SOLUTION: The electrical resistance of a wire varies directly with its length (L) and varies inversely with the square of its diameter (D). If a certain wire is 6.25" long and has a diameter
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Question 1158658: The electrical resistance of a wire varies directly with its length (L) and varies inversely with the square of its diameter (D). If a certain wire is 6.25" long and has a diameter of .25", it has a resistance of 25 ohms. What is the resistance in ohms of a similar wire 40" long and with a diameter of .4 inches? Found 2 solutions by Theo, josmiceli:Answer by Theo(13342) (Show Source):
your problem is:
The electrical resistance of a wire varies directly with its length (L) and varies inversely with the square of its diameter (D). If a certain wire is 6.25" long and has a diameter of .25", it has a resistance of 25 ohms. What is the resistance in ohms of a similar wire 40" long and with a diameter of .4 inches?
let R = the electrical resistance of the wire.
let L equal its length.
let D equal its diameter.
the formula becomes R = k * L / D^2
when L = 6.25 inches and D = .25 inches, it has a resistance of 25 ohms,.
the formula becomes:
25 = k * 6.25 / .25^2
solve for k to get:
k = 25 * .25^2 / 6.25 = .25
k is the constant of variation.
it stays the same regardless of the values of R and L and D in the same formula.
you want to know what is the resistance when L = 40 inches and D = .4 inches.
the formula is still:
R = k * L / D^2
since k = .25 and L = 40 and D = .4, the formula becomes:
R = .25 * 40 / .4^2
solve for R to get:
R = 62.5 ohms.
You can put this solution on YOUR website! Let = resistance in ohms
Let = constant of proportionality
---------------- ohms in in
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----------------- ohms
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check my math & get a 2nd
opinion if needed
