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Answer:
52N
Explanation:
v=SQRT(T/μ), μ=pA, A=πr^2
v=SQRT(T/pπr^2)
v^2=T/pπr^2
v^2*pπr^2=T
34^2*2700*π*0.0023^2=T
T=52N
The tension on the aluminum wire at the given density is 52.02 N.
The tension in the wire is calculated using the following formulas;
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
where;
A = πd²/4
A = π x (4.6 x 10⁻³)²/4
A = 1.66 x 10⁻⁵ m²
μ = ρA
μ = 2700 kg/m³ x 1.66 x 10⁻⁵ m²
μ = 0.045 kg/m
[tex]34 = \sqrt{\frac{T}{0.045} } \\\\34^2 = \frac{T}{0.045}\\\\T = (34^2)(0.045)\\\\T = 52.02 \ N[/tex]
Thus, the tension on the aluminum wire at the given density is 52.02 N.
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