A wire with cross sectional area 1.00 cm2 stretches 0.500% of its length when it is stretched with a tension of 20.0 N. What is the Young's modulus of this wire?
E= (force/area) /(dL/L)
= (20N/1E-4m^2)/(.005)
E= 20E4*2E2=40E6 N/m^2
1 cm^2 = 10^-4 m^2
delta L = L F/(EA)
.005 = 20 /(E*10^-4)
E = 4*10^7 Pascals
Think material is like rubber or you have a typo.
To find the Young's modulus of the wire, we need to use Hooke's Law, which relates the stress and strain of a material. Stress is defined as the force per unit area, and strain is the elongation per unit length.
The formula for stress is given by:
Stress = Force / Area
In this case, the force is 20.0 N and the area is 1.00 cm2. Let's convert the area to square meters:
Area = 1.00 cm2 = 1.00 x 10^(-4) m2
So, the stress is:
Stress = 20.0 N / 1.00 x 10^(-4) m2 = 2.00 x 10^5 N/m2
The formula for strain is given by:
Strain = (Change in length) / Original length
In this case, the wire stretches by 0.500% of its original length. Let's call the original length L:
Strain = (0.500/100) x L
The Young's modulus (E) is defined as the ratio of stress to strain:
Young's modulus = Stress / Strain
Substituting the values we found:
Young's modulus = 2.00 x 10^5 N/m2 / [(0.500/100) x L]
Simplifying:
Young's modulus = (2.00 x 10^5 N/m2) / (0.005 x L)
Now, we need to know the value of L, the original length of the wire, to calculate the Young's modulus. If you provide the value of L, we can calculate the final result.