Answer:
the diameter of the wire is 5 mm.
Explanation:
Given;
tensile force of the wire, F = 60 N
stress on the wire, δ = 3,060,000 Pa = 3,060,000 N/m²
Let A be the cross sectional area of the wire
The cross sectional area of the wire is calculated as follows;
[tex]\sigma = \frac{F}{A} \\\\A = \frac{F}{\sigma} \\\\A = \frac{60}{3,060,000} = 1.961 \times 10^{-5} \ m^2[/tex]
The diameter of the wire is calculated as follows;
[tex]A = \frac{\pi d^2}{4} \\\\d = \sqrt{\frac{4A}{\pi} } \\\\d = \sqrt{\frac{4\times 1.961 \times 10^{-5}}{\pi} } \\\\d = 5.0 \times 10^{-3} \ m\\\\d = 5 \ mm[/tex]
Therefore, the diameter of the wire is 5 mm.
