an aluminium alloy bar with a Young's modulus of 70GPa is under a tensile stress of 10MPa, Which of thefollowing is the equal to the strain in the bar?
A 0.0014
B 7.0x10raise to the power3
c 700
D 1.4x10 raise to the power -4
To find the strain in the aluminum alloy bar, we need to use Hooke's Law, which states that the strain is equal to the stress divided by the Young's modulus.
The formula for strain is:
Strain = Stress / Young's modulus
Given:
Young's modulus = 70 GPa = 70 × 10^9 Pa
Stress = 10 MPa = 10 × 10^6 Pa
Let's substitute the values into the formula:
Strain = 10 × 10^6 Pa / (70 × 10^9 Pa)
Now, let's simplify the equation:
Strain = (10/70) × (10^6/10^9)
Since the unit of strain is dimensionless, we can simplify further:
Strain = (1/7) × 10^(-3)
Expressing it in scientific notation:
Strain = 1.4 × 10^(-4)
So, the correct answer is option D: 1.4 × 10^(-4).