Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 25, Problem 46P
(a)
To determine
The resistance of the wire.
(b)
To determine
Electric field in the wire.
(c)
To determine
Time taken by the electron to travel through the wire.
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Chapter 25 Solutions
Physics for Scientists and Engineers
Ch. 25 - Prob. 1PCh. 25 - Prob. 2PCh. 25 - Prob. 3PCh. 25 - Prob. 4PCh. 25 - Prob. 5PCh. 25 - Prob. 6PCh. 25 - Prob. 7PCh. 25 - Prob. 8PCh. 25 - Prob. 9PCh. 25 - Prob. 10P
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- I need correct answer not chatgptarrow_forwardWhat is the resistance (in (2) of a 27.5 m long piece of 17 gauge copper wire having a 1.150 mm diameter? 0.445 ΧΩarrow_forwardFind the ratio of the diameter of silver to iron wire, if they have the same resistance per unit length (as they might in household wiring). d. Ag dFe = 2.47 ×arrow_forward
- Find the ratio of the diameter of silver to iron wire, if they have the same resistance per unit length (as they might in household wiring). d Ag = 2.51 dFe ×arrow_forwardShow that the units 1 v2/Q = 1 W, as implied by the equation P = V²/R. Starting with the equation P = V²/R, we can get an expression for a watt in terms of voltage and resistance. The units for voltage, V, are equivalent to [? v2 v2 A, are equivalent to J/C ✓ X . Therefore, 1 = 1 = 1 A V1 J/s Ω V-A X = 1 W. . The units for resistance, Q, are equivalent to ? The units for current,arrow_forwardPlease solve and answer the question correctly please. Thank you!!arrow_forward
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