%.*@
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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.
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:
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Self-critique (if necessary):
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Self-critique rating:
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&#Good responses. See my notes and let me know if you have questions. &#