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3 Chapter 3 Problem Assignment

# 3 Chapter 3 Problem Assignment - Chapter 3 Problem...

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Chapter 3 Problem Assignment 1. Work: = = . / Sin 45 x15 10 6 m s Vo = 10.6 m/s a = -9.8 m/ s2 = + + x xo Vot 12at2 0m = 0m + (10.6m/s)t + ½ (-9.8m/ s2 ) t2 0 = 10.6m/s t - . 4 9t2 . / = . / 4 9m s2t2 10 6 m s t 4.9m/ s2 t = 10.6 t = 2.16 s t = 2.16s/2 t = 1.08s Answer: 1.08 s 2. Vo = 100 m/s t = 2 s x = vt x = (100 m/s)2s x = 200m Answer: 200 m 3. + = a2 b2 c2 + = 1002 1002 c2 10,000 + 10,000 = c2 √20,000 = c

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c = 141.4213562 N a = net force/mass a = 141.42N/50kg a = 2.8284 N/ kg2 Answer: 2.82 N/ kg2 4. Magnitude 5. + = a2 b2 c2 + = 52 42 c2 25 + 16 = c2 √41 = c c = 6.40 m Answer: 6.40 m 6. = + + x xo Vot 12at2 0m = 0m + (19 m/s)t + ½ (-9.8m/ s2 ) t2 0m = 19m/s t – 4.9m/ s2 t2 4.9m/ s2 t2 = 19m/s t 4.9m/ s2 t = 19m/s t = 3.88 s t = 3.88s/2 t = 1.94 s x = 0 +19m/s(1.94s) + ½(-9.8m/ s2 )( . ) 1 94 2 x = 18.4m Answer: 18.4 m
7. Vo = 0 m/s, xo = 25m, x = 0m, a = -9.8m/ s2 0m = 25m + 0m/s t + ½(-9.8m/ s2 ) t2 0 = 25 – 4.9 t2 4.9 t2 = 25 t = 2.26 s x = 0 + 35(2.26) + ½ (0)( . ) 2 26 2 x = 35(2.26) x = 79.1 m Answer: 2.26s for the ball to strike the water; 79.1 m to land horizontally from the bridge

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