A heating element is made by maintaining a potential difference of 77.0 V across the length of a Nichrome wire that has a 3.30 × 10-6 m? cross section. Nichrome has a resistivity of 5.00 × 10-7 Q•m. (a) If the element dissipates 7490 W, what is its length? (b) If 114 V is used to obtain the same dissipation rate, what should the length be?

To solve the problem, we can use the following concepts:

1. **Power Dissipated by a Resistor**:  
   The power \( P \) dissipated by a resistor is given by:
   \[
   P = \frac{V^2}{R}
   \]
   where \( V \) is the potential difference (voltage) and \( R \) is the resistance.

2. **Resistance of a Wire**:  
   The resistance \( R \) of a wire is related to its resistivity \( \rho \), length \( L \), and cross-sectional area \( A \) by:
   \[
   R = \rho \frac{L}{A}
   \]

### Part (a): Finding the Length of the Wire for 77.0 V

We are given:
- Power, \( P = 7490 \, \text{W} \)
- Voltage, \( V = 77.0 \, \text{V} \)
- Cross-sectional area, \( A = 3.30 \times 10^{-6} \, \text{m}^2 \)
- Resistivity of Nichrome, \( \rho = 5.00 \times 10^{-7} \, \Omega \cdot \text{m} \)

First, solve for the resistance using the power formula:
\[
P = \frac{V^2}{R}
\]
Rearrange to find \( R \):
\[
R = \frac{V^2}{P} = \frac{(77.0 \, \text{V})^2}{7490 \, \text{W}} = \frac{5929 \, \text{V}^2}{7490 \, \text{W}} = 0.7915 \, \Omega
\]

Next, use the resistance formula to solve for the length \( L \):
\[
R = \rho \frac{L}{A}
\]
Rearrange to find \( L \):
\[
L = \frac{R A}{\rho} = \frac{(0.7915 \, \Omega)(3.30 \times 10^{-6} \, \text{m}^2)}{5.00 \times 10^{-7} \, \Omega \cdot \text{m}}
\]
\[
L = \frac{2.61195 \times 10^{-6} \, \Omega \cdot \text{m}^2}{5.00 \times 10^{-7} \, \Omega \cdot \text{m}} = 5.224 \, \text{m}
\]

So, the length of the wire is approximately **5.22 meters**.

---

### Part (b): Finding the Length of the Wire for 114 V

For the second part, we want the same power dissipation (7490 W), but now with a voltage of 114 V. Using the same power formula:
\[
R = \frac{V^2}{P} = \frac{(114 \, \text{V})^2}{7490 \, \text{W}} = \frac{12996 \, \text{V}^2}{7490 \, \text{W}} = 1.735 \, \Omega
\]

Now solve for the new length \( L \):
\[
L = \frac{R A}{\rho} = \frac{(1.735 \, \Omega)(3.30 \times 10^{-6} \, \text{m}^2)}{5.00 \times 10^{-7} \, \Omega \cdot \text{m}}
\]
\[
L = \frac{5.7255 \times 10^{-6} \, \Omega \cdot \text{m}^2}{5.00 \times 10^{-7} \, \Omega \cdot \text{m}} = 11.451 \, \text{m}
\]

So, the length of the wire should be approximately **11.45 meters** to achieve the same power dissipation with 114 V.

---

### Summary of Results:
- **(a)** The length of the wire for 77.0 V is approximately **5.22 m**.
- **(b)** The length of the wire for 114 V is approximately **11.45 m**.

Do you have any questions or want more details on any step?

---

Here are 5 follow-up questions you might find interesting:
1. How does the length of the wire affect its resistance?
2. What happens to the power dissipation if the cross-sectional area changes?
3. How would the dissipation rate change if the material's resistivity were different?
4. What would be the length if the power dissipation was reduced to 5000 W at 77.0 V?
5. How does temperature affect the resistivity of Nichrome wire?

**Tip**: Resistance increases linearly with the length of a wire, meaning doubling the length will double the resistance.

A heating element is made by maintaining a potential difference of 77.0 V across the length of a Nichrome wire that has a 3.30 × 10-6 m² cross section. Nichrome has a resistivity of 5.00 × 10-7 Ω•m. (a) If the element dissipates 7490 W, what is its length? (b) If 114 V is used to obtain the same dissipation rate, what should the length be?

Calculate the length of a Nichrome wire heating element when given the power dissipation, voltage, cross-sectional area, and resistivity. Includes step-by-step solutions for two scenarios with different voltages: 77.0 V and 114 V, using resistivity and power formulas. Suitable for high school physics students.

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