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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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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: ->->->->->->->->->->->-> :
The length of the vector can be solved using the Pythagorean theorem. The coordinates are connected in a straight line, the hypotenuse. The change in the x-component is the base, 5 cm ( 10 - 5 ). The change in the y-value is the y-component, the length of the long side, 8 cm ( 17 - 9 ). The length of the vector can be found using the Pythagorean theorem. If a^2 + b^2 = c^2, then, 5^2 + 8^2 = c^2. Thus, 89 ( 25 + 64 ) = c^2. And since the square root of 89, the length of a side, cannot be negative, the answer will be roughly a positive 9.4 cm, the length of the rubber band. 9.4-7.5 = 1.9, the additional length of the rubber band. Thus, the tension will be about 1.33 Newtons ( 1.9 * .7 ).
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What is the vector from the first point to the second?
answer/question/discussion: ->->->->->->->->->->->-> :
The points of the vector are ( 5 cm, 9 cm ) and ( 10 cm, 17 cm ).
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What is the magnitude of this vector?
answer/question/discussion: ->->->->->->->->->->->-> :
The magnitude of this vector is about 9.4, as solved from the Pythagorean theorem.
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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: ->->->->->->->->->->->-> :
The unit vector has a magnitude of 1, in the same direction as the original vector. Thus, if the original vector has a magnitude of about 9.4 cm, then to make a unit vector, you divide all of the legs of the triangle by 9.4 to be proportional as well.
5 cm( 10 cm - 5 cm )/ 9.4 cm = .53
8 cm ( 17 cm - 9 cm )/ 9.4 cm = .85
( .53, .85 )
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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: ->->->->->->->->->->->-> :
You get a new vector, because you are multiplying a scalar quantity by a vector quantity. The vector has the same alignment/position in space ( angle of direction as compared to the unit vector ) in space, yet its new magnitude is 1.33.
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What are the x and y components of the new vector?
answer/question/discussion: ->->->->->->->->->->->-> :
I simply multiply them by the tension force to get the new x and y coordinates of the vector of magnitude 1.33.
Original Unit Vector: ( .53, .85 )
1.33 N * Original Vector ( .705, 1.13 )
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