A wire initially has length `L_0` and resistance `5.00 Omega.` The resistance is to be increased to `45.0 Omega` by stretching the wire. Assuming that the resistivity and density of the material are unaffected by the stretching, find the ratio of the new length to `L_(0)`.

Answer

To solve the problem, we need to find the ratio of the new length of the wire (L) to its original length (L₀) after the resistance has been increased from 5.00 Ω to 45.0 Ω by stretching the wire.

### Step-by-Step Solution:

1. **Understanding Resistance Formula**:
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.

2. **Initial Conditions**:
Let the initial resistance \( R_0 = 5.00 \, \Omega \), initial length \( L_0 \), and initial cross-sectional area \( A_0 \). Thus, we can write:
\[ R_0 = \frac{\rho L_0}{A_0} \]

3. **Final Conditions**:
After stretching, the new resistance \( R = 45.0 \, \Omega \), new length \( L \), and new cross-sectional area \( A \). So we have:
\[ R = \frac{\rho L}{A} \]

4. **Volume Conservation**:
When the wire is stretched, its volume remains constant. Therefore, we can write:
\[ A_0 L_0 = A L \]
From this, we can express the ratio of the areas:
\[ \frac{A_0}{A} = \frac{L}{L_0} \]

5. **Setting Up the Resistance Ratio**:
Now, we can set up the ratio of the resistances:
\[ \frac{R}{R_0} = \frac{\frac{\rho L}{A}}{\frac{\rho L_0}{A_0}} = \frac{L}{L_0} \cdot \frac{A_0}{A} \]
Substituting \( \frac{A_0}{A} = \frac{L}{L_0} \) into the equation gives:
\[ \frac{R}{R_0} = \frac{L}{L_0} \cdot \frac{L}{L_0} = \left(\frac{L}{L_0}\right)^2 \]

6. **Calculating the Resistance Ratio**:
Now substituting the values of \( R \) and \( R_0 \):
\[ \frac{45.0 \, \Omega}{5.00 \, \Omega} = \left(\frac{L}{L_0}\right)^2 \]
This simplifies to:
\[ 9 = \left(\frac{L}{L_0}\right)^2 \]

7. **Taking the Square Root**:
Taking the square root of both sides gives:
\[ \frac{L}{L_0} = 3 \]

8. **Final Result**:
Therefore, the ratio of the new length to the original length is:
\[ \frac{L}{L_0} = 3 \]

### Conclusion:
The ratio of the new length to the original length \( L/L_0 \) is 3.