To determine the amount by which the wire stretches under the weight, we can use the formula derived from Young's modulus:

$\Delta L = \frac{F L}{A Y}$

where:

• $F$ is the force applied (5.44 kN),
• $L$ is the original length of the wire (2.25 m),
• $A$ is the cross-sectional area of the wire (5.00 mm²),
• $Y$ is Young’s modulus (92.0 GPa).

Let’s solve step-by-step:

1. Convert units as needed:

   • Convert $F$ from kN to N:
     $F = 5.44 \, \text{kN} = 5440 \, \text{N}$
   • Convert $A$ from mm² to m²:
     $A = 5.00 \, \text{mm}^2 = 5.00 \times 10^{-6} \, \text{m}^2$
   • Young's modulus is already given in Pascals (92.0 GPa = $92.0 \times 10^9 \, \text{Pa}$).

2. Substitute values into the equation: $\Delta L = \frac{5440 \times 2.25}{(5.00 \times 10^{-6}) \times (92.0 \times 10^9)}$

3. Calculate: $\Delta L = \frac{12240}{4.6 \times 10^5} = 0.0266 \, \text{m}$

So, the wire stretches by 0.0266 meters, or 2.66 cm.

Would you like further details on any part of this solution?

Related Questions

1. What is Young's modulus and how does it impact material behavior?
2. How would the stretch change if the weight were doubled?
3. What happens to the stretch if the cross-sectional area is reduced?
4. How does material length affect the amount of stretch?
5. What are other common materials with similar Young's modulus values?

Tip

When using units in physics calculations, always ensure all units are converted to SI units (meters, Newtons, etc.) for consistency and accuracy.