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In 1684 Dr Halley came to visit him at Cambridge, after they had been some time together, the Dr asked him what he thought the curve would be that would be described by the Planets supposing the force of attraction towards the Sun to be reciprocal to the square of their distance from it. What do you expect to see if the center of mass of the container is offset from the geometric center O of the container? Which of the following equalities is true: 2. Its position is spread out to fill the whole string, but its frequency is one of a certain set of quantized values.

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Even if one overcomes this problem, and could observe evidence of the discreteness of space, so many approaches involve such discreteness that one would face a further problem in using this new data to decide between the discrete approaches. Really it increases as the process goes on. 342 P11.45 Angular Momentum (a) I = ∑ mi ri2 =m FG 4d IJ H3K 2 +m FG d IJ H 3K 2 +m FG 2d IJ H3K 2 m m 1 d2 = 7m 3 2d 3 2 d Think of the whole weight, 3mg, acting at the center of gravity. τ =r×F= FG d IJ e− ij × 3mge− jj = bmgdgk H 3K 3g τ 3mgd = = counterclockwise 2 I 7md 7d (c) α= (d) a = αr = FG 3 g IJ FG 2d IJ = H 7d K H 3 K 2g up 7 The angular acceleration is not constant, but energy is. a K + U f + ∆E = a K + U f F dI 1 0 + a3m f g G J + 0 = Iω H 3K 2 i (e) maximum kinetic energy = mgd (f) ωf = (g) L f = Iω f = (h) vf =ω fr = 6g 7d 7md 2 3 6g = 7d 6g d = 7d 3 FG 14 g IJ H3K 2 gd 21 12 md 3 2 f 2 f 3 d FIG.

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I was reminded of the importance of being nitpicky in physics at a press conference yesterday on experimental attempts by Eric Adelberger's group at the University of Washington to find violations in one of the most fundamental aspects of special relativity: Lorentz invariance. (For more specific detail about this experiment, and several others, go here .) That's the bit about the laws of physics being the same for all observers, regardless of frame of reference.

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Since the hollow sphere is uncharged, we require q Q σ2 = 1 2 =. q1 + q 2 = 0 and 4π c 4π c 2 P24.58 z e j E ⋅ dA = E 4π r 2 = (a) e−3.60 × 10 3 qin ∈0 j a f N C 4π 0.100 m 2 = Q 8.85 × 10 −12 C 2 N ⋅ m2 aa < r < bf Q = −4.00 × 10 −9 C = −4.00 nC (b) We take Q ′ to be the net charge on the hollow sphere. This paper establishes for the first time the true foundation of the calculus. Doesn’t matter if you’re Bullock, Penelope Cruz or Nicole Kidman, you would not be looking your best.

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The source can be moving in a plane perpendicular to the line between it and the observer. Easier: do this on your palm, with a dollar bill under the glass. The first person to realize something strange was going on when objects fell was Sir Isaac Newton. LC ∆VL, max = X L I max = ω 0 LI max = 2.24 kV 1 The resonance frequency is ω 0 = XL = ω L = FG H IJ L = 2 LC K 2 b Z = R 2 + X L − XC g 2 LC ∆Vmax = 5.00 A. Assuming the floor had just enough friction to allow movement, it would be easy for almost anyone to accelerate the shopping cart.

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Gravitational field strength at a point is defined as the gravitational force per unit mass at that point. or, the gravitational force of between two point masses is proportional to the product of their masses & inversely proportional to the square of their separation. Instead of casting doubt on the Planck length or other solid physics, it is further support for membrane gravity that photons are not traveling through “pixilated” space. Thus, the gap between the Sun and its blast wave has opened at 1.80c, and the time we calculate to have elapsed since the Sun exploded is 3.60 ly = 2.00 yr. 1.80 c We see Tau Ceti as moving toward us at 0.800c, while its light approaches at 1.00c, only 0.200c faster.

