Physics, asked by tanisha2088, 1 year ago

A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

Answers

Answered by jack6778
8

Answer:

Length of the piece of copper, l = 19.1 mm = 19.1 × 10–3 m

Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10–3 m

Area of the copper piece:

A = l × b

= 19.1 × 10–3 × 15.2 × 10–3

= 2.9 × 10–4 m2

Tension force applied on the piece of copper, F = 44500 N

Modulus of elasticity of copper, η = 42 × 109 N/m2

Modulus of elasticity, η = Stress / Strain

= (F/A) / Strain

∴ Strain = F / Aη

= 44500 / (2.9 × 10-4 × 42 × 109)

= 3.65 × 10–3.

Answered by Aastha6878
4

Answer:

Length of the piece of copper, l = 19.1 mm = 19.1 × 10–3 m

Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10–3 m

Area of the copper piece:

A = l × b

= 19.1 × 10–3 × 15.2 × 10–3

= 2.9 × 10–4 m2

Tension force applied on the piece of copper, F = 44500 N

Modulus of elasticity of copper, η = 42 × 109 N/m2

Modulus of elasticity, η = Stress / Strain

= (F/A) / Strain

∴ Strain = F / Aη

= 44500 / (2.9 × 10-4 × 42 × 109)

= 3.65 × 10–3.

Similar questions