Query 00

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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.

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Question: This question, related to the use of the TIMER program in an experimental

situation, is posed in terms of a familiar first-semester system.

Suppose you use a computer timer to time a steel ball 1 inch in diameter rolling down a

straight wooden incline about 50 cm long. If the computer timer indicates that on five

trials the times of an object down an incline are 2.42sec, 2.56 sec, 2.38 sec, 2.47 sec

and 2.31 sec, then:

Are the discrepancies in timing on the order of 0.1 second, 0.01 second, or 0.001

second?

****

I would think that the discrepancies are plus or minus .1 second.

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To what extent do you think the discrepancies in the time intervals could be explained

by each of the following:

· The lack of precision of the TIMER program. Base your answer on the precision of the

TIMER program as you have experienced it. What percent of the discrepancies in timing

do you think are due to this factor, and why do you think so?

****

I would not say that it is a lack of precision with the TIMER program but rather with

user error. I don't have the eye-hand coordination that I used to, so I do not trust

that I'm clicking on the timer at the exact moments that I should be. I think that I am

close, but I don't feel confident that I am extremely accurate with my clicking.

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· The uncertainty associated with human triggering (uncertainty associated with an

actual human finger on a computer mouse). What percent of the discrepancies in timing

do you think are due to this factor, and why do you think so?

****

I would expect any problems with accuracy to be almost 100% due to human triggering. I

have a daughter who is 26 years old. When we play similar games on the computer that

require fast thinking and clicking, she out-clicks me at least 2 times for every one of

mine, sometimes faster. I am attributing this to the fact that she grew up with

computers, her brain works faster than mine, and she has better eye-hand coordination.

Also, you have to click accurately both at the beginning and the end, so there are two

opportunities to not be accurate.

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· Actual differences in the time required for the object to travel the same

distance. What percent of the discrepancies in timing do you think are due to this

factor, and why do you think so?

****

I would imagine that, despite the fact that you try to release the objects in an

identical way, that you don't always line it up exactly the same way. I would think

that you would also, despite best efforts, not always let go of it exactly the same way.

The book could have had little dents in it, not been completely smooth, so friction

would be a factor. Also, not being accurate in the ""final click"" for the drop off the

edge.

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· Differences in positioning the object prior to release. What percent of the

discrepancies in timing do you think are due to this factor, and why do you think so?

****

I think that is a big factor along with the manner of release, so I'm saying at least

50%.

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· Human uncertainty in observing exactly when the object reached the end of the incline.

What percent of the discrepancies in timing do you think are due to this factor, and

why do you think so?

****

It's hard to put numbers to these thing. After I just said 50% for the last one, I want

to say 50% for this, too, but that would not leave any room for all of the

possibilities. Therefore, I will say at least one-third.

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Question: If you had carefully timed the ball and obtained the results given above, how

confident would you be that the mean of those five intervals was within 0.1 seconds of

the actual mean? (Note that the mean of the given intervals is 2.43 seconds, as rounded

to three significant figures)? Briefly explain your thinking.

****

Because I do not trust my ability to keep up with all of these things happening at the

same time, and getting things released and timed properly, I would not be terribly

confident that my mean was within 0.1 seconds. However, the experiments that we are

working with are over quickly, so it is possible that the accuracy would be off by less

than half a second, which I know is a long time for these experiments.

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How confident would you be that the 2.43 second mean is within .01 second? Briefly

explain your thinking.

****

I admit that I don't feel confident with any of this yet. I have repeated the

activities several extra times to see if I am on the right track. I would not be

confident with 0.01 second.

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How confident would you be that the 2.43 second mean is within .03 second?

****

I am not sure I'd be confident within 0.03 second. To me, 0.01 and 0.03 are so small

that they are impossible to distinguish between. I just tried to click on my stop

watch. I click and then click right away. It is about .33 second just to do that.

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At what level do you think you can be confident of the various degrees of uncertainty?

Do you think you could be 90% confident that the 2.43 second mean is within 0.1 second

of the actual mean?

