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PHY 201
Your 'cq_1_01.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_01.1_labelMessages **
Phy 201 Asmt Q1
The problem:
Here is the definition of rate of change of one quantity with respect to another:
The average rate of change of A with respect to B on an interval is
average rate of change of A with respect to B = (change in A) / (change in B)
Apply the above definition of average rate of change of A with respect to B to each of the following. Be sure to identify the quantity A, the quantity B and the requested average rate.
If the position of a ball rolling along a track changes from 10 cm to 20 cm while the clock time changes from 4 seconds to 9 seconds, what is the average rate of change of its position with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> n/discussion (start in the next line):
If A equals the change in position and B equals the change in clock time, then the change in A with respect to B equals A/B, Equals change in A, (20cm - 10cm) = 10cm / change in B ( 9sec - 4 sec) = 5sec
The average rateq of of chaneg of a with respect to B is (+ 10cm / 5 sec) = 2cm / sec
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If the velocity of a ball rolling along a track changes from 10 cm / second to 40 cm / second during an interval during which the clock time changes by 3 seconds, then what is the average rate of change of its velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
If A equals the change in velocity, and B equals the change in clock time, then the change in A with respect to B equals A/B, Equals change in A, (40cm/s - 10cm/s) = 30cm/s, divided by the change in B ( 3 sec) = +10 cm / s / s.
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If the average rate at which position changes with respect to clock time is 5 cm / second, and if the clock time changes by 10 seconds, by how much does the position change?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
If the average rate of change in position (A) equals 5 cm/s, and the clock time change (B) equals 10 s then, The change in A with respect to the change in B equals 5cm/s * 10 s - 50 cm change in position.
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You will be expected hereafter to know and apply, in a variety of contexts, the definition given in this question. You need to know this definition word for word. If you try to apply the definition without using all the words it is going to cost you time and it will very likely diminish your performance. Briefly explain how you will ensure that you remember this definition.
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
I will remember that the change in A, (position), will always be in tespect of the change in B ( time reference),
#$&*
You are asked in this exercise to apply the definition, and given a general procedure for doing so. Briefly outline the procedure for applying this definition, and briefly explain how you will remember to apply this procedure.
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
I will figure what the dependant variable is and put it in respect to the independant variable ( tiem).
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*#&!
&#Very good responses. Let me know if you have questions. &#
#$&*
PHY 201
Your 'cq_1_01.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_01.1_labelMessages **
Phy 201 Asmt Q1
The problem:
Here is the definition of rate of change of one quantity with respect to another:
The average rate of change of A with respect to B on an interval is
average rate of change of A with respect to B = (change in A) / (change in B)
Apply the above definition of average rate of change of A with respect to B to each of the following. Be sure to identify the quantity A, the quantity B and the requested average rate.
If the position of a ball rolling along a track changes from 10 cm to 20 cm while the clock time changes from 4 seconds to 9 seconds, what is the average rate of change of its position with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> n/discussion (start in the next line):
If A equals the change in position and B equals the change in clock time, then the change in A with respect to B equals A/B, Equals change in A, (20cm - 10cm) = 10cm / change in B ( 9sec - 4 sec) = 5sec
The average rateq of of chaneg of a with respect to B is (+ 10cm / 5 sec) = 2cm / sec
#$&*
If the velocity of a ball rolling along a track changes from 10 cm / second to 40 cm / second during an interval during which the clock time changes by 3 seconds, then what is the average rate of change of its velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
If A equals the change in velocity, and B equals the change in clock time, then the change in A with respect to B equals A/B, Equals change in A, (40cm/s - 10cm/s) = 30cm/s, divided by the change in B ( 3 sec) = +10 cm / s / s.
#$&*
If the average rate at which position changes with respect to clock time is 5 cm / second, and if the clock time changes by 10 seconds, by how much does the position change?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
If the average rate of change in position (A) equals 5 cm/s, and the clock time change (B) equals 10 s then, The change in A with respect to the change in B equals 5cm/s * 10 s - 50 cm change in position.
#$&*
You will be expected hereafter to know and apply, in a variety of contexts, the definition given in this question. You need to know this definition word for word. If you try to apply the definition without using all the words it is going to cost you time and it will very likely diminish your performance. Briefly explain how you will ensure that you remember this definition.
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
I will remember that the change in A, (position), will always be in tespect of the change in B ( time reference),
#$&*
You are asked in this exercise to apply the definition, and given a general procedure for doing so. Briefly outline the procedure for applying this definition, and briefly explain how you will remember to apply this procedure.
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
I will figure what the dependant variable is and put it in respect to the independant variable ( tiem).
