#$&*
Phy 201
Your 'cq_1_16.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
A rubber band has no tension until it reaches a length of 7.5 cm. Beyond that length its tension increases by .7 Newtons for every additional centimeter of length.
What will be its tension if its endpoints are at the points (5 cm, 9 cm) and (10 cm, 17 cm) as measured on an x-y coordinate system?
answer/question/discussion: ->->->->->->->->->->->-> :
tens=sqrt( (5 - 10)^2 + (9 - 17) ^ 2)
=sqrt( (5 - 10)^2 + (9 - 17) ^ 2)
= sqrt( 25 + 64)
= sqrt(89)
=9.4cm
1.9 * .7=1.33N
#$&*
What is the vector from the first point to the second?
answer/question/discussion: ->->->->->->->->->->->-> :
sqrt( (5 cm)^2 + (8 cm)^2 ) = sqrt( 89 cm^2) = 9.4 cm
arctan(8 cm / (5 cm) ) = arctan(1.6) = 58 degrees
#$&*
What is the magnitude of this vector?
answer/question/discussion: ->->->->->->->->->->->-> :
sqrt( (5 cm)^2 + (8 cm)^2 ) = sqrt( 89 cm^2) = 9.4 cm
#$&*
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: ->->->->->->->->->->->-> :
(5 cm, 8 cm) / sqrt(89 cm^2) = 5 cm / sqrt(89 cm^2), 8 cm / sqrt(89 cm^2) = (.53, .83)
#$&*
The new vector should have magnitude 1. When you divide a vector by its magnitude the result is a vector with magnitude 1. We call a vector of magnitude 1 a unit vector. What vector do you get when you multiply this new vector (i.e., the unit vector) by the tension?
answer/question/discussion: ->->->->->->->->->->->-> :
.53, .83 * 1.35 N = .71 N, 1.13 N,
#$&*
What are the x and y components of the new vector?
answer/question/discussion: ->->->->->->->->->->->-> :
cos (theta) = tan (8 / 5) = 58 degrees
vx = v cos (theta) = 9.43 cm * cos 58 degrees= 4.99 cm
sin (58 degree) = 0.85, so .85 * 9.43 cm = 7.99 cm
#$&*
*#&!
&#Very good responses. Let me know if you have questions. &#