Question

use a variation model to solve for the unknown value the resistance of a wire varies directly as its length and inversely as the square of its diameter a 50 foot wire with a .2 in diameter has a resistance of 0.0125 Ohms
find the resistance of a 40 foot wire withe the diameter of .1 inches? round to two decimals if needed

Answer 1

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:

\[
\boxed{0.04 \text{ Ohms}}
\]