121ss06m1_sol

# 121ss06m1_sol - Solutions Exam A 1(1(D m/s4 x s4 = m(2(E...

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Solutions: Exam A 1. (1) (D) m/s 4 x s 4 = m (2) (E) The initial velocity is (–3 m/s ) (term linear in t ) (3) (B) 4 5 3 2 4 (2) 63 x m = - p + = (4) (C) 4 (5) ( ) ( ) 3 3 ( ) 3 4 0 3 4 4 3 16 dx d d t d t v t t t dt dt dt dt = = - + = - + = - + (5) (A) The position vs time graph curves upwards with increasing time. 2. (6) (C) (7) (D) 0.7 1.5 cg v = m/s (8) (A) 2 2 ( ) ( ) cb bg cg v v v = + (9) (D) tan bg cg v v θ = (10) (B) 2 2 2 2 0.7 0.467 / ; ( ) ( ) (0.467) (0.6) 0.760 1.5 cg cb cg bg v m s v v v m s = = = + = + = 1 1 0 0.6 tan ( ) tan ( ) 52.1 ; 0.467 bg cg v v - - = = = 3. (11) (B) | | cos W mg = P (for the angle drawn) (12) (C) sin fr F N mg μ = = (for the angle drawn) (13) (D) net F ma = (14) (A) cos sin mg mg ma - = (15) (E) 0 0 cos 9.81 cos60 1.8 cos sin 0.365 sin 9.81 sin60 g a mg mg ma g - - - =# = = = 4. (16) (B) The acceleration of gravity acting on the satellite is 2 ) /( h R GM + (17) (C) The centripetal force is 2 /( ) GMm R h + (18) (D) 2 / mv GMm r = 1 cg v r v r bg v r C) 0 60 θ=

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(19) (C) 2 v r f π = (20) (B) 2 2 / / / mv r GmM r v GM r = = = 11 24 6 6 6.67 10 6.0 10 /(6.4 10 5.5 10 ) 5799 / m s - = p + = 1 6 6 5 5799 1 24 3600 2 7.76 10 6.70 2 2 . 10 5.5 10 ) v s v r f f day r s day - - =↵ = = = = ♦(6 4♠ + 5. (21) (C) average velocity = 0, since the change in position is 0, but the distance moved is non-zero. (22) (B) both arrive after the same time on the ground (23) (A) 2 2 4 4 4 (0.301 0.1) 1.19 / g m g m m s 2 2 = ∆ = = = 6. (24) (C) Head To Tail method gives for the vectors B A / r and shown: (25) (B) t = trailor, p = passenger, g = ground, call east positive:
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121ss06m1_sol - Solutions Exam A 1(1(D m/s4 x s4 = m(2(E...

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