HW_14_Solutions_F05 - Physics 112 - HW #14 Solutions 13-54...

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- 1 - Physics 112 - HW #14 Solutions Spring 2005 13-54 [Driven Oscillations] If k = m ω d , then A 1 = F max (k - m ω d 2 ) 2 + b 1 2 ω d 2 = F max b 1 ω d (a) A = F max (3b 1 ) ω d = A 1 3 (b) A = F max (b 1 /2) ω d = 2A 1 13-90 [L-rod Oscillator] The CM of the swinging system (two rods combined) is located a distance from the pivot equal to: d = L/2 2 = L 2 2 . The system's moment of inertia about the pivot point is: I = 1 3 mL 2 + 1 3 mL 2 = 2 3 mL 2 = 1 3 ML 2 [where M = 2m is the total mass of the system] So the freq. of its SHM oscillation is: f = 1 2 π Mgd I f = 1 2 π Mg L 2 2 1 3 ML 2 = 1 2 π 3g 2 2L 14.11 [Diving Bell] (a) Gauge pressure p = ρ g h = (1.03 x 10 3 kg/m 3 )(9.81 m/s 2 )(250 m) = 2.52 x 10 6 Pa (b) Force on window F = p A = p π r 2 = (2.52 x 10 6 Pa)(3.142)(0.150 m) 2 = 1.78 x 10 5 N CM d L/2 L/2
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- 2 - 14.54 [Hot Air Balloon] Total weight of balloon + gases inside is W = ρ g V g + 900 N + 1700 N + 3200 N = ρ g V g + 5800 N Bouyancy force is equal to the weight of air displaced by the balloon. F b = ρ a V g = (1.23 kg/m 3 )(2200 m 3 )(9.81 m/s 2 ) = 2.655 x 10 4 N For balloon to be just able to lift, F b = W, so 2.655 x 10 4 N = ρ g V g + 5800 N ρ g V g = 2.075 x 10 4 N ρ g = 2.075 x 10 4 N (2200 m 3 ) (9.81 m/s 2 ) = 0.961 kg/m 3 14.64 [Block and Beaker on Scales] (a) The liquid in the beaker is pushing up on the block with force f b = ρ l V block g, the bouyant force. So the block must be pushing DOWN on the liquid with a force of the same magnitude (N’s 3 rd law.) Then the reading on balance E, times g, is the weight of the liquid plus the beaker, plus (the
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HW_14_Solutions_F05 - Physics 112 - HW #14 Solutions 13-54...

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