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BartSMP [9]
3 years ago
13

A copper wire has a radius of 3.5 mm. When forces of a certain equal magnitude but opposite directions are applied to the ends o

f the wire, the wire stretches by 5.0×10−3 of its original length. What is the tensile stress on the wire? Young's modulus for copper is 11×1010Pa.
Physics
1 answer:
denpristay [2]3 years ago
8 0

Answer:

The tensile stress on the wire is 550 MPa.

Explanation:

Given;

Radius of copper wire, R = 3.5 mm

extension of the copper wire, e =  5.0×10⁻³ L

L is the original length of the copper wire,

Young's modulus for copper, Y =  11×10¹⁰Pa.

Young's modulus, Y is given as the ratio of tensile stress to tensile strain, measured in the same unit as Young's modulus.

Y =\frac{Tensile \ stress}{Tensile \ strain} \\\\Tensile \ stress = Y*Tensile \ strain\\\\But, Tensile \ strain = \frac{extension}{original \ Length} = \frac{5.0*10^{-3} L}{L} = 5.0*10^{-3}\\\\Tensile \ stress = 11*10^{10} *5.0*10^{-3} \ = 550*10^6 \ Pa

Therefore, the tensile stress on the wire is 550 MPa.

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We can use again the equation

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a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

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F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

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