Help Solving Physics Word Problem
Physics

Tutor: None Selected  Time limit: 1 Day 
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
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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)
answer 3 is incorrect
i have checked ... it looks correct
sorry there is mistake in calc
answer is 19
19.05 will be more correct
hello is that 19.05 is not correct for Question 3 rd ???
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