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April 17, 2015

Posted by **John** on Monday, February 25, 2013 at 5:50pm.

Thank you...please help i don't get this

- PLEASE COULD SOMEONE HELP ME OUT -
**Ms. Sue**, Monday, February 25, 2013 at 5:54pmIf you really want help, you'd type in your subject name so that an expert would see your post.

- PLEASE COULD SOMEONE HELP ME OUT -
**Elena**, Monday, February 25, 2013 at 6:10pmSeparation between two wires is

b=2Lsinα=20.5sin3⁰=5.2310⁻² m

The mass of 1 meter of the wire is

m₀=ρπd²/4,

where ρ is the density of Al

The force between two wires (per 1 meter) is

F=μ₀I²/2πd.

F=Tsinα

mg=Tcosα

F/m₀g=tanα

μ₀I²/2πbm₀g= tanα

I=sqrt{2πbm₀g tanα/ μ₀}=

=sqrt{2πbρπd²g/44π10⁻⁷)=..

- PLEASE COULD SOMEONE HELP ME OUT Physics -
**John**, Monday, February 25, 2013 at 7:33pmHi Elena thank you, but this is what I did and i got it wrong

sqrt(2π*5.23x10^-2*2700*π*2.25x10^-6*9.8/4*4π*10^-7)= 1.38x10^-4 but it is wrong

what did i do wrong aluminum density is 2700 kg/m^3 and 2.25x10^-6 is .30 mm divided by 100 to get meters an then divided by 2 to get r and square that so gives me 2.25x10^-6

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