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It must be emphasized that the truss is not orbiting i.e. not rotating. One final issue we might mention here is whether there is a conflict between the superposition principle and general relativity. For a car moving in a straight line relative to the laboratory, g@V and g@L will be the same. It can't be different density on one side, and on the other side. All depends on what exactly books that you are looking for. When sunlight falls on the water surface at a low angle, a spectrum will be cast on the ceiling.

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For a continuous-transfer process we may divide by time to have (b) Q h Weng Q c = + ∆t ∆t ∆t Weng Q h Q c Useful power output = = − ∆t ∆t ∆t F 7.89 × 10 J − 4.58 × 10 J I 2 500 rev 1 min = 1.38 × 10 GH revolution revolution JK 1 min 60 s F 1 hp IJ = 185 hp P = 1.38 × 10 WG H 746 W K P 1.38 × 10 J s F 1 rev I P = τω ⇒ τ = = ω b2 500 rev 60 sg GH 2π rad JK = 527 N ⋅ m Q 4.58 × 10 J F 2 500 rev I = G J = 1.91 × 10 W ∆t revolution H 60 s K 3 = 5 eng 3 c 5 e je j Q c = mL f = 15 × 10 −3 kg 1.18 × 10 4 J kg = 177 J The heat to melt 15.0 g of Hg is The energy absorbed to freeze 1.00 g of aluminum is e je j Q h = mL f = 10 −3 kg 3.97 × 10 5 J / kg = 397 J Weng = Q h − Q c = 220 J and the work output is e= Section 22.2 Weng Qh = 220 J = 0.554, or 55.4% 397 J Th − Tc 933 K − 243.1 K = = 0.749 = 74.9% 933 K Th The theoretical (Carnot) efficiency is P22.7 W eng (d) P22.6 5 5 eng (c) 3 Heat Pumps and Refrigerators b g COP refrigerator = Qc W (a) If Q c = 120 J and COP = 5.00, then W = 24.0 J (b) Heat expelled = Heat removed + Work done.

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As one student observed, when asked to explain this, "If you apply a force quickly enough to something, it doesn't notice." This curbside balancing act will serve as a simple but useful test to see how changing arm position or holding a pole affects how long you can stay on the curb. If done indoors, from a balcony or high ladder, have it fall into a plastic wastebasket. [DES] Richard DeLombard suggests a very nice extension of this demo.

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First, we know that gravity is a force that attracts all physical objects towards each other (but why this happens is largely unknown!). When you do this correctly on polar coordinate paper, the path is a segment of an ellipse. This property of space, the greater the mass-energy density, the slower the wave velocity, causes 'force' of gravity, thus as mass-energy density is always positive, gravity is always positive. 4. M.s�), and their �elegant equations�. Any other use requires prior permission of the author and Elsevier.

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Choose something small – if that pen you had earlier is still around, use that. These new developments were probably spurred on by the mother of invention, since without a wireless energy system, the prior conception would be rather useless. And if the separation distance (r) is tripled (increased by a factor of 3), then the force of gravity is decreased by a factor of nine (3 raised to the second power). When B is pulled slowly, A breaks, because the tension in A is larger than that in B by an amount equal to the weight of the ball.

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The frame of the building and your legs exert forces, small relative to the hammer blow, to bring the partition, block, and you to rest again over a time large relative to the hammer blow. The new Bullet 2.82 SDK is available for download. Rovelli: I was young in the sixties and seventies, and shared the dream of my generation: changing the world and make it more just and gentle. The Speed (velocity) of light is the velocity of the spherical waves which make up matter, and is thus determined by the wave medium, i.e. by the Wave Amplitude and mass-energy density of Space.

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The path to understanding it begins with one question: Has it remained the same through the universe’s history? However, if one were to fire a rocket at 17,700 MPH (28,500 km/h), at every instant of time the projectile is falling toward Earth with the force of gravity—but the curved Earth would be falling away from it at the same moment as well. Everything in our universe can be explained by these processes! Applying ∑ F = ma 2T − 480 = ma, where m = 480 = 49.0 kg. 9.80 FIG.

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Think of a time where you've almost fallen into a pool and you swung your arms in a giant circle over your head and bowed forward. We eliminate t = and d by substitution: vw bv − v g15 min + 2 km = bv + v gFGH 2vkm − 15 minIJK v va15 minf − v a15 minf + 2 km = 2 km + 2 km − va15 minf − v a15 minf v v va30 minf = 2 km v w w w w w w w 2 km vw = = 4.00 km h. 30 min (b) In the reference frame of the water, the chest is motionless.

