A copper wire 2 mm in diameter was stretched by 1 mm under the action of a force of 100 N
| February 1, 2021 | education
A copper wire 2 mm in diameter was stretched by 1 mm under the action of a force of 100 N. What is the initial length of the wire? (E = 120 GPa)
d = 2 mm = 2 * 10 ^ -3 m.
F = 100 N.
Δ l = 1 mm = 1 * 10 ^ -3 m.
E = 120 GPa = 120 * 10 ^ 9 Pa.
l0 -?
F / S = E * Δ l / l0.
l0 = S * E * Δ l / F.
S = P * d ^ 2/4.
l0 = P * d ^ 2 * E * Δ l / 4 * F.
l0 = 3.14 * (2 * 10 ^ -3 m) ^ 2 * 120 * 10 ^ 9 Pa * 1 * 10 ^ -3 m / 4 * 100 N = 3.768 m.
Answer: l0 = 3.768 m.
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