Lista Relatividade - Física 3 D

UNIDADE IV - RELATIVIDADE\n1 - a) 0.00254m = 1s 2.72.10^(-7)m 0.5.10^(-3)m l = 1.67.10^(-12)\n31.104.10^2 = 2.983792438m l 299792458m\nfazer o mesmo com os outros\n* 2 - luz - 6 anos mesmo d.\npartícula - 8 anos\n6.c = 8.v\nv = 0.75c\n3 - a) 1.01/(1-β²)=1 β=0.1404\n1.0201/(1-β²)=1\n1.0201-1.0201β²=1\n1000.0.1=0.1\n3 - 2.2.10^6 s = Δt\nΔt=γ.Δto\n58.9-52.9β²=(c)\n 52.9.1\n1+β²=0.9805=β\n4 - v=0.99c\nΔt=7.09.26s=184.3μs\nΔto=26.10^-8s\n c\n y=1/(1-0.9926)=7.09\nL=v.Δt\nL0=0.99m.298793243.184.3.10^5\nL=54.7m 6 - Δt=1000 anos a) 1000=γΔt Δto=0.5 anos ida √(1-β²) + 0.5 anos voltando 1000000-1000000β²=1\n0.9999985=β²=v\nb) 0.888856=c\n7\nL0=4.70m v=0.63c\nL=\nL=Lo/√(1-β²)\nL=1.71=1.32m\n1.288\n8 - β=0.999987 L=Lo γ=1/√(1-β²)\nLo=3.0m\nL=0.0153m=1.53cm\n9 - L0=130m\nv=0.740 \na) L1=130 - L=87.44m\n1/0.7402\nb) relógio -> Δto?\n87.44=0.74c.Δto\n3.94.10^-6 = Δto 10 - L0=206000 luz a)Δs=0.99c.Δt\nv=0.99c\nΔs=6.3 anos ida + 6 anos volta = 52.3 anos\nΔt?\nb) 26.3 anos ida + 26 anos volta = 52.3 anos\nΔt=Δto/γ\n52.3 = 3.7 anos\n(1/√(1-0.99²)\n11 - | -> depois rever os passos\n12 - L0=230000000 luz\n a) 230000=Δt (supendo Δt=100 anos)\n230000=γ.100\n230000=γ/√(1-β²)\n0=Δt\nlogarítmia previsível viáve loge=previsível\nb) Δt=230000 230000=1\nΔto=80\n80/√(1-β²) β=0.99998c\n13 - x=100km α=t=200.10^-6\nv=0.95c\n-------------------------------------------------------------------\n y=1/√(1-β²)=3.2 x'=3.2.(100°.95c.200.10^-6)\n x' =3.2.413.04043m\n x' =137.79 km\n Δt' = - 3.74.10^-6 14 - y = 0.6c evento 1s t=0 x=0\n ls \nj: t = 4.10^ - 3 s x = 3km\n 1? a) evento 1: t' = f(0 - 0.6c.0) = 0\n j: 2) t' = 1.25(4.10^ - 6 - 0.6c)3$ - 29.10^-6 s\n x - 1\n \n \\ - 0.62\n * b) o s' tem o evento O ocorrendo antes devido à diferença de velocidade relativa entre s e e'\n\n15 - A x= 0 y=30 s=0.85c\n s' -> \n y/f =1 - 1.03\n \\ 1 -0.95\n a) \\ Δt' = 1.03( 0 - 0.25c.3.10^ - 4) = - 2.576.10^-5\n * b) t'' = 1.03 (0-0.25c.0) = 0\n t' = 0.0 + 2.576.10^-5 = 2.576.10^-5\n \n Δt = t'\n 16 - para o vermel ino afastamento\n Q1 -> afasta-se 0.8c^3\n Q2 -> afasta-se 0.4c\n \nmá maior perto Terra\n Lorentz: u^2 = 0.8c^2 + 0.4c\n ↔ w = 0.4c ↔ u' = 0.528c\n (1-0.8)(0.4)\n (1-0.32) galileu: 0.8c + u' = 0.4c mesmo sentido, subtroi\n P' + u' = 0.4c\n \n 18 - v = 0.9c\n ρ = 10^34kg/i\n ? eterno? \n ρ' = 10^8 \n 1 - 0.9 = 1100^3.0.72/9 = 2.394MH'z\n 1+0.9\n \n19 - Afasto-vermellino λo= 450.10^-9\n v = 0.4c\n b'λ' λ terra? \n\n c = c √(1 - 0.8)\n λ = 55.1 + 3 nm -> amarelo λ.