P150_Exam3_F07sol - Physics 150 Exam #3 Solution 12/3/07 1....

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6. Answer = C 16. Answer = B 7. Answer = D 17. Answer = E 8. Answer = C 18. Answer = C 9. Answer = E 19. Answer = C 10. Answer = D 20. Answer = C Physics 150 Exam #3 Solution 12/3/07 1. Answer = A 11. Answer = B 2. Answer = B 12. Answer = E 3. Answer = E 13. Answer = E 4. Answer = B 14. Answer = C 5. Answer = D 15. Answer = E
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F h 657.854N = F h T 1 cos θ 1 () := Then to the horizontal forces equation: F v 1176N = F v m scaffold g m man g + T 1 sin θ 1 := Returning to the vertical forces equation: T 1 1023.439 N = T 1 L 2 m scaffold g x man m man g + L sin θ 1 := τ L 2 m scaffold g x man m man g + T 1 sin θ 1 L := 0 := Taking torques about the right end of the scaffold: T 1 cos θ 1 F h + 0N := Total horizontal forces; T 1 sin θ 1 F v + m scaffold g m man g := Total vertical forces: θ 1 50 deg := m man 120 kg := x man 4m := m scaffold 80 kg := L1 2 m := Problem #1: Physics 150 Exam #3 Solution 12/3/07
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E 3.842 10 28 × J = EKU + := U 7.685 10 28 × J = U G M E m moon r := K 3.842 10 28 × J = K 1 2 m moon v 2 := Energies: D) v 1.022 km sec =
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P150_Exam3_F07sol - Physics 150 Exam #3 Solution 12/3/07 1....

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