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Phy 201
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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Copy the problem below into a text editor or word processor.
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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: ->->->->->->->->->->->-> :
= [sqrt] 8^2 + 5^2
= [sqrt] 89
= + - 9.4 cm = 9.4 cm
T = 0.7 N * 9.4 cm
= 6.6 Newtons
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• What is the vector from the first point to the second?
answer/question/discussion: ->->->->->->->->->->->-> :
9.4 – 7.5 cm = 1.9 cm
1.9 cm * 0.7 Newtons = 1.33 Newtons
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• What is the magnitude of this vector?
answer/question/discussion: ->->->->->->->->->->->-> :
= [sqrt] 8^2 + 5^2
= + - [sqrt] 89
= 9.4 cm
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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: ->->->->->->->->->->->-> :
We get the unit position vector.
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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: ->->->->->->->->->->->-> :
Tension vector = 1 * 9.4 magnitude. = 9.4 Newtons
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• What are the x and y components of the new vector?
answer/question/discussion: ->->->->->->->->->->->-> :
T_x_ = T cos 0-
T_y_ = T sin 0-
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1 hour
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10/19 3 pm
You're on the right track, but you should submit a revision on this one.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#