Respuesta :
Answer:
7.07 x 10⁸ N/m²
Explanation:
F = Force applied to the wire = 68 N
d = diameter of the wire = 0.7 mm = 0.7 x 10⁻³ m
r = radius of the wire = (0.5) d = (0.5) (0.7 x 10⁻³) = 0.35 x 10⁻³ m
Area of cross-section of wire is given as
A = (0.25) πr²
A = (0.25) (3.14) (0.35 x 10⁻³)²
A = 9.61625 x 10⁻⁸ m²
Tensile stress is given as
[tex]P = \frac{F}{A}[/tex]
[tex]P = \frac{68}{9.61625\times 10^{-8}}[/tex]
P = 7.07 x 10⁸ N/m²
Given:
Applied Force on wire = 68 N
Diameter of wire, d = 0.7 mm = [tex]0.7\times 10^{-3}[/tex] m
Radius of wire, r = [tex]\frac{d}{2}[/tex] = 0.35 mm = [tex]0.35\times 10^{-3}[/tex] m
Formula used:
Stress = [tex]\frac{Applied Force}{cross-sectional area}[/tex]
Explanation:
Cross-sectional area, A = [tex]\pi r^{2}[/tex] = [tex]\pi (0.35\times 10^{-3})^{2}[/tex]
A = [tex]3.84\times 10^{-7} m^{2}[/tex]
Using the formula for stress:
Stress = [tex]\frac{68}{3.84\times 10^{-7}}[/tex] = [tex]1.76\times 10^{8}[/tex]
