A4052 resiskos is made from a coch of Capper wire shas that the Length and the Cross sectimal area be written as and A whose bokel mass is 10:29. may =me V Rd resisfiviky of tHie material,d is the densiky, m is L=IRm Pd P is the the mass and R is the resisfance.
A4052 resiskos is made from a coch of Capper wire shas that the Length and the Cross sectimal area be written as and A whose bokel mass is 10:29. may =me V Rd resisfiviky of tHie material,d is the densiky, m is L=IRm Pd P is the the mass and R is the resisfance.
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![### Determining Length and Cross-sectional Area of a Copper Wire Resistor
A 40Ω resistor is made from a coil of copper wire whose total mass is 10.2g. Show that the length and the cross-sectional area may be written as:
\[ L = \sqrt{\frac{Rm}{\rho d}} \]
\[ A = \sqrt{\frac{mP}{Rd}} \]
where
- \( \rho \) (rho) is the resistivity of the material,
- \( d \) is the density,
- \( m \) is the mass, and
- \( R \) is the resistance.
### Explanation:
To solve the given problem, we need to derive the formulas for the length (L) and cross-sectional area (A) of the copper wire used in the resistor. The known parameters are:
- Resistance (R) = 40Ω
- Mass (m) = 10.2g
- Resistivity (\(\rho\)) of copper
- Density (d) of copper
### Detailed Steps:
1. **Relate the resistance to resistivity:**
The resistance \( R \) of a wire is given by the formula:
\[
R = \rho \frac{L}{A}
\]
where \( L \) is the length of the wire and \( A \) is its cross-sectional area.
2. **Relate mass to volume:**
The mass \( m \) of the wire can be related to its density \( d \) and volume \( V \):
\[
m = dV
\]
3. **Relate volume to length and cross-sectional area:**
The volume \( V \) of the wire can also be expressed in terms of its length and cross-sectional area:
\[
V = AL
\]
Therefore, the mass \( m \) becomes:
\[
m = dAL
\]
4. **Combining the Equations:**
From Steps 1 and 3, solve for \( L \) and \( A \):
Step 1:
\[
R = \rho \frac{L}{A}
\]
\[
A = \frac{\rho L}{R}
\]
Step 3:
\[
m = dAL
\]
Substitute \( A \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5db72156-5f58-48bc-bfda-30e04998aaf5%2Ff50aacdb-8b43-49a7-9cc2-fea62d01b79c%2Fi0fgo8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Determining Length and Cross-sectional Area of a Copper Wire Resistor
A 40Ω resistor is made from a coil of copper wire whose total mass is 10.2g. Show that the length and the cross-sectional area may be written as:
\[ L = \sqrt{\frac{Rm}{\rho d}} \]
\[ A = \sqrt{\frac{mP}{Rd}} \]
where
- \( \rho \) (rho) is the resistivity of the material,
- \( d \) is the density,
- \( m \) is the mass, and
- \( R \) is the resistance.
### Explanation:
To solve the given problem, we need to derive the formulas for the length (L) and cross-sectional area (A) of the copper wire used in the resistor. The known parameters are:
- Resistance (R) = 40Ω
- Mass (m) = 10.2g
- Resistivity (\(\rho\)) of copper
- Density (d) of copper
### Detailed Steps:
1. **Relate the resistance to resistivity:**
The resistance \( R \) of a wire is given by the formula:
\[
R = \rho \frac{L}{A}
\]
where \( L \) is the length of the wire and \( A \) is its cross-sectional area.
2. **Relate mass to volume:**
The mass \( m \) of the wire can be related to its density \( d \) and volume \( V \):
\[
m = dV
\]
3. **Relate volume to length and cross-sectional area:**
The volume \( V \) of the wire can also be expressed in terms of its length and cross-sectional area:
\[
V = AL
\]
Therefore, the mass \( m \) becomes:
\[
m = dAL
\]
4. **Combining the Equations:**
From Steps 1 and 3, solve for \( L \) and \( A \):
Step 1:
\[
R = \rho \frac{L}{A}
\]
\[
A = \frac{\rho L}{R}
\]
Step 3:
\[
m = dAL
\]
Substitute \( A \
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