College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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## Item 18

**Scenario:**
You apply a force of magnitude \( F_1 \) to one end of a wire and another force \( F_1 \) in the opposite direction to the other end of the wire. The cross-sectional area of the wire is \( 8.00 \, \text{mm}^2 \). You measure the fractional change in the length of the wire, \( \Delta l / l_0 \), for several values of \( F_1 \). When you plot your data with \( \Delta l / l_0 \) (in units of N) on the vertical axis and \( F_1 \) (in units of N) on the horizontal axis, the data lie close to a line that has a slope \( 8.0 \times 10^{-7} \, \text{N}^{-1} \).

### Part A
**Question:**
What is the value of Young's modulus for this wire?

**Instructions:**
Express your answer with the appropriate units.

### Input Field
- **Value:** Text box to enter the numerical value of Young's modulus.
- **Units:** Drop-down menu to select the appropriate units.

### Submit Button:
- **Submit:** Button to submit the answer.
- **Request Answer:** Button to request the correct answer if unsure.

### Additional Options:
There are various tools and icons designed to help the user format their answer, including equation editors and help functions.

### Feedback:
A link to provide feedback on the problem is available at the bottom of the interface.

**Graphical Interpretation:**
- The question implies a linear relationship between \( \Delta l / l_0 \) and \( F_1 \), with the given slope \( 8.0 \times 10^{-7} \, \text{N}^{-1} \).

### Calculation Tip:
To find Young's modulus (\( Y \)), use the formula:
\[ Y = \frac{F/A}{\Delta l / l_0} \]
Given the problem's slope as \( 8.0 \times 10^{-7} \, \text{N}^{-1} \) and the cross-sectional area \( A = 8.00 \, \text{mm}^2 \).

Ensure proper conversion of units as needed and input the value accordingly in the fields provided.
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Transcribed Image Text:## Item 18 **Scenario:** You apply a force of magnitude \( F_1 \) to one end of a wire and another force \( F_1 \) in the opposite direction to the other end of the wire. The cross-sectional area of the wire is \( 8.00 \, \text{mm}^2 \). You measure the fractional change in the length of the wire, \( \Delta l / l_0 \), for several values of \( F_1 \). When you plot your data with \( \Delta l / l_0 \) (in units of N) on the vertical axis and \( F_1 \) (in units of N) on the horizontal axis, the data lie close to a line that has a slope \( 8.0 \times 10^{-7} \, \text{N}^{-1} \). ### Part A **Question:** What is the value of Young's modulus for this wire? **Instructions:** Express your answer with the appropriate units. ### Input Field - **Value:** Text box to enter the numerical value of Young's modulus. - **Units:** Drop-down menu to select the appropriate units. ### Submit Button: - **Submit:** Button to submit the answer. - **Request Answer:** Button to request the correct answer if unsure. ### Additional Options: There are various tools and icons designed to help the user format their answer, including equation editors and help functions. ### Feedback: A link to provide feedback on the problem is available at the bottom of the interface. **Graphical Interpretation:** - The question implies a linear relationship between \( \Delta l / l_0 \) and \( F_1 \), with the given slope \( 8.0 \times 10^{-7} \, \text{N}^{-1} \). ### Calculation Tip: To find Young's modulus (\( Y \)), use the formula: \[ Y = \frac{F/A}{\Delta l / l_0} \] Given the problem's slope as \( 8.0 \times 10^{-7} \, \text{N}^{-1} \) and the cross-sectional area \( A = 8.00 \, \text{mm}^2 \). Ensure proper conversion of units as needed and input the value accordingly in the fields provided.
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