Do you think you could be 90% confident that the 2.43 second mean is within 0.01 second

of the actual mean?

Do you think you could be 90% confident that the 2.43 second mean is within 0.03 second

of the actual mean?

Give your three answers and briefly explain your thinking:

****

I would not be 90% confident for 0.1, but I would be more confident of that than the

others. I would maybe be 50% confident.

I would not be 90% confident for 0.01 because it's so small of a time. I may be 20%

confident.

For 0.03 second, I would not be 90% confident. I may be 5% confident.

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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.

****

I believe that the TIMER program will do exactly what I tell it, with minimal

inaccuracy.

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· The uncertain precision of human triggering (uncertainty associated with an

actual human finger on a computer mouse)

****

I have repeated the trials more than required to ensure that I am clicking more

accurately. I have tried to use a stopwatch rather than always use the TIMER program

for some of the trials, just to compare.

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· Actual differences in the time required for the object to travel the same

distance.

****

Making sure that the surface is smooth. Making sure that the surface is level. Making

sure that the spot of release and type of release are consistent.

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· Differences in positioning the object prior to release.

****

For the last trials where I was checking the first half vs. last half speed, I actually

drew a line on my book so that I was releasing if from the same spot, or as close as

possible.

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· Human uncertainty in observing exactly when the object reached the end of the

incline.

****

Again, when I don't think that I was observing it accurately, I just re-did it.

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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:

I took the length of the path the object traveled and divided it by the amount of time

it took to travel. This gave me mm/sec speed.

confidence rating #$&*: 2

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Question: If an object travels 40 centimeters down an incline in 5 seconds then what is

its average velocity on the incline? Explain how your answer is connected to your

experience.

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Your solution:

40 centimeters in 5 seconds would be 8 cm per second (dividing 40 by 5). This is similar

to what we just did because we measured the length of the path of our object and timed

it to find out how fast it was moving.

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:

Half of the incline would be 20 cm. It would go 20 cm in 3 seconds. This would result

in 6.7 cm per second.

For the second half of the incline, it would be going 20 cm in 2 seconds. This would

result in 10 cm/second.

The total trip was 5 seconds, which is why the 2nd half had to be completed in 2

seconds.

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:

Doubling the length of the pendulum, resulted in less cycles. This number was more than

half of the previous length. For example, 8 cm length resulted in 94 cycles; 16 cm

length resulted in 71 cycles; 32 cm in length resulted in 50 cycles. 71 is more than

half of 94, and 50 is more than half of 71.

confidence rating #$&*: 2

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Question: `qNote that for a graph of y vs. x, a point on the x axis has y coordinate

zero and a point on the y axis has x coordinate zero. In your own words explain why

this is so.

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Your solution:

This is where the line crosses the x and y axes. The point (3,0) is three over and zero

up. This is where the line crosses the x axis. The point (0,3) is over zero, up three.

This is where the line would cross the y axis. These 2 points would allow you to draw

the line and find other ordered pairs that satisfy an equation.

confidence rating #$&*: 2

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Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the

vertical axis and length on the horizontal), what would it mean for the graph to

intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its

behavior, if the line or curve representing frequency vs. length goes through the

vertical axis)? What would this tell you about the length and frequency of the

pendulum?

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Your solution:

The vertical axis would the the y axis. This is where we counted number of cycles.

This would mean that for a given length, there would be no cycles. The shorter the

length, the more cycles. The place where the curve gets closest to the y axis would be

when the pendulum is extremely short. It would have to be too short to swing at all.

confidence rating #$&*: 2

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Question: `qOn a graph of frequency vs. pendulum length, what would it mean for the

graph to intersect the horizontal axis (i.e., what would it mean, in terms of the

pendulum and its behavior, if the line or curve representing frequency vs. length goes

through the horizontal axis)? What would this tell you about the length and frequency

of the pendulum?

