An aluminum bar 125 mm long with a square cross section 16 mm on an edge is pulled in tension with a load of 66,700 N and experiences an elongation of 0.43 mm. Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum. (Please leave a space between the number and the unit, and use correct capitalization and lower case for units - for example: MPa, not Mpa.)

the modulus of elasticity of the aluminum is 75740.37 MPa

Explanation:

Given that;

Length of Aluminum bar L = 125 mm

square cross section s = 16 mm

so area of cross section of the aluminum bar is;

A = s² = 16² = 256 mm²

Tensile load acting the bar p = 66,700 N

elongation produced Δ = 0.43

so

Δ = PL / AE

we substitute

0.43 = (66,700 × 125) / (256 × E)

0.43(256 × E) = (66,700 × 125)

110.08E = 8337500

E = 8337500 / 110.08

E = 75740.37 MPa

Therefore, the modulus of elasticity of the aluminum is 75740.37 MPa

answer:

the answer is d

explanation:

answer: a

a is the answer