Ch 3 Solutions

# Ch 3 Solutions - 1 P3.1 Physics 195 Chapter 3 Homework...

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1. P3.1 xr == ° = = cos cos . θ 550 240 05 275 . m m . m a f a fa f yr ° = −= sin sin . 0866 476 m m a f a fa f 2. P3.6 We have rx y =+ 22 and = F H G I K J tan 1 y x . (a) The radius for this new point is −+= += xy x y r af 2 2 and its angle is tan F H G I K J 1 180 y x . (b) () −+ 2 r . This point is in the third quadrant if , bg is in the first quadrant or in the fourth quadrant if is in the second quadrant. It is at an angle of 180 °+ . (c) () ( ) 33 3 r +− = . This point is in the fourth quadrant if is in the first quadrant or in the third quadrant if is in the second quadrant. It is at an angle of . 3. P3.11 (a) di =− = 10 0 10 0 . ± . m since the displacement is in a straight line from point A to point B. (b) The actual distance skated is not equal to the straight-line displacement. The distance follows the curved path of the semi-circle ( ACB ). sr = 1 2 25 1 5 7 ππ . m C B A 5.00 m d FIG. P3.11 (c) If the circle is complete, d begins and ends at point A . Hence, d = 0. 4. P3.15 To find these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor.

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Ch 3 Solutions - 1 P3.1 Physics 195 Chapter 3 Homework...

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