Aspecimen of aluminum having a rectangular cross section 9.5 mm × 13.0 mm (0.3740 in. × 0.5118 in.) is pulled in tension with 35600 n (8003 lbf) force, producing only elastic deformation. the elastic modulus for aluminum is 69 gpa (or 10 × 106 psi). calculate the resulting strain.

〖4.1x10〗^(-3)

Explanation:

From the question, the parameters given are:

We can solve this using the formula for elastic modulus ( E ) = stress/strain  

making strain the subject of the formula the equation above becomes strain= stress/ elastic modulus

Where strain= change in length/ length

And  

Stress= force/ area

now let’s calculate value for stress

Stress= 35600/0.0001235=288,259,109  N/m^2

substituting the value for stress into the equation strain= stress/ elastic modulus we have

Strain= 288,259,109 / 69 x 〖10〗^9 = 〖4.1x10〗^(-3)

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.