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April 16, 2014

April 16, 2014

Posted by **John** on Monday, February 25, 2013 at 10:28pm.

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

PLEASE COULD SOMEONE HELP ME OUT - Ms. Sue, Monday, February 25, 2013 at 5:54pm

If 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 6:10pm

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)=..

PLEASE COULD SOMEONE HELP ME OUT Physics - John, Monday, February 25, 2013 at 7:33pm

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

- PLEASE COULD SOMEONE HELP ME OUT Physics -
**Elena**, Tuesday, February 26, 2013 at 4:57amSeparation between two wires is

b=2Lsinα=20.5sin3⁰=5.2310⁻² m

The mass of 1meter of the wire is

m₀=ρπd²/4=27003.14(0.310⁻³)²/4=1.910⁻⁴ kg

where ρ =2700 kg/m³ is the density of Al

The force between two wires (per 1 meter) is

F=μ₀I²/2πd

F=Tsinα

mg=Tcosα

F/mg=tanα

μ₀I²/2πbm₀g= tanα

I=sqrt{2πbm₀g tanα/ μ₀}=

=sqrt{2π 5.2310⁻²1.910⁻⁴9.85.210⁻²/4π10⁻⁷}=

=5.03 A

- PLEASE COULD SOMEONE HELP ME OUT Physics -
**John**, Tuesday, February 26, 2013 at 8:41amThank you Elena your a life saver :)

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