Posted by John on Monday, February 25, 2013 at 5:50pm.
If you really want help, you'd type in your subject name so that an expert would see your post.
Separation 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)=..
Hi 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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