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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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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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
The tension will be approximately 1.4N because the length of the line is 9.43cm, as found by the Pythagorean theorem (5^2+8^2=89 and sqrt(89)=1.93).
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What is the vector from the first point to the second?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
The vector from the first point is 9.43cm.
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That is the magnitude of the vector.
You also need to specify its direction, either by giving its components or its angle with the positive x axis.
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&&&&The angle of this vector is 53.13 degrees in relation to the positive x axis.&&&&&
What is the magnitude of this vector?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
The magnitude of this vector is 9.43cm.
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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 vector divided by its magnitude is as follows:
9.43cm/9.43cm=1cm.
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The vector has a magnitude and a direction. If you divide a vector by a positive number, its direction does not change.
&&&This vector will be in the positive 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 you multiply the unit vector by 1.4N, you get 1.4N because 1*1.4N = 1.4N.
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If you have found the components of the unit vector, you can multiply those components by 1.4 Newtons to find the components of the force vector.
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What are the x and y components of the new vector?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
The component of this vector is .9 and the y component is 1.2.
?????I really dont understand how you find this?????
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Your length vector had magnitude 9.43 cm. Your force is 1.4 Newtons.
The vector has some direction in the plane. The x and y axes have positive directions, but no direction in the plane can be considered positive or negative.
If you can answer the following everything should become clear. Give me your best answers:
What were the magnitude and angle of the vector you divided by its magnitude to get the unit vector?
What were the x and y components of the vector you divided by its magnitude to get the unit vector?
What do you get when you divide the x and y components of that vector (i.e., the length vector) by its magnitude (which was 9.43 cm)?
What therefore are the x and y components of your unit vector?
What therefore is the direction of your unit vector?
What are the components of the vector you get when you multiply the components of the unit vector by 1.4 Newtons?
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Mark your insertions in the usual manner.
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