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Your solution:

This would be the x axis. The x axis shows length. The longer the pendulum, the less

number of cycles. It doesn't make sense that the pendulum could get so long it didn't

swing unless it was so long that it just rested on the ground. I don't think that,

unless the pendulum was resting on the ground, it would ever cross the x axis.

confidence rating #$&*: 2

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Question: `qIf a ball rolls between two points with an average velocity of 6 cm / sec,

and if it takes 5 sec between the points, then how far apart are the points?

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Your solution:

The points would be 30 cm apart. This is because the ball travels 6 cm in a second and

it took 5 seconds to get from one point to the other. You would multiply 6 cm/sec times

5 seconds to get 30 cm.

confidence rating #$&*: 3

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Given Solution:

`aOn the average the ball moves 6 centimeters every second, so in 5 seconds it will move

30 cm.

The formal calculation goes like this:

We know that vAve = `ds / `dt, where vAve is ave velocity, `ds is displacement and `dt

is the time interval.

It follows by algebraic rearrangement that `ds = vAve * `dt.

We are told that vAve = 6 cm / sec and `dt = 5 sec. It therefore follows that

`ds = 6 cm / sec * 5 sec = 30 (cm / sec) * sec = 30 cm.

The details of the algebraic rearrangement are as follows:

vAve = `ds / `dt. We multiply both sides of the equation by `dt:

vAve * `dt = `ds / `dt * `dt. We simplify to obtain

vAve * `dt = `ds, which we then write as{}`ds = vAve *`dt

Be sure to address anything you do not fully understand in your self-critique.

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Your solution:

I understand the concepts but when they get into these formulas, I've been getting

confused. It just will take time and repetition to get them through my head.

confidence rating #$&*: 2

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Question: `qYou were asked to read the text and some of the problems at the end of the

section. Tell your instructor about something in the text you understood up to a point

but didn't understand fully. Explain what you did understand, and ask the best question

you can about what you didn't understand.

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Your solution:

I got confused when it talked about percent error. It made sense that, if you divide 97

by 92 you will get 1.05. It also made some sense that both 97 and 92 have 2 significant

figures, so the answer needs to have 2 signficant figures as well, so you round to 1.1.

Where it got confusing was to say that 97 is plus or minus 1 and 92 is plus or minus

one, and the implied uncertainty for the answer of 1.1 is 0.1. And then it went on to

take the 0.1/1.1 and get an uncertainty of 0.1 or 10%. One reason I'm having trouble

with this concept is that I don't think of numbers in terms of kind of being in the

ballpark with uncertainty attached to each number, which then attaches an uncertainty to

the answer.

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@&

I think you're articulating the issues pretty clearly.

The working assumption is that you can measure accurately to within +- one unit of measurement.

For example if you count 97 pendulum swings in a minute, it's pretty certain that that last swing did not end precisly on the 60th second. There's a very good probability that there's at least some extra fraction of a swing in there, and some possibility that you counted the 97th cycle because it was almost, but not quite, completed.

*@

STUDENT QUESTION

I understand that we cannot measure to exact precision, but when we are dealing with

estimated uncertainty, do we always

just increment our lowest unit by one and that is our uncertainty? Is there a standard

that is used to figure out this?

INSTRUCTOR RESPONSE

The standard answer is that we assume an uncertainty of +- 1 of our smallest unit of

precision. However, depending on how well we can 'see' that smallest unit, we can get

pretty close to +- 1/2 of a unit.

A more sophisticated answer can be given in terms of the statistics of the normal

distribution, but in this course we're not going to go into a whole lot of depth with

that. A calculus background would be just about required to understand the analysis well

enough to apply it meaningfully.

STUDENT QUESTION

I fully understand how to calculate uncertainty, but what if the uncertainty isn’t

given? For example,

problem 6 asks us for the uncertainty of 1.67. Do we just use .01 as the uncertainty?

INSTRUCTOR RESPONSE

Depending on the nature of the instrument and the observation, +- .01 might be

necessary, but we could go to +-.005 if can regard 1.67 as an accurate roundoff.

Without very good reason, though, +-.01 would be the safer assumption.

STUDENT QUESTION

I had trouble grasping the uncertainty. I understand the bit about significant figures,

but I’m not sure how that applies

to the uncertainty. Is it just the last digit of the significant figure that could be

wrong?

