Old P1112 Exam 1 Solns.pdf - Prelim Exam#1 Solutions Physics 1112#1 D Spring 2019#2 B E-1 pt each incorrect#3(a E(b A#4(a#5 D g❖ = GM❖/R❖ = gE/2

Old P1112 Exam 1 Solns.pdf - Prelim Exam#1 Solutions...

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Physics 1112 Prelim Exam #1 Solutions Spring 2019 #1. D #2. B, E [-1 pt. each incorrect] #3. (a) E (b) A #4. (a) H (b) C (c) G #5. D [ g = GM /R 2 = g E /2 = (GM E /R E 2 )/2 M /M E = (R /R E ) 2 /2 = (1/2) 2 /2 = 1/8 ] #6. A [-1 each incorrect] #7. B, D [-1 each incorrect] #8. (a) A, C (b) C, D [-1 each incorrect] #9. (a) (20.0 m/s) ˆ i (b) 0 (c) (20.0 m/s)( ˆ i - ˆ j ) (d) (- 1210 m/s 2 ) ˆ j [a = v 2 /R] (e) (- 1210 m/s 2 ) ˆ j [ ! a P / R = ! a P / C + ! a C / R ] #10. (a) (b) Δ x = v x ( t ) dt 0 50 s = area under v x vs t = (20 m/s)(10 s)/2 + (20 m/s)(10 s) + (20 m/s)(10 s)/2 = 800 m v av = Δ x/ Δ t = (800 m)/(50 s) = 16.0 m/s #11. (a) x ( t ) = ( v o cos θ o ) t y ( t ) = h + ( v o sin θ o ) t 1 2 gt 2 (b) t = D v o cos θ o H = h + ( v o sin θ o ) D v o cos θ o 1 2 g D v o cos θ o 2 Simplify H = h + D tan θ o gD 2 2 v o 2 cos 2 θ o (c) Solve for: v o = D cos θ o g /2 h H + D tan θ o = 50.0 m cos45 ° (9.80 m / s 2 )/2 2.00 m 32.0 m + (50.0 m )tan45 ° = 35.0 m/s
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Unformatted text preview: 0 #12. (a) ( ⊥ path) (b) (from A to B) (c) (at 45°) (d) | a av | = | Δ v | Δ t = 2 v (C / 4) / v = 4 2 v 2 C #13. (a) F (b) T (c) F (d) F (e) F #14. (a) ! F ∑ = m ! a ⇒ m 2 g - T = m 2 a m 2 = T g − a = 1.47 N 9.80 m/s 2 − 2.45 m/s 2 = 0.200 kg (b) ! F ∑ = m ! a ⇒ T = (M + m 1 ) a M = T a − m 1 = 1.47 N 2.45 m/s 2 − 0.100 kg = 0.500 kg (c) ! F ∑ = m ! a x: N sin θ = m 1 a y: N cos θ- m 1 g = 0 ⇒ tan θ = a g = 2.45 m/s 2 9.80 m/s 2 = 0.250 θ = 14.0° m 2 m 2 g T a M+m 1 (M+m 1 )g N s a T m 1 g N +y +x θ m 1 a Nsin θ Ncos θ = m 1 a m 1 g v A v B Δ v...
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