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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.
• 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 vector is an arrow, with magnitude and direction. The length of the original vector is the slope between the two points. You use Pythagorean theorem to find its length. Tension, is a vector of length 1.33 , starting at the head of the original vector of length 9.4. The resultant vector is the adding of the two. Likewise, we can use the Pythagorean theorem again. 9.4^2 + 1.33^2 = c^2. c is the length of the vector. Thus, the length of the new vector is about 9.50. It's Cartesian coordinates are ( 5, 8 ) and ( 6.33 ( + 1.33 ), 17.4 ( + 9.4 ))
####&&&& ( I am not quite sure what the question is asking for. Could you please clarify which vector you are talking about ??? )
The two points are (5 cm, 9 cm) and (10 cm, 17 cm), as indicated at the beginning.
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• What is the magnitude of this vector?
answer/question/discussion: ->->->->->->->->->->->-> :
The magnitude of this vector is about 9.5, 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: ->->->->->->->->->->->-> :
You get a unit vector, a vector with simply a magnitude of one in the same direction; it just shows the direction. Its coordinates are approximately ( 5.707, 8.707 ). .707^2 + .707^2 = 1 ( approx. ). ( .707, .707 ) are unit vector sides.
There is no constant number you can multiply by the vector ( 5.707, 8.707 ) to get the vector ( .707, .707 ).
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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 allignment/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: ->->->->->->->->->->->-> :
It's new components, x and y, are now ( 5.815, 8.815 ), because it is a 45-45-90 triangle.
( 5.815, 8.815 ) is not part of a 45-45-90 triangle whose legs are parallel to the coordinate axes.
1.33 = a^2 + b^2
1.33/2 = a^2
1.33/2 = b^2
a & b = .815 ( est. )
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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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40 Minutes
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Solution
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