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Then t = = t t v 2 A ′ A ′ 2 gd a0.5 mf 0.5 m = m 2e9.8 m s j10 m 2 (b) *P14.75 t= (a) For diverging stream lines that pass just above and just below the hydrofoil we have 2 × 10 −4 2 2 44.6 s Pt + ρgy t + 1 2 1 2 ρv t = Pb + ρgy b + ρv b. 2 2 Ignoring the buoyant force means taking y t ≈ y b b g 1 1 2 2 ρ nv b = Pb + ρv b 2 2 1 2 2 Pb − Pt = ρv b n − 1 2 Pt + e b g The lift force is Pb − Pt A = (b) j 1 2 2 ρv b n − 1 A. 2 e j For liftoff, 1 2 2 ρv b n − 1 A = Mg 2 e vb j F 2Mg I =G GH ρen − 1jA JJK 12 2 The speed of the boat relative to the shore must be nearly equal to this speed of the water below the hydrofoil relative to the boat. (c) e j v 2 n 2 − 1 Aρ = 2 Mg A= b g b9.5 m sg e1.05 − 1j1 000 kg m 2 800 kg 9.8 m s 2 2 2 3 = 1.70 m 2 438 Fluid Mechanics ANSWERS TO EVEN PROBLEMS P14.2 ~ 10 18 kg m3; matter is mostly empty space P14.4 P14.6 (a) 1.01 × 10 7 Pa; (b)7.09 × 10 5 N outward 12.8 kg s P14.40 1.92 × 10 4 N P14.38 (a) 27.9 N; (b) 3.32 × 10 4 kg; (c) 7.26 × 10 4 Pa P14.10 P14.44 (a) 2.28 N toward Holland; (b) 1.74 × 10 6 s (a), (b) 28.0 m s; (c) 2.11 MPa P14.48 6.80 × 10 4 Pa P14.50 347 m s P14.52 (a) 489 N outward; (b) 1.96 kN outward P14.54 255 N (a) see the solution; (b) 616 MW P14.46 P14.8 P14.42 2.25 m above the level where the water emerges (a) 65.1 N; (b) 275 N P14.12 6 5.88 × 10 N down; 196 kN outward; 588 kN outward P14.14 (a) 29.4 kN to the right; (b) 16.3 kN ⋅ m counterclockwise P14.16 (a) 10.3 m; (b) zero P14.18 (a) 20.0 cm; (b) 0.490 cm P14.56 455 kPa P14.20 12.6 cm P14.58 709 kg m3 P14.22 (a) 444 kg; (b) 480 kg P14.60 8.01 km; yes P14.62 (a) see the solution; (b) 2.58 × 10 4 N P14.64 top scale: 1 − m ρw − ρs h P14.24 b P14.26 (a) see the solution; (b) 25.0 N up; (c) horizontally inward; (d) tension increases; see the solution; (e) 62.5%; (f) 18.7% g P14.28 ~ 10 4 balloons of 25-cm diameter P14.30 (a) 6.70 cm; (b) 5.74 cm FG H IJ K ρ0 m Fe g; ρ Fe FG H bottom scale: m b + m 0 + P14.66 (a) 0.461 m s 2; (b) 4.06 s P14.68 IJ K ρ 0 m Fe g ρ Fe see the solution 3 P14.32 (a) 11.6 cm; (b) 0.963 g cm; (c) no; see the solution P14.70 (a) 18.3 mm; (b) 14.3 mm; (c) 8.56 mm P14.34 0.611 kg P14.72 (a) 2.65 m s; (b) 2.31 × 10 4 Pa P14.36 2.67 × 10 3 kg P14.74 (a) see the solution; (b) 44.6 s 15 Oscillatory Motion CHAPTER OUTLINE 15.1 Motion of an Object Attached to a Spring Mathematical Representation of Simple Harmonic Motion Energy of the Simple Harmonic Oscillator Comparing Simple Harmonic Motion with Uniform Circular Motion The Pendulum Damped Oscillations Forced Oscillations ANSWERS TO QUESTIONS 15.3 15.4 15.5 15.6 15.7 Q15.4 Q15.1 Neither are examples of simple harmonic motion, although they are both periodic motion.

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Once the external voltage source is removed—provided that there is no external resistance through which the capacitor can discharge—the capacitor can hold onto this energy for a very long time. H 2K 2 2n 2 or 2 2 2 2 2 2 Chapter 37 2nt = mλ 2 2nt = m ′ + for λ 1 > λ 2, FG m′ + 1 IJ < m H 2K so m′ = m − 1. P29.23 From τ = µ × B = IA × B, the magnitude of the torque is IAB sin 90.0°. (a) Each side of the triangle is 40.0 cm. 3 Its altitude is 13.3 2 − 6.67 2 cm = 11.5 cm and its area is 1 A = 11.5 cm 13.3 cm = 7.70 × 10 −3 m 2. 2 Then τ = 20.0 A 7.70 × 10 −3 m 2 0.520 N ⋅ s C ⋅ m = 80.1 mN ⋅ m. a fa a (b) f fe jb Each side of the square is 10.0 cm and its area is 100 cm 2 = 10 −2 m 2. a fe ja f τ = 20.0 A 10 −2 m 2 0.520 T = 0.104 N ⋅ m (c) 0.400 m = 0.063 7 m 2π A = π r 2 = 1.27 × 10 −2 m 2 r= a fe ja f τ = 20.0 A 1.27 × 10 −2 m 2 0.520 = 0.132 N ⋅ m (d) g The circular loop experiences the largest torque. 3.97°.

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The photon energy is inversely proportional to its wavelength, so for positronium, a f λ 32 = 2 656.3 nm = 1.31 µm (in the infrared region). (b) For He +, µ ≈ m e, q1 = e, and q 2 = 2 e, so the transition energy is 2 2 = 4 times larger than hydrogen. λ 32 = Then, P42.17 (a) ∆x∆p ≥ (b) Choosing ∆p ≈ U= (c) 2 so if ∆x = r, ∆p ≥ r ,K= FG 656 IJ nm = H4K 2r 164 nm (in the ultraviolet region).. b g 2 2 ∆p p2 ≈ = 2m e 2m e 2m e r 2 2 −kee2 k e2 − e, so E = K + U ≈. r r 2m e r 2 To minimize E, 2 2 k e2 dE =− + e2 = 0 → r = = a0 3 dr mer r meke e2 Then, E = 2 2m e Fm k e I GH JK e e 2 2 2 − kee2 (the Bohr radius).

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Newton died in 1727, and was buried with much pomp and circumstance in Westminster Abbey, despite his well-known reservations about the Anglican faith. P30.8 x R 0.00 1.00 2.00 3.00 4.00 5.00 *P30.9 B B0 1.00 0.354 0.0894 0.0316 0.0143 0.00754 Wire 1 creates at the origin magnetic field B1 = (a) µ0I µ I right hand rule = 0 1 2π r 2π a = If the total field at the origin is according to B 2 = µ 0 I1 j 2π a µ I 2µ 0 I1 j = 0 1 j + B 2 then the second wire must create field 2π a 2π a µ 0 I1 µ I j= 0 2 2π a 2π 2 a. a f Then I 2 = 2 I 1 out of the paper = 2 I 1 k. (b) The other possibility is B1 + B 2 = B2 = *P30.10 µ I 2 µ 0 I1 − j = 0 1 j + B 2.