Phy 231
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.
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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: ->->->->->->->->->->->-> scussion:
length: (17-9)^2 + (10-5)^2 = c^2, and c =~ 9.4 cm
and 9.4 - 7.5 = 1.9 cm, and 1.9cm(.7) =~ 1.3 N
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What is the vector from the first point to the second?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
5i + 8j + 0k from subtracting from the coordinates,
Or we can get the angle by knowing that for the triangle the sides are 8 cm and 5 cm and the hypotenuse is 9.4 cm, so you could use
tan^-1(rise/run) = 58 deg
The x-comp is 9.4 cm cos(58 deg) = 5 cm
The y-comp is 9.4 cm sin(58 deg) = 8 cm
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What is the magnitude of this vector?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
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: ->->->->->->->->->->->-> scussion:
5i/9.4cm + 8j/9.4cm + 0k = .53i + 0.85j + 0k
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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: ->->->->->->->->->->->-> scussion:
0.69i + 1.12j + 0k
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What are the x and y components of the new vector?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
0.69 and 1.12
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This new vector is called the tension vector. It is a force vector which represents the tension. A force vector can be specified by its components, or equivalently by its magnitude and direction.
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10 min
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&#This looks very good. Let me know if you have any questions. &#