- #1

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so i am using the equation R=rho(L/A)

A=pi(d2-d1)^2/4 which is 1.29E-4 m

and for copper rho is 1.7E-8

so 8.05E-3=(1.7E-8)(L/1.29E-4)

then L=60.93 m

but this isnt right.

please help me where i am going wrong.

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- Thread starter GravityGirl
- Start date

- #1

- 29

- 0

so i am using the equation R=rho(L/A)

A=pi(d2-d1)^2/4 which is 1.29E-4 m

and for copper rho is 1.7E-8

so 8.05E-3=(1.7E-8)(L/1.29E-4)

then L=60.93 m

but this isnt right.

please help me where i am going wrong.

- #2

Dick

Science Advisor

Homework Helper

- 26,260

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Where did you get the funny area formula? Don't you want pi*r2^2-pi*r1^2 where r's are the radii?

- #3

- 29

- 0

- #4

Dick

Science Advisor

Homework Helper

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But (d2-d1)^2 is not equal to d2^2-d1^2.

- #5

- 29

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so what did you get for the area...i still get 1.29E-4 m

- #6

Dick

Science Advisor

Homework Helper

- 26,260

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0.00147 m^2. Throw out that formula and subtract the area of the two circles!!!!!

- #7

- 29

- 0

ok ok....i will do that

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