Answers to posted final exam:
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(a) 2.4 X 10-3 kg-m/s
(b) For the raindrop, vf - vi = 0-(-12) = +12. Dp (change in momentum) = +, force on the raindrop is +, force is UP. By Newton's Third Law, force on the roof is down.
(c) number = 1.8 X 104ADt
(d) Force = Dmomentum/Dt; pressure = force/area, pressure = 43.2 Pa.
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(a) Q and Uint; Q = ND[(1/2)mv2]
(b) Q = CnDT
(c) C = (3/2)R = 12.5 J/mole-deg
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(a)cneon = 617 J/kg-deg; cargon = 312 J/kg-deg
(b) 15.2 g
- (a) 30 J
(b) 20 J
(c) 0.025 deg. C
- (a) Ff = 10.8 N; Wf = -4.31 J
(b) -7.84 J
(c) 0.839 m/s
- (a)3.33 rad/s
(b) 0.751 J
(c) 0.038 m, down
- (a) x: -0.193 kg-m/s; y: +0.052 kg-m/s
(b) x: 0.493 kg-m/s; y: -0.052 kg-m/s
(c) 0.450 m/s
(d) 5.99 deg.
- (a) -5.39 X 107 J
(b) -0.84 X 107 J
(c) 4.1 X 103 m/s
- (a) 2.67 X 104 m/s
(b) 3.96 m/s2
(c) 0.727 m/s2
- (a) 4.56 kg-m2/s
(b) 2.22 rad/s
(c) Positive; he exerts inward force as masses move inward.
- (a) Fx = Fsin30o; Fy = Fcos30o
(b) Fy = mg
(c) Fx = mv2/r
(d) tan30o = v2/rg, v = 1.064 m/s