The Young’s modulus for steel is 20.7*1010 N/m2. How much will a 2.5 mm diameter wire 12 cm long, be strained when it supports a load of 450 N? Aluminum’s Young’s modulus is 6.9*1010 N/m2.

Express the strain as a decimal, not a percentage

What diameter of aluminum wire would be strained the same amount?

Express your answer in mm:

I found the strain 5.23*10^-4

and the diameter 3.98 mm but it garded wrong even though i think that is true

same here... Strange...

I got the same solution...

Me too...

a) 4.4E-4

b)4.33 mm

To find the strain in the steel wire, we can use the formula:

Strain (ε) = Stress (σ) / Young's Modulus (E)

First, we need to calculate the stress applied to the wire. Stress is defined as the force applied per unit area.

Given:
Young's Modulus of steel (E) = 20.7 * 10^10 N/m^2
Diameter of steel wire (d) = 2.5 mm (radius, r = d/2)
Length of steel wire (L) = 12 cm = 0.12 m
Load (F) = 450 N

We can find the stress using the formula:
Stress (σ) = Load (F) / Area (A)

The area of a wire can be calculated using the formula:
Area (A) = π * radius^2

Substituting the given values, we have:
radius (r) = (2.5 mm) / 2 = 1.25 mm = 1.25 * 10^-3 m
Area (A) = π * (1.25 * 10^-3 m)^2

Now we can calculate the stress:
Stress (σ) = 450 N / [π * (1.25 * 10^-3 m)^2]

Next, we can calculate the strain using the given Young's Modulus for steel:
Strain (ε) = Stress (σ) / Young's Modulus (E)
Strain (ε) = [450 N / [π * (1.25 * 10^-3 m)^2]] / (20.7 * 10^10 N/m^2)

Calculating this expression will give us the strain in decimal form.

To find the diameter of an aluminum wire that would be strained the same amount, we can use the same formula with the given Young's Modulus for aluminum (E = 6.9 * 10^10 N/m^2) and solve for the diameter.

Let's calculate both the strain in steel and the diameter of the aluminum wire:

First, calculate the strain in the steel wire:
Strain (ε) = [450 N / [π * (1.25 * 10^-3 m)^2]] / (20.7 * 10^10 N/m^2)

Next, calculate the diameter of the aluminum wire using the same strain formula:
Strain (ε) = Stress (σ) / Young's Modulus (E) for aluminum
Stress (σ) = Load (F) / Area (A), and Area (A) = π * radius^2
Solve for diameter (d), given Young's Modulus (E) for aluminum and strain (ε).

The result will be the diameter expressed in mm.