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conehate0712
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The electrical resistance of a wire varies directly as the square of the diameter . if a wire 9.45m long and 0.09 cm in diameter has a resistance of a wire of the same material which is 15.5m in length and 0.16cm in diameter?
 one year ago
 one year ago
conehate0712 Group Title
The electrical resistance of a wire varies directly as the square of the diameter . if a wire 9.45m long and 0.09 cm in diameter has a resistance of a wire of the same material which is 15.5m in length and 0.16cm in diameter?
 one year ago
 one year ago

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hba Group TitleBest ResponseYou've already chosen the best response.0
This goes in the physics portion :)
 one year ago

conehate0712 Group TitleBest ResponseYou've already chosen the best response.0
so sorry ... actually its my first time here ... and ive got a lots of papers to pass this week :((
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
No worries c: Just close this up and post it in the physics portion :)
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
This is a standard direct variation problem. Such problems are encountered often in algebra, not just physics. resistance varies directly as the square of the diameter MEANS \(Resistance = k\cdot Diameter^{2}\) We are given: wire 9.45m long and 0.09 cm in diameter has a resistance of a wire of the same material which is 15.5m in length and 0.16cm in diameter? This is a little odd for two reasons: 1) the length is not important. Throw it out. 2) We have two partial hints and no complete hint. Discarding the length will be more clear. We are given: wire 0.09 cm in diameter has a resistance of a wire of the same material which is 0.16cm in diameter? Thus: \(k\cdot (0.09\;cm)^{2} = k\cdot (0.16\;cm)^{2}\) We see immediately that this is no good, since k = 0 is the only solution. I have to believe we DID need the length and the variation description provided is incorrect. I would guess that it is this: The resistance of a wire varies directly with its length and inversely with the square of its diameter Now, we can solve it with the original information given, Let's see what you get. @conehate0712
 one year ago
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