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PHYS218_CHP01_41-50

# PHYS218_CHP01_41-50 - 1.41 The total northward displacement...

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1.41: The total northward displacement is km, 75 . 1 km 50 . 1 km 3.25 = - , and the total westward displacement is km 4.75 . The magnitude of the net displacement is ( 29 ( 29 km. 06 . 5 km 75 . 4 km 1.75 2 2 = + The south and west displacements are the same, so The direction of the net displacement is 69.80 West of North. 1.42: a) The x - and y -components of the sum are 1.30 cm + 4.10 cm = 5.40 cm, 2.25 cm + (–3.75 cm) = –1.50 cm. b) Using Equations (1-8) and (1-9), 2 2 ) cm 50 . 1 ( ) cm 0 4 . 5 ( - = 5.60 cm, arctan ( 29 40 . 5 50 . 1 + - = 344.5 o ccw. c) Similarly, 4.10 cm – (1.30 cm) = 2.80 cm, –3.75 cm – (2.25 cm) = –6.00 cm. d) 2 2 ) cm 0 . 6 ( ) cm 80 . 2 ( - + = 6.62 cm, arctan ( 29 80 . 2 00 . 6 - = 295 o (which is 360 o – 65 o ). 1.43: a) The magnitude of B A + is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 cm 48 . 2 60.0 sin cm 90 . 1 0 . 60 sin cm 2.80 60.0 cos cm 90 . 1 60.0 cos cm 80 . 2 2 2 = - + + and the angle is ( 29 ( 29 ( 29 ( 29 18 0 . 60 cos

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PHYS218_CHP01_41-50 - 1.41 The total northward displacement...

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