28_Chapter 16 HomeworkCH17 Current and Resistance

28_Chapter 16 HomeworkCH17 Current and Resistance - 96...

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96 CHAPTER 17 (b) The mass may be written as ( ) ( ) md e n s i t yV o l u m ed e n s i t yA L = ⋅= ⋅ From RL A ρ = , the cross-sectional area is A LR = , so the expression for the mass becomes () ( )( ) 2 2 3 1.7 1 500 m 8 920 kg m 76 kg L e n s i t y R =⋅ = = 8 10 m 4.5 ×Ω 17.66 (a) At temperature T , the resistance is L R A = , where ( ) 00 1 TT ρρ α =+− , ( ) 1 LL   , and AA 2 0 11 2 A αα ′′ 0  ≈+ −  Thus, () () R + − == +− 0 0 12 +−⋅ + 0 0 L R A    (b) ( )( ) 8 0 2 -3 0 1.70 10 m 2.00 m 1.082 0.100 10 L A π = × ( R Then ) 1 RR gives 3 10 C 80.0 C 1.420 R × ° ° = 1.082 1 3.90 =Ω + The more complex formula gives ( ) 6 6 1.420 1 17 10 C 80.0 C 1.418 1 2 17 10 C 80.0 C R Ω⋅ + × ° ° +×° ° Note : Some rules for handing significant figures have been deliberately violated in
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