#$&*
course Phy 201
6/4 7:30 pm
Question: This question, related to the use of the TIMER program in an experimental situation, is posed in terms of a familiar
first-semester system.
Suppose you use a computer timer to time a steel ball 1 inch in diameter rolling down a straight wooden incline about 50 cm long. If the computer
timer indicates that on five trials the times of an object down an incline are 2.42sec, 2.56 sec, 2.38 sec, 2.47 sec and 2.31 sec, then to what extent
do you think the discrepancies could be explained by each of the following:
• The lack of precision of the TIMER program.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I believe that the discrepancies in the time due to the timer program are minimal, because it reads up to 7 decimal places. I understand that
eventually the timer must round off to give you those final 7 decimal places, but the difference is so minimal, that I don’t believe it effects the
time that much.
#$&*
• 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 believe that this is what causes the majority of the discrepancies between the times. Human error is unpredictable and always a factor in
experiments run by human calculations. It takes so long for us to see the cue, send it to our brain to move our finger, and then actually touch the
mouse. Our reaction time is fast, but not perfect.
#$&*
• Actual differences in the time required for the object to travel the same distance.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The time required to the object to travel the same distance could change speeds, however I believe that this is mainly due to other human error
factors. For instance: is the marble placed in the same starting position every trial, does it follow the same path every trial, is the friction
becoming less as the wood heats up during trials. However, as far as doing the same distance goes, if it has taken the same route starting from the
same place, it should take the same amount of time.
#$&*
• Differences in positioning the object prior to release.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I know that I didn’t mark on my book where to release my object, and that would have played into the discrepancies in my times. If your object is
starting at a different position every time, it will travel a different route, and therefore have longer or shorter times.
#$&*
• Human uncertainty in observing exactly when the object reached the end of the incline.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
This again is a human error, and I believe to be one of the major factor in causing discrepancies between times. I stopped timing when the object
rolled completely off the book, however sometimes I am sure I stopped it too soon and other too late. Again, human reaction time between sight and
muscle movement is not simultaneous and can cause variation in time.
#$&*
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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
I believe that this uncertainty is there, but does not actually effect the times enough to be factored in to the timing variations.
#$&*
• The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I believe that the human triggering can contribute to about 20% of the error in timing variation. Humans can trigger at ~.2 rate or about 5 triggers
in a second. This is really fast, however that is just if you let your muscles do the work. If you have to see the object then tell your muscles to
twitch the time is slower.
#$&*
• 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 believe this error is so minimal that here is not real difference required for the object to travel the same distance. I believe that these errors
would again come from human error, and if that was taken out the times would be identical
#$&*
• Differences in positioning the object prior to release.
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I believe that positioning could be about 10 % of the actual uncertainty. The surface should be flat, however if it has even the slightest
differences (rough, bumpy etc) then it will allow the object to move more quickly or more slowly based on the route that it is started on.
#$&*
• 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 can account for ~20 % of the timing error. If the object it moving fast enough, it is hard to tell when it completely rolls off of
the incline or when it is still on.
#$&*
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Question: What, if anything, could you do about the uncertainty due to each of the following? Address each specifically.
• The lack of precision of the TIMER program.
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
You could increase the number of decimal places that the timer runs too ( to 10). However, I don’t think that a few more decimal places would really
make that much of a difference in how accurate the time is.
#$&*
• 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 take out the human touch altogether. There are swimming timers that start when the beeper goes off and stop when the swimmer reaches the
other end and touches the pad placed on the wall. This takes out much of the timing error (but has some errors of its own), and could be used here.
You could set up a timer that would begin timing when the object is released, and stop when the object rolls over the edge of the incline. You could
also have multiple people timing the object and average out the times. This would make a more accurate experiment.
#$&*
• Actual differences in the time required for the object to travel the same distance.
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
For actual differenced in the time required for the object to travel the same distance I don’t believe there is much that can be done to stop the
error. You can make sure everything from wind, heat, gravity and such measurements are all the same, but that wont make much of a difference to the
times.
#$&*
• 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
The difference in positioning can be easily fixed, by marking the starting position, and making a grove in the board. These two things would ensure,
that the object to starting in the same positions and traveling the same path every time.
#$&*
• Human uncertainty in observing exactly when the object reached the end of the incline.
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
In this case you could again take out the humans and run the experiment on a computer system for the timing, or you would again have multiple timers.
These would both minimize the human triggering error.
#$&*
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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 time, then you simply divide the distance by the time and you get cm per second traveled. This will tell you the
average speed of the object.
confidence rating #$&*: 3
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Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer
is connected to your experience.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
-40cm/5 sec= 8 cm per second
My object traveled 26.6 cm in about 2.3 seconds, Therefore it was traveling about 11 cm per sec. My object was moving much faster than this
hypothetical object.
confidence rating #$&*: 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:
1st -20cm/ 3 sec= 6.67 cm/sec
2nd -20cm/2 sec=10 cm/sec
The second half is moving at a much greater velocity than the first half.
confidence rating #$&*: 3
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Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half
the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
I believe that doubling the length of my pendulum will result in about half of the frequency. As the length gets longer the pendulum cycles occur at
a slower rate.
confidence rating #$&*: 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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
This is because they are the x and y intercepts. I am actually not sure, but I believe this could be true.
confidence rating #$&*: 2
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Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean
for the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing
frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
It would mean that the pendulum length was started at 0 length.
confidence rating #$&*: 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:
If the pendulum graph intersected the horizontal axis, then this would mean that it was no longer able to oscillate. This would mean that the length
was so long that it would not be able to complete one oscillation.
confidence rating #$&*: 3
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Question: `qIf a ball rolls down between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far
apart are the points?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
-6cm/sec *5sec= then the points are 30 cm apart.
confidence rating #$&*: 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: Ok
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Question: `qYou were asked to read the text and some of the problems at the end of the section. Tell your instructor about something in the text you
understood up to a point but didn't understand fully. Explain what you did understand, and ask the best question you can about what you didn't
understand.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
The book asked me to take the age of the universe (14,000,000,000 years) and write it in powers by years and then by seconds. I understood how to get
the years, by just counting over all the zeros and 4 and you get 1.4 X 10^10. However, I didn’t understand how to get the same number in seconds. I
know if have to do some calculations, but I’m not sure what they are.
#$&*
@& A year is 365 days; a day is 24 hours; an hour is 60 minute; and a minute is 60 seconds.
One way to reason it out:
So 14 000 000 000 years is 14 000 000 000 * 365 days = 14 000 000 000 * 365 * 24 hours = ... etc..
Or more formally
14 000 000 000 years * 365 days / year * 24 hours / day * 60 minutes / hour * 60 seconds / minute leaves you with units of minutes and the number 14 000 000 000 * 365 * 24 * 60 * 60, which you can multiply out.*@
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 #$&*: 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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
What is the area and its approximate uncertainty of a circle of radius 3.8 x 10 ^4 cm? I understand that I have to use the area formula, but I have
no clue what to do after that.
#$&*
@& This will occur in the Query for the corresponding assignment, and if it's still troubling you at that time we'll address it then.*@
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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