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Sagot :
The elastic modulus of magnesium bar is 3.14 GPa.
We need to know about the modulus of elasticity to solve this problem. The elastic modulus of material can be measured by strain and stress. It can be determined as
B = σ / e
B = (F/A) / (ΔV/V)
where B is modulus of elasticity, σ is stress, e is strain, F is force, A is area, V is initial volume, ΔV is volume stretched.
From the question above, we know that:
F = 22000 N
A = 1cm x 1cm = 1 cm²
V = A x 10cm = 10 cm³
ΔV = (10.07-10) x A =0.07 cm³
Convert to SI unit
A = 1 x 10¯⁴ m²
V = 10¯⁵ m³
ΔV = 7 x 10¯⁴ m³
By substituting the following parameters, we get
B = (F/A) / (ΔV/V)
B = (22000 / (1 x 10¯⁴) / (7 x 10¯⁴ / 10¯⁵)
B = 3.14 x 10⁶ Pa
B = 3.14 GPa
For more on elastic modulus at: https://brainly.com/question/19551488
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