Question

a wire 2.5 m long with a cross-sectional area of 6mm^2, street den by 1.27mm when a mass of 45 kg is suspended from it. Find the stress on the wire, the resulting strain of the value of young's modulus, Y of the wire?

Answer 1

Answer:

Here, l=4.0m;Δl=2×10−3m;a=2.0×10−6m2,Y=2.0×1011N/m2

(i) The energy density of stretched wire

u=21× stress × strain

=21×Y×(strain)2

=21×2.0×1011×(2×10−3)/4)2

=0.25×105=2.5×104J/m3.

(ii) Elastic potential energy = energy density × volume

=2.5×104×(2.0×10−6)×4.0J=20×10−2=0.20J.

Explanation:

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