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bariátrica 25. Bouchard, C. Atividade física e obesidade. Barueri: Manole, 2003. 26. Hotamisligil GS. Inflammatory pathways and insulin action. Internacional Journal of Obesity. 2003; 27 (Suppl 3):S53-5. 37. Fortes RC, Guimarães NG, Haack A, Torres AAL, Carvalho KMB. Orlistat e sibutramina: bons coadjuvantes para perda e manutenção de peso? Revista Brasileira de Nutrição Clínica 2006; 21(3): 244251. 27. Shulman GI. Cellular mechanisms of insulin resistance. The Journal of Clinical…

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m3 )(0.820 A) E = ρJ = = = 5.16 × 10− 3 V m . −3 2 A (π 4)(3.26 × 10 m) b) aluminum: ρI (2.75 × 10 −8 Ω m3 )(0.820 A) E = ρJ = = = 2.70 × 10− 3 V m . −3 2 A (π 4)(3.26 × 10 m) 25.14: R Al = RCu ⇒ 2 πd ρ Al L ρCu L πd = ⇒ A l = Cu ⇒ d Cu = d Al AAl ACu 4 ρ Al 4 ρCu 2 ρCu ρ Al ⇒ d Al = (3.26 mm) 1.72 × 10−8 Ω ⋅ m = 2.6 mm. 2.75 × 10−8 Ω ⋅ m 25.15: Find the volume of one of the wires: ρL ρL R= so A = and A R ρL2 1.72 × 10−8 Ohm ⋅ m)(3.50m) 2 volume = AL = = = 1.686 ×…

m3 )(0.820 A) E = ρJ = = = 5.16 × 10− 3 V m . −3 2 A (π 4)(3.26 × 10 m) b) aluminum: ρI (2.75 × 10 −8 Ω m3 )(0.820 A) E = ρJ = = = 2.70 × 10− 3 V m . −3 2 A (π 4)(3.26 × 10 m) 25.14: R Al = RCu ⇒ 2 πd ρ Al L ρCu L πd = ⇒ A l = Cu ⇒ d Cu = d Al AAl ACu 4 ρ Al 4 ρCu 2 ρCu ρ Al ⇒ d Al = (3.26 mm) 1.72 × 10−8 Ω ⋅ m = 2.6 mm. 2.75 × 10−8 Ω ⋅ m 25.15: Find the volume of one of the wires: ρL ρL R= so A = and A R ρL2 1.72 × 10−8 Ohm ⋅ m)(3.50m) 2 volume = AL = = = 1.686 ×…

m3 )(0.820 A) E = ρJ = = = 5.16 × 10− 3 V m . −3 2 A (π 4)(3.26 × 10 m) b) aluminum: ρI (2.75 × 10 −8 Ω m3 )(0.820 A) E = ρJ = = = 2.70 × 10− 3 V m . −3 2 A (π 4)(3.26 × 10 m) 25.14: R Al = RCu ⇒ 2 πd ρ Al L ρCu L πd = ⇒ A l = Cu ⇒ d Cu = d Al AAl ACu 4 ρ Al 4 ρCu 2 ρCu ρ Al ⇒ d Al = (3.26 mm) 1.72 × 10−8 Ω ⋅ m = 2.6 mm. 2.75 × 10−8 Ω ⋅ m 25.15: Find the volume of one of the wires: ρL ρL R= so A = and A R ρL2 1.72 × 10−8 Ohm ⋅ m)(3.50m) 2 volume = AL = = = 1.686 ×…

m3 )(0.820 A) E = ρJ = = = 5.16 × 10− 3 V m . −3 2 A (π 4)(3.26 × 10 m) b) aluminum: ρI (2.75 × 10 −8 Ω m3 )(0.820 A) E = ρJ = = = 2.70 × 10− 3 V m . −3 2 A (π 4)(3.26 × 10 m) 25.14: R Al = RCu ⇒ 2 πd ρ Al L ρCu L πd = ⇒ A l = Cu ⇒ d Cu = d Al AAl ACu 4 ρ Al 4 ρCu 2 ρCu ρ Al ⇒ d Al = (3.26 mm) 1.72 × 10−8 Ω ⋅ m = 2.6 mm. 2.75 × 10−8 Ω ⋅ m 25.15: Find the volume of one of the wires: ρL ρL R= so A = and A R ρL2 1.72 × 10−8 Ohm ⋅ m)(3.50m) 2 volume = AL = = = 1.686 ×…

m3 )(0.820 A) E = ρJ = = = 5.16 × 10− 3 V m . −3 2 A (π 4)(3.26 × 10 m) b) aluminum: ρI (2.75 × 10 −8 Ω m3 )(0.820 A) E = ρJ = = = 2.70 × 10− 3 V m . −3 2 A (π 4)(3.26 × 10 m) 25.14: R Al = RCu ⇒ 2 πd ρ Al L ρCu L πd = ⇒ A l = Cu ⇒ d Cu = d Al AAl ACu 4 ρ Al 4 ρCu 2 ρCu ρ Al ⇒ d Al = (3.26 mm) 1.72 × 10−8 Ω ⋅ m = 2.6 mm. 2.75 × 10−8 Ω ⋅ m 25.15: Find the volume of one of the wires: ρL ρL R= so A = and A R ρL2 1.72 × 10−8 Ohm ⋅ m)(3.50m) 2 volume = AL = = = 1.686 ×…

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