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obj-9 - 25.1 IDENTIFY I = Q/t SET UP 1.0 h = 3600 s EXECUTE...

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Unformatted text preview: 25.1. IDENTIFY: I = Q/t . SET UP: 1.0 h = 3600 s EXECUTE: Q = It 2 (3.6 A)(3.0)(3600 s) = 3.89x104 C. EVALUATE: Compared to typical charges of objects in electrostatics, this is a huge amount of charge. 25.13. IDENTIFY: E = p] , where J = I / A. SET UP: For tungsten p = 5.25><10'8 9. ‘ m and for aluminum ,0 = 2.75><10'8 (2‘ m. ‘: t t :E= =—-———~—————~————=5.16x10‘3Vm. EXECUTE (a) ungs en ,0] A (7r/4)(3.26><10‘3 m)2 / p1 __ (2.75 x10"8 Q-m)(0.820 A) A _ (Ir/4)(3.26x10'3 my EVALUATE: A larger electric field is required for tungsten, because it has a larger resistivity. (b) aluminum: E = p] = = 2.70><10'3 V/m. 25.17. IDENTIFY: = '07" SETUP: Forcopper, p:l.72><10‘8 Q-m. A=7rr2. —s EXECUTE: R = W 7r(1.025x10‘3 ml The resistance is proportional to the length of the piece of wire. = 0.125 9 g EVALUATE: 25.24. IDENTIFY: Apply R = pf and V = [R . SET UP: A = Itr2 -4 2 EXECUTE: ,0 = fl : V_A = (4.50 V)7r(6.54 x10 m) L [L (17.6 A)(2.50 m) EVALUATE: Our result for p shows that the wire is made of a metal with resistivit =1.37><10—7 Q-m. y greater than that of good metallic conductors such as copper and aluminum. 25.42. IDENTIFY and SET UP: For a resistor, P = VI = V2/ R and V 2 IR. 2 2 EXECUTE: (a) R = Y— = 91—0!)- = 0.688 (2 P 327 W (01:3: 15")" =21.8A R 0.688 9 EVALUATE: We could also write P = VI to calculate I =5 = 327 W = 21.8 A. V 15.0 V . I' ' m 25.46. IDENTIFY: P =VI . Energy 2 Pt. SET UP: P = (9.0 V)(0.13 A) = 1.17 W EXECUTE: Energy = (1.17 W)(l.5 h)(3600 s/h) = 6320 J . EVALUATE: The energy consumed is proportional to the voltage, to the current and to the time. ...
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