Query 0

course Phy 121

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.

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

If the timer is only accurate to the 0.1 of a second, then the discrepancies of the time differences of 2.31 to 2.56 could be because the timer is imply not able to take measurements as accurately as the times portray

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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

The human finger is of course controlled by a human and inherent time delays or differences will exist when the human tries to click on the mouse the second the ball begins to roll. This is all based on the accuracy of the humans perception and is far less precise than a digital or laser the ball rolls through to start or stop the timer. The differences are actually not that far apart when considering a human was hitting the start/stop button for the timer

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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

Depending on the smoothness of the marble, or any deviation in the straightness of the ruler, this could be the case. If the ball rolls down the incline, but hits a snag, divot, or splinter, this could significantly slow down the ball for the longer time recorded.

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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 ball could be millimeters to several cm off the original starting position in subsequent attempts, making it very possible the times could be shorter or longer. Starting at the same place, or as close to there as possible, makes the experiment more accurate. Attention to detail is important.

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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

This is kind of like the reaction time experiment in high school, where a yard stick is dropped between a waiting persons thumb and forefinger, and when the person realizes it has been released, they are supposed to catch it. Reaction time varies greatly with each person. Perceiving when the ball reaches the end, and then hitting the stop button on the stop watch are all subject to human error and reaction time, which are so unreliable, professional experiments use digital devices to measure, record time, and judge starting and stopping of an object (They don’t even rely on humans that much on Myth Busters)

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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 definitely adds to the error analysis at the end of a lab or experiment. The less precise an instrument of measure was, the more likely, the experiment really isn’t all that accurate and should be repeated with more precise equipment

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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

Again, humans are actually pretty unreliable when it comes to perceiving and subsequently causing a timing device to start or stop at the exact spot during every run of an experiment.

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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

I have no doubt there are actual time differences in the same object traveling the same distance under the same conditions numerous times. It just becomes easy to blame the differences, especially large time differences, on variables such as reaction time, starting positions, and non-precise experimentation.

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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:

That depends on how precise the experimenter was on positioning the object. Was the starting position clearly marked, measured, or otherwise double checked before each run? If not, chances are, the time differences are very probably dependant on that the starting positions were different, meaning the distances traveled were different, meaning a time comparison cannot accurately be made.

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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

I think the “Human Factor” is the greatest problem in an experiment conducted by people. Reaction time, Precision, and general accuracy and attention to detail differ not only from person to person, but also from run to run in the same experiment with the same person. We all get tired and at some point sloppy.

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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

With the exception of getting a different timer that is more accurate, just being aware of the limits of the timer and recording the time differences within the limits of the timer would be all that could really be done.

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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

Here, something could be rigged (a laser starting line and finish line) that could be directly linked to a computer, measuring with much more accuracy exactly when an object rolled through each point. The start could also be made more precise through some sort of releasing mechanism that allows the object to be released from the exact same point every time.

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• Actual differences in the time required for the object to travel the same distance.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Making sure the same distance is traveled each time is key. Making the starting point as precise as possible and making sure the same path is traveled each time would make the experiment more accurate, but some variables may still exist that could account for differences in time it takes to travel the same distance.

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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

Here, something could be done, like some sort of trap door release that would make sure the object is released from the same position and in the same manner each time. Less variation in the release or position, simply leads to more accuracy.

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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:

The best thing here is to simply take out the human factor and replace it with something like a motion sensor or laser set at the same spot at the end of the incline that can measure the time taken for the object to travel down the incline.

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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:

If you know the distance and the total time, then you can figure out the time taken to travel certain distance intervals. The total distance can be divided into several interval, the more the better, and then the time taken to travel each of these needs to be figured out. They can be averaged afterward to find the average for the distance intervals which can be used to find the average for the total distance.

confidence rating #$&* 2

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Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.

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Your solution:

The object travels an average of 40 cm/ 5 seconds, or 8cm/ second. This makes sense if the object is not gaining speed, which from the videos seen so far I this course, objects increase speed and velocity when traveling down an incline. So the 8cm/sec in definitely an average, not necessarily the rate at which the object is traveling in any one time interval.

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:

The halfway point is 20cm. So 3 seconds for 20 cm is an average velocity of 6.66 cm/second. The second half is 20cm in 2 seconds with an average velocity of 10cm/second.

confidence rating #$&*: 3

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Question: According 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:

NO. In my experiment, 6 cm pendulum length lead to 104 oscillations. At 12 cm, with the above theory, the oscillations SHOULD have been 52, but they were actually 79. At 24cm length, the oscillations should have been half of the 79 at 12 cm, but they were not 39 or 40, they were actually 57. SO doubling the length of your pendulum will result in MORE than half of the oscillations of half the length.

confidence rating #$&*: 3

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Question: `Note 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:

The origin, or point where the x and y axis cross, is (0,0). Moving along the x axid, without moving up or down results in a value for x, but y remains 0 (because you are literally on the x axis). Moving along the y axis, has the same affect on x, or a value of 0.

confidence rating #$&*: 3

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Question: On 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 length of the pendulum is represented on the x axis, and an intersection of the y axis means a value of 0 for the x axis, it means the pendulum literally has no length and the weight o the end of the pendulum is attached directly to the point of attachment of the pendulum. This either means the weight has a huge amount of oscillations, since the shorter it got the more it oscillated, or it would mean the pendulum or weight would not move at all. Not sure on that one…..

confidence rating #$&*: 2

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Question: On 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 number of frequencies is represented by the y axis, an intersection with the x axis would mean a value of 0 for the number of frequencies, meaning the pendulum was simply sitting still in the position of equilibrium, or not moving at all.

confidence rating #$&*: 3

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Question: If 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:

At a rate of 6cm/sec, traveling 5 seconds, (6*5), the points would be 30 cm 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.

Self Critique not necessary

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Question: You were asked to read the text and some of the problems at the end of the section. Tell your instructor about something in the text you understood up to a point but didn't understand fully. Explain what you did understand, and ask the best question you can about what you didn't understand.

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Your solution:

I think I pretty much understood the point of everything. I was intrigued by the difference between a law and a principle, though. The book states that a law is a certain concise but general statement. A principle is a less general statement…. So where is the line drawn between law and principle. Is there a value of x % before something is a law and if its less, it is regarded as a principle?

There is no fixed threshold value for a law vs. a principle. General acceptance of universality, and simplicity, might be the basic criteria.

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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.

confidence rating #$&* 3

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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:

I understood converting SI units, although I had to look them up and compare what I thought was right to a table. What I got lost on was when the author asked for conversions of units I was not familiar with… like the ps or the fm…… I’m guessing there are picoseconds and femtometers. I guess what I’m asking is, it doesn’t matter if the unit is “pieces” or “bucks”, if they are assumed to be SI units, I can use the table on page 9 to convert them right?

All the SI units for first-semester physics are based on the meter, the kilogram and the second. When we get to electricity and magnetism we add the Ampere.

The table will do for conversions, but I advocate just knowing and applying a few simple equivalencies (1 inch = 2.54 cm, a 1 kg mass weighs 2.204 pounds, and a few other).

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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

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:

&#Good work. See my notes and let me know if you have questions. &#

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