College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Can you please type out all of your solution so that they are easy for me to read I have really bad eyesight
![### Physics Problem: Resistance Ratio Calculation
#### Problem Statement
**14. (II)** Calculate the ratio of the resistance of 10.0 m of aluminum wire 2.2 mm in diameter, to 24.0 m of copper wire 1.8 mm in diameter.
---
This problem requires an understanding of the physical principles behind electrical resistance in conductors. To solve it, consider the formula for resistance:
\[ R = \rho \frac{L}{A} \]
Where:
- **R** is the resistance,
- **\(\rho\)** (rho) is the resistivity of the material,
- **L** is the length of the conductor,
- **A** is the cross-sectional area of the conductor.
For a wire with a circular cross-section, the area **A** can be determined using the formula for the area of a circle:
\[ A = \pi \left(\frac{d}{2}\right)^2 \]
Here, **d** is the diameter of the wire.
To find the ratio of the resistances of the aluminum and copper wires, follow these steps:
1. **Calculate the cross-sectional area** for each wire:
\[ A_{\text{Al}} = \pi \left(\frac{2.2 \, \text{mm}}{2}\right)^2 = \pi (1.1 \, \text{mm})^2 = \pi (1.21 \, \text{mm}^2) = 3.80 \, \text{mm}^2 \]
\[ A_{\text{Cu}} = \pi \left(\frac{1.8 \, \text{mm}}{2}\right)^2 = \pi (0.9 \, \text{mm})^2 = \pi (0.81 \, \text{mm}^2) = 2.54 \, \text{mm}^2 \]
2. **Calculate the resistance** for each wire:
\[ R_{\text{Al}} = \rho_{\text{Al}} \frac{L_{\text{Al}}}{A_{\text{Al}}} \]
\[ R_{\text{Cu}} = \rho_{\text{Cu}} \frac{L_{\text{Cu}}}{A_{\text{Cu}}} \]
Given the resistivity values (at room temperature):](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5d3f54ef-f765-47f8-9b56-6f713dd53ab6%2F4593da28-b564-4320-a970-bf43dbf19919%2Fvnv2h6_processed.png&w=3840&q=75)
Transcribed Image Text:### Physics Problem: Resistance Ratio Calculation
#### Problem Statement
**14. (II)** Calculate the ratio of the resistance of 10.0 m of aluminum wire 2.2 mm in diameter, to 24.0 m of copper wire 1.8 mm in diameter.
---
This problem requires an understanding of the physical principles behind electrical resistance in conductors. To solve it, consider the formula for resistance:
\[ R = \rho \frac{L}{A} \]
Where:
- **R** is the resistance,
- **\(\rho\)** (rho) is the resistivity of the material,
- **L** is the length of the conductor,
- **A** is the cross-sectional area of the conductor.
For a wire with a circular cross-section, the area **A** can be determined using the formula for the area of a circle:
\[ A = \pi \left(\frac{d}{2}\right)^2 \]
Here, **d** is the diameter of the wire.
To find the ratio of the resistances of the aluminum and copper wires, follow these steps:
1. **Calculate the cross-sectional area** for each wire:
\[ A_{\text{Al}} = \pi \left(\frac{2.2 \, \text{mm}}{2}\right)^2 = \pi (1.1 \, \text{mm})^2 = \pi (1.21 \, \text{mm}^2) = 3.80 \, \text{mm}^2 \]
\[ A_{\text{Cu}} = \pi \left(\frac{1.8 \, \text{mm}}{2}\right)^2 = \pi (0.9 \, \text{mm})^2 = \pi (0.81 \, \text{mm}^2) = 2.54 \, \text{mm}^2 \]
2. **Calculate the resistance** for each wire:
\[ R_{\text{Al}} = \rho_{\text{Al}} \frac{L_{\text{Al}}}{A_{\text{Al}}} \]
\[ R_{\text{Cu}} = \rho_{\text{Cu}} \frac{L_{\text{Cu}}}{A_{\text{Cu}}} \]
Given the resistivity values (at room temperature):
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