\n \n 20 - p = 1/p\n a) I - 1 1 - β\n p' = ρ √(1 - β) \n \n p = ρ' √(1 + v/c) = ρ - ρ' √c + v\n c - v\n (ρ1 ± V/c) c = c + v = \n (ρ1 ± V/c)1 c - v\n \n d) pois o observador A sofre, juntamente com doppler normal, a doppler transversal, que assemelha-se a uma dilatação temporal 21 - diâmetro Terra - 12800km\n a) Circunferência = 1.28.10^7 m\n v = 128.10^7m 1.28.10^7m = 0.134c\n b) K = m0c^2 (y-1)\n y = 1 + 0.009 m0 = 9.109.10^-31kg\n \n k = 3.108.10^31c^2 (0.009) = 7.45.10^-16 J\n 1ev = 1.602.176634.10^-19J\n y = 4.6486.62 = √4.64KeV\n\n c) E = 1/2mv^2 = 8.10^10 (0.134c)² = 7.35.10^-6\n\n erro 4.64 - 4.58 - 0.0129 = 1.29%\n 4.64\n\n 22 - P=10^4 W 1ms = 2.10^30kg\n P.E {= 10^6 J} Lh 1000 = 31536000s\n 10^4 = m0c² m0 = 1.11.10^24kg\n x = - 1.11.10^24kg\n 1s = 5.56.10^-7 m/s x2 = 1.753 m/ppm\n\n 23 - a) K = m0c²(y-1) m0 = 9.11.10^-31kg\n Ki = 9.11.10^-31c²(0.004)\n y = 1.1 1.07 \n ki = 1.39.10^-15 J\n √(1 - 0.18) Kf = 8.11.10^{-1} \cdot 2 (0.018)\nKf = 1.47.10^{15} J\n\nΔk = 0.08c^{195} → 511.03 eV\nUsar masa\nDiferente de\n\nb) Ki = 511.10^{-2} (4.085)\nKi = 3.89.10^{-13}\n\nKf = 8.11.10^{-31} (6.088)\nYf = 1 = 5.088\nKf = 4.98.10^{-13}\n\nΔK = 1.69.10^{-13} J = 1.0582.226.74 eV - 105.64 NeV\n\n24-a) Física clásica\nK = mv^{2} = 9.11.10^{-31} (0.8973924582) = 4.09.10^{-14} J\n\n2 = 255.51 keV\n\nb) 4.08.10^{-18} = 9.11.10^{-31} (x)(1)\n0.499 = x - 1\nj = 1.5\n\n1.5.1.1-β^{2} = 1\nB = v = 0.747c\nc\n\n2.25 - 2.25β^{2} - 1\nβ = 0.745\n\n25 - p = mc\n1/p = γmv = mc\nc = 1/v√(1 - β^{2})\n\nm/ a) j = c/v t + 1\nβ^{2} = 1 - β^{2}\nβ = 1\nγ = 1/\sqrt{β}\n\nb) β = v/\sqrt{c}\n\nc) k = mc^{2} (γ - 1)\nK = 0.414 mc^{2}\n\n26 - Fotón: E = 0 eV\na) Fotón, una vez que nos presentó masa y se mueve en la luz\nV = 10^{10} eV\n\nb) K = mv^{2}\nZ = 0.4: me = ve^{2}\n\nE: 0.4: me = ve^{2}\n\np. pβ = mαv \nFotón preci. p = 0\n\np = p = 2000 me√{10^{3}} = 2000000 me^{1/2}\nme\ne: p = me √(2.10^{5}) = 284 me^{1/2}\nO fotón!\n\nd) El fotón, pues p = 0 27 - m = 207 me\na) Δt = γ Δto\nγ = 3.13\nΔto = 2.10^{-6}\nΔt = 6.9.10^{-6}\n\nb) K = mo c^{2} (β + 1)\nβ = 0.947\nK = 207.9.11.10^{-31} (2.13)\nK = 2.853198.8143 eV = 825.31 MeV\n\nc) p = mo γ v\n= 207.9.11.10^{-31} 3.13 0.947c + 67.10^{-15} kg.m/s\n\n= 5.00.10^{-n} (kg.m/s.c)\n= 3.12.48MeV/c\n\n28 - AB kg – OJ é convertido em E\nE liberado = E0 = 1.3. c = 2.7.10^{-4} J/100\n\nb) 1 mol = 3.4.10^{6} J\nx = 79411764.71 moles\nα = 2.7.10^{-5} = x/227.139/mol = 18.10^{7} Kg/100\n\nc) E = 0.4 1.81.10^{-2} = 462.10^{-23} J\n462.16^{23} J = 6.10^{6} J\n2.7.10^{14} J v = 0.98c\nd = 0.6 anos - l0z = l0\nl0 = 26 anos - l0z\nl0 = v. Δt\nb) 26.3 anos * 26 anos = 52.3 anos\n26.6 - 0.89c. Δt\n26.3 = Δt\n\nc) Δt. y = l0\ny = 1 = 7.08\nΔt. y = 26.3\nΔt. 7.08 = 26.3\nΔt = 3.7 anos