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Consider a cylindrical nickel wire 1.8 mm in diameter and 2.6 × 104 mm long. Calculate its elongation when a load of 290 N is applied. Assume that the deformation is totally elastic and that the elastic modulus for nickel is 207 GPa (or 207 × 109 N/m2).

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klefenz

Answer:

e = 3.97*10^-4

Explanation:

1.8 mm = 0.0018 m

2.6*10^4 mm = 26 m

Elongation is The ratio between the stretched length and the original length.

e = L/L0

This is calculated with Hooke's law:

e = σ/E

Where

σ: normal stress

E: elastic constant

σ = P/A

Where

P: normal load

A: cross section

A = π/4 * d^2

Therefore:

e = P / (A * E)

e = 4 * P / (π * d^2 * E)

e = 4 * 290 / (π * 0.0018^2 * 207*10^9) = 3.97*10^-4

Cricetus

When a load of 290 N is applied, its elongation will be "3.97 × 10⁻⁴".

Elastic modulus

According to the question,

Length of wire, L = 1.8 mm or,

                            = 0.0018 m

Diameter, d = 2.6 × 10⁴ mm or,

                    = 26 m

We know,

The cross-section be:

A = [tex]\frac{\pi}{d}[/tex] × d²

By Hooke's law, we get

→ e = [tex]\frac{L}{L_0}[/tex]

or, the elongation be:

→ e = [tex]\frac{P}{A\times E}[/tex]

     = 4 × [tex]\frac{P}{\pi\times d^2\times E}[/tex]

By substituting the values, we get

     = 4 × [tex]\frac{P}{\pi\times (0.0018)^2\times 207\times 10^9}[/tex]

     = 3.97 × 10⁻⁴

Thus the above response is correct.

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