Prelim_Solutions__1_F05 - 0 T = mg cos = (0.50 kg)(9.8 m/s...

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Physics 112 - Exam #1 Solutions Fall 2005 1. (a) C (b) B 2. Point #1: Point #2: Point #3: (1 point for each correctly identified force, direction, and magnitude) 3. A 4. C (1 point for B or D) 5. (a) (b) D 6. 7. (a) (b) (1 point for each arrow in correct quadrant but not correct direction) 8. G (1 point) & g sin 2 θ (2 points) 9. (a) v x = D D 2 + H 2 v o = 4.00 m 5.00m (10.0 m/s) = 8.00 m/s v x t = D/2 t = D/2 v x = 2.00 m 8.00 m/s = 0.250 s (b) v x = H D 2 + H 2 v o = 3.00 m 5.00m (10.0 m/s) = 6.00 m/s y = v yo t - 1 2 gt 2 = (6.00 m/s)(0.250 s) - 1 2 (9.80 m/s 2 )(0.250 s) 2 = 1.19 m OR y = H + v yo t - 1 2 gt 2 = 3.00 m + (- 6.00 m/s)(0.250 s) - 1 2 (9.80 m/s 2 )(0.250 s) 2 = 1.19 m w F buoy f d w F buoy w F buoy f d B/S v B/R v R/S v E A B C D F G
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10. (a) v y (t) (c) v y = v yo - gt = 0 - (9.8 m/s 2 )(3.0 s) = - 27.4 m/s max. speed = 27.4 m/s (d) area under a y (t) vs. t graph = Δ v y y(t) = 0 (from t = 0 to 5 s) = - g (3.0 s) + 1 2 a max (2.0 s) = 0 a max = 3g = 27.4 m/s 2 (b) Scale Reading 11. (a) (b) Σ F y = T cos θ - mg =
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Unformatted text preview: 0 T = mg cos = (0.50 kg)(9.8 m/s 2 ) cos75 = 19 N (c) F x = T sin = ma x = m v 2 r = m v 2 L sin v = T L sin 2 m = (19 N)(2.0 m) sin 2 75 0.50 kg = 8.4 m/s 12. (a) C (c) F y = N - mg cos + F sin = ma y = (b) N = mg cos - F sin (d) f k = k N = k (mg cos - F sin ) Fx = mg sin + F cos - f k = ma x = = mg sin + F cos - k (mg cos - F sin ) = F = mg k cos - sin cos + k sin = mg k- tan 1 + k tan (e) E t t t w +x +y T w (= mg) +y w grav Earth F tension string N contact ramp f k ramp friction +x (1 point for each force correctly identified by interaction and body providing it AND in the correct direction )...
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This note was uploaded on 03/27/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Spring '07 term at Cornell University (Engineering School).

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Prelim_Solutions__1_F05 - 0 T = mg cos = (0.50 kg)(9.8 m/s...

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