Help Solving Physics Word Problem

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The length of the minute hand on a clock is
11cm.
A) Through what angle did this hand turn between 3:00PM and 3:20PM? Answer in units of degrees

B) What distance was actually traveled by the tip of the minute hand during this time interval? Answer in units of cm.

C) What was the magnitude of the displacement vector of the minute hand’s tip during this time period ?Answer in units of cm

Jun 23rd, 2015

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Jun 23rd, 2015

60 minutes of minute hand rotation = 360°

therefore for 20° it will move 360 /60 *20 = 120°   (answer 1)

(2) in one rotation (360°) traveled distance = 2 * pi * r = 2*3.14 *11 =69.08 cm

therefore for 120° distance traveled = 69.08/3 = 23.0266 cm    (answer 2) 

(3) for displacement vector find the distance between the starting tip point, and the end tip point. Draw the minute hand at 3.00 and a minute hand at 3:20, join their tips and notice that they form a triangle with angle 120° with legs 11 cm long.

To find the long side (the displacement magnitude), use cosine law:

c² = a² + b² - 2abcos(Θ)
c = √( 11² + 11² - 2(11)(11) cos(120°) )
c = 11 cm     (answer 3) 


Jun 23rd, 2015

answer 3 is incorrect


Jun 23rd, 2015

i have checked ... it looks correct

Jun 23rd, 2015

sorry there is mistake in calc

answer is 19



Jun 23rd, 2015

19.05 will be more correct

Jun 23rd, 2015

hello is that 19.05 is not correct for Question 3 rd ???

Jun 23rd, 2015

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