An aluminum wire 1.0 m long is stretched 0.02 cm when under stress.....?

If a brass wire with an identical diameter were used to support the same load, what would its initial length have been for it to have had the same extension?

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An aluminum wire 1.0 m long is stretched 0.02 cm when under stress.....?binoculars

Assume elastic deformation.

For elastic deformation strain and stress are related as:

蟽 = E路蔚

E is Young's modulus

Stress is defined as force per coss-sectional area of the object

蟽 = F/A = F / (蟺路R2) for a wire.

Strain equals is elongation divided by initial length

蔚 = 螖L/L?

The force on the wire is the same as well as the diameter and the cross sectional area. Hence both experience the same stress:

蟽_a = 蟽_b (a aluminum; b brass)

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E_a 路 蔚_a = E_b 路 蔚_b

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E_al路 螖L_a / L?_a = E_b 路 螖L_b / L?_b

Moreover the wires have the same elongation:

螖L_a = 螖L_b

Therefore:

E_al路 螖L_a / L?_a = E_b 路 螖L_a / L?_b

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L?_b = L?_a 路 E_b / E_a

= 1m路 113.5GPa / 69Gpa

= 1.645m

( i took Young's modulus from wikipedia, for brass i took the arithmetic average of the range given)