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A current-carrying gold wire has a diameter of 0.90 mm. The electric field in the wire is 0.51 V/m. Submit Request Answer Par

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0.26 ohms

R=fl = se Td2 = use - lux 2.44X108 rm) 16.7m) T (0,9x103)2 (0.257 r as - 10.26 2 sig. figures

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