#$&*
Copy and paste your work into the box below and submit as indicated:
Your instructor is trying to gauge the typical time spent by students on these questions. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
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30 mon
&#Very good responses. Let me know if you have questions. &#
course phy 122
Professor I have inserted multiple questions throughout the document. Any information you can give me will be greatly appreciated.
018. `Query 16*********************************************
Question: `qPrinciples of Physics and General College Physics 23.28 A light beam exits the water surface at 66 degrees to vertical. At what angle was it incident from under the water?
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Your solution:
Refraction of air= 1
Refraction of water=1.3
1/1.3= .769
Sin(angle of incidence)/ sin (angle of refraction)
Sin(angle of incidence)= 1.3*sin(angle of refraction0=.769 * sin(66degrees)= .7*.9=.6
Angle of incidence= arc sin (.6)= 37 degrees
confidence rating #$&*:
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Poor
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Given Solution:
`a**** The angle of incidence and the angle of refraction are both measured relative to the normal direction, i.e., to the direction which is perpendicular to the surface. For a horizontal water surface, these angles will therefore be measured relative to the vertical.
66 degrees is therefore the angle of refraction.
Using 1.3 as the index of refraction of water and 1 as the index of refraction of air, we would get
• sin(angle of incidence) / sin(angle of refraction) = 1 / 1.3 = .7, very approximately, so that
• sin(angle of incidence) = 1.3 * sin(angle of refraction) = .7 * sin(66 deg) = .7 * .9 = .6, very approximately, so that
• angle of incidence = arcSin(.6) = 37 degrees, again a very approximate result.
You should of course use a more accurate value for the index of refraction and calculate your results to at least two significant figures.
MORE ACCURATE SOLUTOIN:
The above says that the angle is 37 deg, very approximately.
A much more accurate solution:
Using theta_i for the angle of incidence your first equation would have been
sin(theta_i) / sin(theta_r) = 1 / 1.33
which gives you
sin(theta_i) / sin(66 deg) = .75
It follows that sin(theta_i) = .75 * sin(66 deg) = .75 * .914 = .685.
This gives us
theta_i = sin^-1(.685) = 43.2 degrees.
We were given the 66 degree angle with only 2 significant figures, so the appropriate answer would be
theta_i = 43 degrees.
STUDENT QUESTION
it should be 1/1.333 right? nb is where its going which is air
sin(66)/sin (theta)=1/1.333=.75
INSTRUCTOR RESPONSE
The incident beam is in the water (call this medium a, consistent with your usage), the refracted beam in the air (medium b).
It's sin(theta_a) / sin(theta_b) = n_b / n_a, so
sin(theta_a) = 1 / 1.333 * sin(theta_b) = .7 sin(66 dec) = .6 and
theta_a = 37 degrees.
Again all calculations are very approximate.
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Self-critique (if necessary):
Professor,
I have been studying this one problem for hours it seems. I have looked at the given solution along with the more accurate solution. I am plugging the same numbers into my calculator and I am still getting different numbers than you. When I try to solve I can NOT get 37 degrees any way that I try. Would you mind to solve this one in greater detail? It seems to me like there are some steps missing, but perhaps not. Is the .75 found from dividing 1/1.333? When I plug the sin-1(.685) into my calculator I get 0.01195, not 42.3. Please explain.
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Self-critique Rating:
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Question: `qPrinciples of Physics and General College Physics 23.46 What is the power of a 20.5 cm lens? What is the focal length of a -6.25 diopter lens? Are these lenses converging or diverging?
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Your solution:
I will first convert cm to meters. 20.5 cm = .205m
1/.205m=4.87 m ^-1= 4.87 diopters
1/-6.25m^-1=-.16m=-16cm
The lens with 20.5 cm is converging, while the -6.25 lens is diverging because of the negative focal length.
confidence rating #$&*:
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I am having some difficulty wrapping my head around these principles, even with performing the labs and studying the introset problems.
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Given Solution:
`aThe power of the 20.5 cm lens is 1 / (.205 meters) = 4.87 m^-1 = 4.87 diopters.
A positive focal length implies a converging lens, so this lens is converging.
A lens with power -6.25 diopters has focal length 1 / (-6.25 m^-1) = -.16 m = -16 cm.
The negative focal length implies a diverging lens.
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Self-critique (if necessary):
Why is the 4.87 to a negative power? Why are the meters in the -6.25 raised to a negative power? I want to understand this and I’m sure with your input that I will.
@&
1 / (.205 meters) = 1/.205 * 1/meters.
1/.205 = 4.87
1 / meters = meters^-1.