INSTRUCTOR RESPONSE

Any measurement is uncertain to some degree.

On some of the initial videos, despite the fact that the ruler was marked in inches and

subdivided to eighths of an inch, the resolution of the image was poor and it wasn't

possible to observe its position within eighths of an inch. Had the videos been very

sharp (and taken from a distance sufficient to remove the effects of parallax), it might

have been possible to make a good estimate of position to within a sixteenth of an inch

or better.

So for the videos, the uncertainty in position was probably at least +- 1/4 inch, very

possibly +- 1/2 inch. But had we used a better camera, we might well have been able to

observe positions to within +-1/16 inch.

The video camera is one instrument, and each camera (and each setup) introduces its own

unique uncertainties into the process of observation.

The same can be said of any setup and any instrument or combination of instruments.

STUDENT QUESTION: I understood the portion discussing the nature of science and felt

familiar with much of the measurement. What I did not fully understand was how do you

know when to write an answer using the powers of 10 or to leave it alone? Several of the

tables had values in powers of 10 for metric prefixes such as centi and mili.

INSTRUCTOR RESPONSE

Whether you use scientific notation or not depends a lot on the context of the

situation.

As a rule of thumb, I would recommend going to scientific notation for numbers greater

than a million (10^6) and less than a millionth (10^-6). When numbers outside this range

are involved in an analysis it's a good idea to put everything into scientific notation.

And when you know that scientific notation is or is not expected by your audience, write

your numbers accordingly.

QUESTION RELATED TO UNIVERSITY PHYSICS (relevant only to University Physics students)

I don’t fully understand the dot product rule

INSTRUCTOR RESPONSE

The dot product of vectors A = a_1 i + a_2 j + a_3 k and B = b_1 i + b_2 j + b_3 k is

a_1 * b_1 + a_2 * b_2 + a_3 * b_3. The dot product is simply a number.

The magnitude of A is | A | = sqrt( a_1 ^ 2 + a^2 ^ 2 + a_3 ^ 2); the magnitude of B is

found in a similar manner.

The dot product is equal to | A | * | B | * cos(theta), where theta is the angle between

the two vectors.

If you have the coefficients of the i, j and k vectors, it is easy to calculate the dot

product, and it's easy to calculate the magnitudes of the two vectors. Setting the two

expressions for the dot product equal to one another, we can easily solve for cos

(theta), which we can then use to find theta.

More importantly for physics, we can find the projection of one vector on another. The

projection of A on B is just the component of A in the direction of B, equal to | A |

cos(theta). The projection of one vector on another is important in a number of

situations (e.g., the projection of the force vector on the displacement, multiplied by

the displacement, is the work done by the force on the interval corresponding to the

displacement).

Dot products are a standard precalculus concept. Check the documents at the links below

for an introduction to vectors and dot products. You are welcome to complete these

documents, in whole or in part, and submit your work. If you aren't familiar with dot

products, it is recommended you do so.

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/pc2/pc2_qa_09.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/pc2/pc2_qa_10.htm

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SOME ADDITIONAL COMMON QUESTIONS:

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QUESTION: I didn’t understand how to calculate uncertainty for a number such as 1.34.

When given examples we had problems such as 1.34 ±0.5 and with that we had a formula

(0.5/1.34)*100. So I do not understand how to compute uncertainty when no estimated

uncertainty is given.

INSTRUCTOR RESPONSE:

The +- number is the uncertainty in the measurement.

The percent uncertainty is the uncertainty, expressed as a percent of the number being

observed.

So the question in this case is simply, 'what percent of 1.34 is 0.5?'.

0.5 / 1.34 = .037, approximately. So 0.5 is .037 of 1.34.

.037 is the same as 3.7%.

I recommend understanding the principles of ratio, proportion and percent as opposed to

using a formula. These principles are part of the standard school curriculum, though

it does not appear that these concepts have been well mastered by the majority of

students who have completed the curriculum. However most students who have the

prerequisites for this course do fine with these ideas, after a little review. It will

in the long run save you time to do so.

