To solve this problem, we can use the relationship described in the prompt:
\[ R = k \cdot \frac{L}{d^2} \]
where:
- \( R \) is the resistance,
- \( L \) is the length of the wire,
- \( d \) is the diameter of the wire,
- \( k \) is a proportionality constant.
Step 1: Find the Constant \( k \)
We know from the information given that:
- For a 50-foot wire with a diameter of 0.2 inches, the resistance \( R \) is 0.0125 Ohms.
Plugging in the values, we can solve for \( k \):
\[ 0.0125 = k \cdot \frac{50}{(0.2)^2} \]
Calculating \( (0.2)^2 \):
\[ (0.2)^2 = 0.04 \]
Now substituting it back in:
\[ 0.0125 = k \cdot \frac{50}{0.04} \]
Calculating \( \frac{50}{0.04} \):
\[ \frac{50}{0.04} = 1250 \]
So, we have:
\[ 0.0125 = k \cdot 1250 \]
Now, solving for \( k \):
\[ k = \frac{0.0125}{1250} = 0.00001 \]
Step 2: Find the Resistance of the 40-Foot Wire with 0.1-Inch Diameter
Now substituting \( k \) back into the equation and using the new values:
- \( L = 40 \) feet
- \( d = 0.1 \) inches
\[ R = 0.00001 \cdot \frac{40}{(0.1)^2} \]
Calculating \( (0.1)^2 \):
\[ (0.1)^2 = 0.01 \]
Now substituting that value in:
\[ R = 0.00001 \cdot \frac{40}{0.01} \]
Calculating \( \frac{40}{0.01} \):
\[ \frac{40}{0.01} = 4000 \]
Substituting this back in:
\[ R = 0.00001 \cdot 4000 \]
Calculating \( 0.00001 \cdot 4000 \):
\[ R = 0.04 \text{ Ohms} \]
Final Answer
The resistance of a 40-foot wire with a diameter of 0.1 inches is:
\[ \boxed{0.04 \text{ Ohms}} \]