The elongation of the wire is [tex]3.96 \times 10^{-3} \ m[/tex]
The given parameters;
- mass of the object = 50 kg
- length of the wire, l = 4 m
- diameter of the wire, d = 3 mm = 0.003 m
- Young's modulus = [tex]7 \times 10^{10 \ Nm[/tex]
The Young's modulus is given as;
[tex]E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae}[/tex]
The area of the wire is calculated as;
[tex]A = \frac{\pi d^2}{4} = \frac{\pi (0.003)^2}{4} = 7.069\times 10^{-6} \ m^2[/tex]
The elongation of the wire is calculated as;
[tex]E = \frac{FL}{Ae} \\\\e = \frac{FL}{AE} = \frac{(50\times 9.8) \times 4}{(7.069\times 10^{-6}) \times 7\times 10^{10}} \\\\e = 3.96 \times 10^{-3} \ m[/tex]
Thus, the elongation of the wire is [tex]3.96 \times 10^{-3} \ m[/tex]
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