There are numerous Web resources available for understanding these concepts. You should

check out these resources and let me know if you have questions.

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QUESTION: I understood the main points of changing the different units, but I’m not

sure when in the problem I should change the number to 10 raised to a certain power. In

example 1-8 I did not understand why they changed 70 beats/min to 2 x 10^9 s.

2 * 10^9 is about the number of seconds in 70 years.

70 beats / min were not changed to 2 * 10^9 seconds; in changing the beats / minute to

beats in a lifetime, there was a step where it was necessary to multiply by 2 * 10^9

seconds.

The example actually used 80 beats / min as a basis for the solution. This was converted

to beats / second by the calculation

80 beats / min * 1 minute / (60 seconds), which would yield about 1.33 beats / second.

This was then multiplied by 2 * 10^9 seconds to get the number of beats in a lifetime:

2 * 10^9 seconds * 1.33 beats / second = 3 * 10^9 beats.

In the given solution 80 beats / min * 1 minute / (60 seconds) was not actually

calculated; instead 80 beats / min * 1 minute / (60 seconds) was multiplied by 2 * 10^9

seconds in one step

80 beats / min * 1 minute / (60 seconds) * 2 * 10^9 seconds = 3 * 10^9 beats.

In your instructor's opinion the unit 'beats' should have been left in the result; the

text expressed the result simply as 3 * 10^9, apparently ignoring the fact that the unit

'beats' was included in the quantities on the left-hand side.

Also the text identified this number as 3 trillion. In the British terminology this

would be correct; in American terminology this number would be 3 billion, not 3

trillion.

COMMENT:

I thought that these problems were pretty basic and felt that I understood them well.

However, when I got to questions 14 (determine your own mass in kg) and 15 (determining

how many meters away the Sun is from the Earth), I did not understand how to complete

these. I know my weight in pounds, but how can that be converted to mass in kilograms? I

can look up how to convert miles to meters, but is this something I should already know?

INSTRUCTOR RESPONSE:

Both of these questions could be answered knowing that an object with a mass of 1 kg has

a weight of 2.2 lb, and that an inch is 2.54 centimeters. This assumes that you know how

many feet in a mile, and that the Sun is 93 million miles away. All these things should

be common knowledge, but it doesn't appear to be so.

For my own weight I would reason as follows:

I weigh 170 lb and every kg of my mass weighs 2.2 lb. I'll have fewer kg of mass than I

will pounds of weight, so it's reasonable to conclude that my mass is 170 / 2.2 kg, or

about 78 kg.

More formally 170 lb * (1 kg / (2.2 lb) ) = 170 / 2.2 kg = 78 kg, approx.. (technical

point: this isn't really right because pounds and kilograms don't measure the same

thing--pounds measure force and kg measure mass--but we'll worry about that later in the

course).

Converting 93 million miles to kilometers:

93 million miles * (5280 feet / mile) * (12 inches / foot) * (2.54 cm / inch) * (1 meter

/ (100 cm) ) = 160 billion meters (approx.) or 160 million kilometers.

QUESTION

What proved to be most tricky in the problems portion was the scientific notation. I am

somewhat familiar with this from

past math classes, but had trouble when dealing with using the powers of 10. I had

trouble dealing with which way to move my decimal according to the problems that were

written as 10^-3 versus 10^3. Which way do you move the decimal when dealing with

negative or positive powers of 10?

INSTRUCTOR RESPONSE

Using your numbers, 10^3 means 10 * 10 * 10 = 1000.

When you multiply a number by 1000 you move the decimal accordingly. For example 3.5 *

1000 = 3500.

10^-3 means 1 / 10^3 = 1 / (10 * 10 * 10) = 1 / 1000.

When you multiply by 10^-3 you are therefore multiplying by 1 / 1000, which is the same

as dividing by 1000, or multiplying by .001.

For example 3.5 * 10^-3 = 3.5 * .001 = .0035.

