Let the resistance \( R \) be modeled by the equation
\[ R = k \frac{L}{d^2} \]
where \( L \) is the length of the wire, \( d \) is the diameter of the wire, and \( k \) is a constant of proportionality.
Given:
- For the first wire, \( L_1 = 32 \) ft, \( d_1 = 0.8 \) in, and \( R_1 = 0.0125 \) ohms.
We can solve for \( k \) using this information:
\[ 0.0125 = k \frac{32}{(0.8)^2} \]
Calculating \( (0.8)^2 \):
\[ (0.8)^2 = 0.64 \]
Substituting that into the equation gives:
\[ 0.0125 = k \frac{32}{0.64} \]
Calculating \( \frac{32}{0.64} \):
\[ \frac{32}{0.64} = 50 \]
Now we can rewrite the equation:
\[ 0.0125 = k \cdot 50 \]
Solving for \( k \):
\[ k = \frac{0.0125}{50} = 0.00025 \]
Now, we want to find the resistance \( R_2 \) of a second wire with \( L_2 = 62 \) ft and \( d_2 = 0.4 \) in:
\[ R_2 = k \frac{L_2}{d_2^2} \]
Calculating \( d_2^2 \):
\[ (0.4)^2 = 0.16 \]
Now substituting \( L_2 \), \( d_2 \), and \( k \):
\[ R_2 = 0.00025 \frac{62}{0.16} \]
Calculating \( \frac{62}{0.16} \):
\[ \frac{62}{0.16} = 387.5 \]
Thus,
\[ R_2 = 0.00025 \cdot 387.5 = 0.096875 \]
Therefore, the resistance of the 62-ft wire with a diameter of 0.4 in is:
\[ \boxed{0.096875} \text{ ohms} \]
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