SOLUTION: the electrical resistance R of a wire varies directly as its length L and inversely as the square of its diameter. A wire 20 meters long and 0.6 centimeters in diameter made from

Algebra ->  Expressions-with-variables -> SOLUTION: the electrical resistance R of a wire varies directly as its length L and inversely as the square of its diameter. A wire 20 meters long and 0.6 centimeters in diameter made from       Log On


   

Question 46551: the electrical resistance R of a wire varies directly as its length L and inversely as the square of its diameter. A wire 20 meters long and 0.6 centimeters in diameter made from a certain alloy has a resistance of 36ohms. What is the resistance of a piece of wire 60 meters long and 1.2 centimenters in diameter made from the same material?
i know that the equation should look something like this:
R=k L/D^2
but from there i get lost when plugging in the values
36ohms=k (20m)/(.6cm)^2
if i can figure out K i can plug in the other values and get my answer.
i figure i have to make m into cm so 20 becomes 2000cm/.36cm if i times by 100
to get rid of the decimal in denominator i get 3600oms=K* 200000/36
multiply both sides by 36 and i get 129600oms=200000K
divide by 200000
k=.648
pluggin in the values that i have from above i get the resistance would be 27ohms.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
R = k*l *(1/(d^2))
R = resistance of wire
k = proportionality constant
l = length of wire
d = diameter of wire
36+=+k%2A2000%2A%281%2F%28.6%29%5E2%29 all values in cm.
36+=+k%2A2000%2F.36
k+=+36%2A.36%2F2000
k+=+1296%2F%282%2A10%5E5%29
k+=+648%2A10%5E%28-5%29
R+=+k%2Al+%2A%281%2F%28d%5E2%29%29
R+=+k%2A6000%2A%281%2F%281.2%29%5E2%29
R+=+648%2A10%5E%28-5%29%2A6%2A10%5E3%2A%281%2F1.44%29
R+=+6%2A648%2A10%5E%28-2%29%2A.6944
R+=+2700%2A10%5E%28-2%29
R+=+27
27 ohms is my answer, too
a quick check
The 2nd wire is 3 times as long and 4 times the
cross-section area of the 1st wire
R2 = 3/4*R1
R2 = (3/4)* 36
R2 = 3*9
R2 = 27 ohms
OK