As another example 5 700 000 * 10^-3 would be 5 700 000 * (1 / 1000) = 5 700.

From these examples you should be able to infer how the decimal point moves.

You can also search the Web under 'laws of exponents', 'arithmetic in scientific

notation', and other keywords.

There isn't a single site I can recommend, and if I did find a good one its URL might

change by the time you try to locate it. In any case it's best to let you judge the

available materials yourself.

When searching under 'arithmetic in scientific notation' using Google, the following

appear as additional suggested search phrases:

scientific notation

exponents

scientific notation metric prefixes

significant digits

multiply with scientific notation

scientific notation decimal

scientific notation lessons

addition and subtraction with scientific notation

scientific notation metric system

'scientific notation lessons' might be a good place to look.

QUESTIONS AND RESPONSES

1)In the text question five asks for the percent uncertainty of a measurement given 1.57

m^2

I think that we figure this by an uncertainty of .01/1.57m^2 = .6369 or approximately

one. ??????Am I correct in how I

calculate this??????? Can I asuume that if the number given was 1.579 then we would

calculate it by .001/1.57 = .1 % approximately or am I incorrect?????

You're on the right track.

There are two ways to look at this.

1.57 m^2 represents a quantity which rounds off to 1.57, so presumably lies between

1.565 and 1.575.

This means that the quantity is within .005 of 1.57.

.005 / 1.57 = .003, approx., so the uncertainty is .003 of 1.57, which is the same as

0.3%, of 1.57.

Another way to look at it:

1.57 could be interpreted to mean a number between 1.56 and 1.58. The uncertainty would

then be .01, which is .01 / 1.57 = .006, or 6%, of 1.57.

2)In the text question number 11 the book asks what is the percent uncertainty in the

volume of a sphere whose radius is

r=2.86 plus or minus .09.

I know that the Volume of a sphere is 4/3 pi r^3, so I calculated the volume to be 4/3

pi (2.86)^3 = 97.99 and to get the

percent uncertainty I tried to divide 0.09/97.99 * 100 =.091846, but the book answer is

9% ??????I am not sure what i am doing wrong here?????????????????

Again there are two ways to approach this.

I believe the book tells you that the uncertainty in the square of a number is double

the uncertainty in the number, and the uncertainty in the cube of the number is trip the

uncertainty in the number.

An uncertainty of .09 in a measurement of 2.86 is .09 / 2.86 = .03, approx., or about

3%. As you state, you cube the radius to find the volume. When 2.86 is cubed, the

resulting number has three times the uncertainty, or about 9%.

Another approach:

Calculate the volume for r = 2.86.

Then calculate the volume for r = 2.86 - .09 = 2.77.

You will find that the resulting volumes differ by about 9%.

You could just as well have calculated the volume for r = 2.86 + .09 = 2.95. Again you

would find that the volume differs from the r = 2.86 volume by about 9%.

STUDENT QUESTION

When reading the section about the scientific notation some of the answers were written

in powers of 10 and some were just

written regularly. How do I know when to turn my answer into a power of 10 or to leave

my answer as is?

INSTRUCTOR RESPONSE

Good question.

Convenience and readability are the main factors. It's a lot less typing or writing to

use 438 000 000 000 000 000 000 than 4.38 * 10^20, and it's easier for the reader to

understand what 10^20 means than to count up all the zeros.

For readability any number greater than 100 000 or less than .001 should probably be

written in scientific notation.

When scientific notation is first used in a calculation or result, it should be used

with all numbers in that step, and in every subsequent step of the solution.

QUESTION

In my problems (I am working from the University Physics text- exercise 1.14) they are

asking for the ratio of length to

width of a rectangle based on the fact that both of the measurements have uncertainty.

?????Is there anything special you

have to do when adding or multiplying numbers with uncertainty?????? I know that there

are rules with significant figures,

but I don’t understand if the same is true for uncertain measurements.

INSTRUCTOR RESPONSE:

For example:

If there is a 5% uncertainty in length and no significant uncertainty in width, then

area will be uncertain by 5%.

