775_Physics ProblemsTechnical Physics

775_Physics - Chapter 27 P27.33 R = R 0 1 αT R − R0 = R0α∆T R − R0 = α∆T = 5.00 × 10 −3 25.0 = 0.125 R0 e P27.34 j a Assuming linear

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Unformatted text preview: Chapter 27 P27.33 R = R 0 1 + αT R − R0 = R0α∆T R − R0 = α∆T = 5.00 × 10 −3 25.0 = 0.125 R0 e P27.34 j a Assuming linear change of resistance with temperature, R = R0 1 + α∆T a ja fe f f R77 K = 1.00 Ω 1 + 3.92 × 10 −3 −216° C = 0.153 Ω . a f 1 FG ρ Hρ W IJ K ρ = ρ 0 1 + α∆T or ∆TW = Require that ρ W = 4ρ 0 Cu so that ∆TW = Therefore, P27.35 TW = 47.6° C + T0 = 67.6° C . Section 27.5 αW −1 0W F1 GH 4.50 × 10 I FG 4e1.70 × 10 j − 1IJ = 47.6° C . JK ° C J G 5.60 × 10 KH −8 −3 Superconductors Problem 48 in Chapter 43 can be assigned with this section. Section 27.6 P27.36 I= Electric Power P ∆V = and R = *P27.37 P27.38 600 W = 5.00 A 120 V ∆V 120 V = = 24.0 Ω . I 5.00 A e b gb g P = 0.800 1 500 hp 746 W hp = 8.95 × 10 5 W b 8.95 × 10 5 = I 2 000 P = I∆V P27.39 j P = I∆V = 500 × 10 −6 A 15 × 10 3 V = 7.50 W g I = 448 A The heat that must be added to the water is b gb f ga Q = mc∆T = 1.50 kg 4 186 J kg ° C 40.0° C = 2.51 × 10 5 J . Thus, the power supplied by the heater is P= W Q 2.51 × 10 5 J = = = 419 W ∆t ∆t 600 s a∆V f = a110 V f 2 and the resistance is R = P 2 419 W = 28.9 Ω . −8 115 ...
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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