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PHY 121
Your 'cq_1_16.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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cq_1_161
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PHY 121
Your 'cq_1_16.1' report has been received. Scroll down through the document to see any
comments I might have inserted, and my final comment at the end.
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A rubber band has no tension until it reaches a length of 7.5 cm. Beyond that length its
tension increases by .7 Newtons for every additional centimeter of length.
What will be its tension if its endpoints are at the points (5 cm, 9 cm) and (10 cm, 17
cm) as measured on an x-y coordinate system?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
I found the length of the rubber band as follows:
a^2 + b^2 = c^2
(10 cm-5 cm)^2 + (17 cm-9 cm)^2 = c^2
(5cm)^2 + (8cm)^2 = c^2
25 cm^2 + 64 cm^2 = c^2
c^2 = 89 cm^2
c = 9.4 cm
Since it doesn't have tension until 7.5 cm, I subtracted 7.5 cm from 9.4 cm to find the
difference. 9.4 cm - 7.5 cm = 1.9 cm.
For each cm, it has .7 Newtons of force, so I multiplied .7 N * 1.9, and got 1.33 N
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What is the vector from the first point to the second?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
The vector is 9.4
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What is the magnitude of this vector?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
The magnitude is 9.4
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What vector do you get when you divide this vector by its magnitude? (Specify the x and
y components of the resulting vector).
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
If I divide the vector by its magnitude, the answer is 1.
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A vector has a magnitude and a direction. You are right about the magnitudes.
What is the direction of your force vector, and what is the direction of this unit
vector?
&&&& I think that the direction of the force vector is tan-1 (8/5) for an angle of 58 degrees, but I am still in the dark about the unit vector and what it actually is.&&&&
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If you divide a vector by its magnitude the resulting vector has magnitude 1. That makes it a unit vector.
If you divide a vector by its magnitude the resulting vector has the same direction as the original vector.
The unit vector is therefore the vector of magnitude 1, in the direction of the original vector.
If you multiply the unit vector by any
quantity, you therefore get a vector in the same direction as the original vector, with magnitude equal to the quantity you multiplied by.
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If you multiply the unit vector by 1.33 Newtons, then, what do you get?
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What are the x and y components of the unit vector?
What do you get when you multiply those components by 1.33 Newtons?
If the resulting components are the components of a new vector, what is its magnitude and what is its direction?
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The new vector should have magnitude 1. When you divide a vector by its magnitude the
result is a vector with magnitude 1. We call a vector of magnitude 1 a unit vector. What
vector do you get when you multiply this new vector (i.e., the unit vector) by the
tension?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
When I multiply this vector by the tension per centimeter, I get .7. If I am
misinterpreting and I'm supposed to multiply by the tension of this rubberband, then I
get 1.33.
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You should multiply by the tension, so the magnitude will be 1.33.
The vector also has a direction, which you must specify.
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What are the x and y components of the new vector?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
I do not know what the x and y components of the new vector would be.
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I've inserted a few questions that should lead you, in 10-15 minutes, to a pretty good picture of what's going on here.
Please submit one more revision, and this time mark your insertions with ####.
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