If there is a 5% uncertainty in length and a 3% uncertainty in width, then it is

possible for the area result to be as much as 1.05 * 1.03 = 1.08 times the actual area,

or as little as .95 * .97 = .92 times the actual area. Thus the area is uncertain by

about 8%.

This generalizes. The percent uncertainty in the product or quotient of two quantities

is equal to the sum of the percent uncertainties in the individual quantities (assuming

the uncertainties are small compared to the quantities themselves).

(optional addition for University Physics students): The argument is a little abstract

for this level, but the proof that it must be so, and the degree to which it actually is

so, can be understood in terms of the product rule (fg) ' = f ' g + g ' f. However we

won't go into those details at this point.

QUESTIONs RELATED TO UNIVERSITY PHYSICS (relevant only to University Physics students)

I understand everything but the part on measuring the individual i j k vectors by using

cosine.

INSTRUCTOR RESPONSE

It's not completely clear what you are asking, but I suspect it has to do with direction

cosines.

The vector A = a_1 i + a_2 j + a_3 k makes angles with the directions of the x axis, the

y axis and the z axis.

Let's consider first the x axis.

The direction of the x axis is the same as the direction of the unit vector i.

The projection of A on the x direction is just a_1. This is obvious, but it can also be

found by projecting the A vector on the i vector.

This projection is just | A | cos(alpha), where alpha is the angle between A and the x

direction.

Now A dot i = A = (a_1 i + a_2 j + a_3 k) dot i = A = a_1 i dot i + a_2 j dot i + a_3 k

dot i = a_1 * 1 + a_2 * 0 + a_3 * 0 = a_1.

It's also the case that A dot i = | A | | i | cos(alpha). Since | i | = 1, it follows

that A dot i = | A | cos(alpha), so that

cos(alpha) = A dot i / | A | = a_1 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ).

Making the convention that alpha is the angle made by the vector with the x direction,

we say that cos(alpha) is the direction cosine of the vector with the x axis.

If beta and gamma are, respectively, the angles with the y and z axes, reasoning similar

to the above tells us that

cos(beta) = a_2 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ) and

cos(gamma) = a_3 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ).

cos(alpha), cos(beta) and cos(gamma) are called the 'direction cosines of the vector A'

with respect to the three coordinate axes.

Recall that alpha, beta and gamma are the angles made the the vector with the three

respective coordinate axes.

If we know the direction cosines and the magnitude of the vector, we can among other

things find its projection on any of the coordinate axes.

STUDENT QUESTION (University Physics)

Chapter 1 wasn’t bad of course I had to read in detail the vector section there is

little confusion on what is meant by

antiparallel. Does that mean that you wouldn’t displace anything if the magnitude was

equal only the direction was different?

Also when handwritten vectors are written above the say A the arrow is only in one

direction (to the right) not the direction

traveled?

INSTRUCTOR RESPONSE

I don't have that reference handy, but my understanding of the word 'antiparallel' is

two vectors, one of which is in the direction exactly opposite the other.

If two vectors are antiparallel, then their dot product would equal negative of the

product of their magnitudes:

The angle theta between antiparallel vectors v and w would be 180 degrees, so v dot w =

| v | * | w | * cos(180 deg) = - | v | * | w | .

STUDENT QUESTION

I do not understand the answer to problem 13b. I do not understand why it is not correct

to write the total distance covered

by the train as 890,010 meters. I do not understand this because 890 km equals 890,000

meters and if you add the 10 meters

the train overshot the end of the track by, it seems to me the answer should be 890,010

meters. I think the answer has

something to do with uncertainty, but I cannot figure out how to apply it to this

problem.

INSTRUCTOR RESPONSE

If the given distance was 890. kilometers instead of 890 km, then the 0 would be

significant and it would be appropriate to consider additional distances as small as 1

km.

Had the given distance been 890 000. meters then all the zeros would be significant and

additional distances as small as 1 meter would be considered.

As it is only the 8 and the 9 are significant, so that distances less than 10 km would

not be considered significant.

Please feel free to include additional comments or questions:

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