*@
@&
Recall that for any quantity a that
a^-n = 1 / a^n.
This is one of the basic rules of exponents.
*@
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Self-critique Rating:
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Question: `qquery gen phy problem 23.32 incident at 45 deg to equilateral prism, n = 1.52; and what angle does light emerge?
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Your solution:
Using Snell’s law: n1sin’thea1=n2sinthea2
Incident ray is in air therefore n1= 1.00 and n2= 1.52
1.00sin(45degrees)= 1.52sin thea2
Thea2= 27.7 degrees
To calculate the angle of incidence at the 2nd surface:
(90-thea2) +(90-thea3)+ angle at top of triangle= 180 degrees
(90-27.7 degrees)=62.3 degrees
62.3 degrees + (90-thea 3) + 60 degrees= 180 degrees-thea=57.7 degrees
Thea=32.3 degrees
All the angles= 180 degrees( 62.3+60+57.7)
Angle of incidence at second surface is 32.3 degrees
N sinthea3= n (air)sinthea4
1.52 sin 32.3=1.00sin(thea 4)
Thea 4=54.3 degrees
confidence rating #$&*:
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Awful!
@&
Note that this is a General College Physics problem (and one of the more challenging problems for that coruse), not a Principles of Physics problem.
This problem requires a solid background in trigonometry.
*@
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Given Solution:
`aSTUDENT SOLUTION: To solve this problem, I used figure 23-51 in the text book to help me visualize the problem. The problem states that light is incident on an equilateral crown glass prism at a 45 degree angle at which light emerges from the oppposite face. Assume that n=1.52. First, I calculated the angle of refraction at the first surface. To do this , I used Snell's Law: n1sin'thea1=n2sin'thea2
I assumed that the incident ray is in air, so n1=1.00 and the problem stated that n2=1.52.
Thus,
1.00sin45 degrees=1.52sin'theta2
'thea 2=27.7 degrees.
Now I have determined the angle of incidence of the second surface('thea3). This was the toughest portion of the problem. To do this I had to use some simple rules from geometry. I noticed that the normal dashed lines onthe figure are perpendicular to the surface(right angle). Also, the sum of all three angles in an equilateral triangle is 180degrees and that all three angles in the equilateral triangle are the same. Using this information, I was able to calculate the angle of incidence at the second surface.
I use the equation
(90-'thea2)+(90-'thea3)+angle at top of triangle=180degrees.
(90-27.7degrees)=62.3 degrees. Since this angle is around 60 degrees then the top angle would be approx. 60 degrees. ###this is the part of the problem I am a little hesitant about. Thus,
62.3 degrees+(90-'thea3)+60 degrees=180 degrees-'thea)=57.7degrees
'thea=32.3 degrees
This is reasonable because all three angles add up to be 180 degrees.62.3+60.0+57.7=180degrees
Now, I have determined that the angle of incidence at the second surface is 32.3 degrees, I can calculate the refraction at the second surface by using Snell's Law. Because the angles are parallel,
nsin'thea3=n(air)sin''thea4
1.52sin32.3=1.00sin (thea4)
'thea 4=54.3 degrees
INSTRUCTOR COMMENT:
Looks great. Here's my explanation (I did everything in my head so your results should be more accurate than mine):
Light incident at 45 deg from n=1 to n=1.52 will have refracted angle `thetaRef such that sin(`thetaRef) / sin(45 deg) = 1 / 1.52 so sin(`thetaRef) = .707 * 1 / 1.52 = .47 (approx), so that `thetaRef = sin^-1(.47) = 28 deg (approx).
We then have to consider the refraction at the second face. This might be hard to understand from the accompanying explanation, but patient construction of the triangles should either verify or refute the following results:
This light will then be incident on the opposing face of the prism at approx 32 deg (approx 28 deg from normal at the first face, the normals make an angle of 120 deg, so the triangle defined by the normal lines and the refracted ray has angles of 28 deg and 120 deg, so the remaining angle is 32 deg).
Using Snell's Law again shows that this ray will refract at about 53 deg from the second face. Constructing appropriate triangles we see that the angle between the direction of the first ray and the normal to the second face is 15 deg, and that the angle between the final ray and the first ray is therefore part of a triangle with angles 15 deg, 127 deg (the complement of the 53 deg angle) so the remaining angle of 28 deg is the angle between the incident and refracted ray. **
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Self-critique (if necessary):
Can you please show me in detail how you are getting these angles?
I keep looking and calculating the answers and I am not getting anything remotely close to the answers above.
"
Self-critique (if necessary):
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Self-critique rating:
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&#Your work looks good. See my notes. Let me know if you have any questions. &#