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Sagot :
The minimum diameter of a steel wire is 1.41mm
STRESS : The restoring force per unit area. unit of stress is N/m²
It is denoted by σ
σ = F/ A
Where σ = stress
F = force
A = area
force = ma
Area of a steel wire is πr²
2r = d
so area becomes
π(d/2)² =πd²/4
here , mass = 80kg
elastic limit = 5.0×10⁸kg
using these values
σ = mg/ πr²
5×10⁸ = 80×10³ × 9.8/ π*d²/4
π *d²/4 = 784 ×10³ / 5×10⁸
= 156.8 ×10⁻⁵
d² = 19.9×10⁻⁵
d = 1.41 mm
The minimum diameter of a steel wire is 1.41mm
To know more about the elasticity :
https://brainly.com/question/28447665
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