Hw09_Solutions

# Hw09_Solutions - Homework 9 Solutions Problem 9.82 Part A...

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Homework # 9 Solutions Problem 9.82 Part A: F thrust f uel = dm f uel dt v f uel F = 5.0 kg s · 510 m s = 2550 N Part B: F thrust air = dm a ir dt v air F = 110 kg s · 510 m s = 56, 100 N Part C: P = dW dt W = Z F dx Since F is constant W = Fx P = d dt ( Fx ) = F dx dt P = 58, 650 N · 290 m s P = 17, 008, 500 J = 2.3 × 10 4 hp 1

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Problem 9.84 Σ F = dp dt mgsin θ = ma + dm dt v rel v rel = 0 since the sand is not moving relative to the sled as it leaks out mgsin θ = ma a = gsin θ d = 1 2 at 2 r 2 d a = s 2 · 150 m 9.8 m s 2 · sin ( 39 ) = 6.97 s Problem 10.21 Part A: In contact without slipping means their linear velocities and linear accelera- tions at the point of contact are equal. α 1 r 1 = α 2 r 2 α 2 = α 1 r 1 r 2 = 7.7 rad s 2 · 7.0 cm 24.0 cm = 2.24 rad s 2 Part B: ω = α t t = ω α 65 rev min · 1 min 60 s · 2 π rad 1 rev = 6.81 rad s t = 6.81 rad s 2.24 rad s 2 = 3.03 s Problem 10.31 2
For a solid sphere, the moment of inertia about an axis through the center is 2 5 MR 2 I = 2 5 MR 2 = 2 5 · 12.5 kg · ( 0.7 m ) 2 = 2.45 kg · m 2 Problem 10.46 Part A: Part B: Part C: Part D: Σ F = m A a F TA - m A gsin θ = m A a F TA = m A ( a + gsin θ ) F TA = 8.0 kg · [ 1.00 m s 2 + 9.8 m s 2 · sin ( 32 )] = 49.5 N 3

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Part E: Σ F = m B a m B gsin φ - F TB = m B a F TB = m B ( gsin φ - a ) F TB = 10.0 kg · [ 9.8 m s 2 · sin ( 61 ) - 1.00 m s 2 ] = 75.7 N Part F: ~ τ = ~ r × ~ F
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