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Q19.4 Rubber contracts when it is warmed. It is often most convenient to work with angular velocity in units of radians/s; doing the conversion gives: Your speed is simply this angular velocity multiplied by your distance from the center of the wheel: (b) We've calculated the initial angular velocity, the final angular velocity is zero, and the angular acceleration is -0.11 rad/s2. One other note on black holes and their information. P18.57 *P18.58 (a) Use the Doppler formula f′= f bv ± v g. bv ∓ v g 0 s With f1′ = frequency of the speaker in front of student and f 2′ = frequency of the speaker behind the student. m + f b343b343sm1s.500m sg = 458 Hz − g b343 m s − 1.50 m sg = 454 Hz f ′ = a 456 Hzf b343 m s + 0g a f1′ = 456 Hz 2 ′ Therefore, fb = f1′ − f 2 = 3.99 Hz. (b) The waves broadcast by both speakers have λ = v 343 m s = = 0.752 m.

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You ask about a way to overcome this separation. The time required to move v dt from 0 to L is From v = ∆t = z z z FGH b L dx L dx 1 = = v 0 T T 0 µ z L af µ x dx 0 I FG µ − µ IJ dxF ∆t = JK H L K GH µ T F µ −µ x I 1 F ∆t = GH µ L µ IJK GH b L g + µ JK 1 − T 2L ∆t = eµ − µ j 3 T bµ − µ g 2Le µ − µ je µ + µ µ + µ j ∆t = 3 T e µ − µ je µ + µ j 2L F µ + µ µ + µ I ∆t = G µ + µ JK 3 TH 1 L 0 g 12 µL − µ0 x + µ0 L 0 L 32 L L 0 0 0 L 32 L L 0 L L L 0 L L L 3 2 0 32 0 L 0 0 0 0 0 0 0 L L L − µ0 IJ K Chapter 16 495 ANSWERS TO EVEN PROBLEMS P16.2 see the solution P16.4 (a) the P wave; (b) 665 s P16.6 0.800 m s P16.8 2.40 m s P16.10 0.300 m in the positive x-direction P16.12 ±6.67 cm P16.14 (a) see the solution; (b) 0.125 s; in agreement with the example P16.16 (a) see the solution; (b) 18.0 m; 83.3 ms; 75.4 rad s; 4.20 m s; (c) 0.2 m sin 18 x + 75.4t − 0.151 P16.40 (a) 15.1 W; (b) 3.02 J P16.42 The amplitude increases by 5.00 times P16.44 see the solution P16.46 f a (a) see the solution; 1 1 2 2 x + vt + x − vt; (b) 2 2 1 1 (c) sin x + vt + sin x − vt 2 2 a a f a f a f P16.48 (a) 0.021 5 m; (b) 1.95 rad; (c) 5.41 m s; (d) y x, t = b g P16.50 (a) 21.0 ms; (b) 1.68 m P16.52 ∆t = P16.20 30.0 N (a) 2Mg; (b) L0 + P16.24 (a) y = 0.2 mm sin 16 x − 3 140t; (b) 158 N P16.26 631 N P16.28 v= P16.30 F m I (a) v = G 30.4 H s ⋅ kg JK mL Mg sin θ (a) see the solution; (b) 3.18 Hz P16.22 a Tg 2π f b g (c) 2 Mg; k 2 Mg 2 Mg L0 + m k P16.56 (a) v = FG H IJ K 14.7 kg P16.58 M m f (a) 0.040 0 m; (b) 0.031 4 m; (c) 0.477 Hz; (d) 2.09 s; (e) positive x -direction f b0.021 5 mg sinb8.38x + 80.0π t + 1.95g P16.32 f (a) y = 0.075 0 sin 4.19 x − 314t; (b) 625 W P16.54 a P16.18 g a b P16.38 e T −7 ρ 10 x + 10 −6 j in SI units; (b) 94.3 m s; 66.7 m s m; (b) 3.89 kg mL tan θ 4Mg P16.34 1.07 kW P16.36 (a), (b), (c) P is constant; (d) P is quadrupled P16.60 see the solution P16.62 (a) 5.00 i m s; (b) −5.00 i m s; (c) −7.50 i m s; (d) 24.0 i m s P16.64 2 (a) µ v 0; (b) v 0; (c) One travels 2 rev and the other does not move around the loop. 496 Wave Motion F 2T IJ = v 2; (a) v = G Hµ K F 2T IJ = v 2; (b) 0.966∆t v′ = G 3 H 3µ K 12 P16.66 0 P16.68 0 0 12 0 0 0 0 130 m s; 1.73 km 17 Sound Waves CHAPTER OUTLINE 17.1 17.2 17.3 17.4 17.5 17.6 Speed of Sound Waves Periodic Sound Waves Intensity of Periodic Sound Waves The Doppler Effect Digital Sound Recording Motion Picture Sound ANSWERS TO QUESTIONS Q17.1 Sound waves are longitudinal because elements of the medium—parcels of air—move parallel and antiparallel to the direction of wave motion.