A student is tesling project. is 10:0 m length is made up a wire for If length of diameter is the wire If the wire ff steel, haw much force is required to stretched this wire by 1 cm? (Ascume Young's modulus af steel 8) = 200 ×10N/m2) 25 cm. p
A student is tesling project. is 10:0 m length is made up a wire for If length of diameter is the wire If the wire ff steel, haw much force is required to stretched this wire by 1 cm? (Ascume Young's modulus af steel 8) = 200 ×10N/m2) 25 cm. p
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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A. 9.8 x 106 N
B. 18 x 106 N
C. 1.1 x 106 N
![**Title:** Calculating the Force Required to Stretch a Steel Wire
**Introduction:**
In this problem, we explore the concept of elasticity and how force is required to stretch a material. A student is conducting an experiment involving a wire. Here's the given scenario:
**Problem Statement:**
A student is testing a wire for his project. The wire has:
- A length of 10.0 meters
- A diameter of 2.5 centimeters
The wire is made of steel. We need to calculate how much force is required to stretch this wire by 1 centimeter. For this calculation, assume Young's modulus (\(Y\)) for steel is \(200 \times 10^9 \, \text{N/m}^2\).
**Solution Approach:**
To solve this, we can use Hooke’s Law in terms of Young’s modulus. The formula is:
\[
F = \frac{Y \cdot A \cdot \Delta L}{L}
\]
Where:
- \(F\) is the force required
- \(Y\) is Young's modulus
- \(A\) is the cross-sectional area of the wire
- \(\Delta L\) is the change in length (1 cm = 0.01 m)
- \(L\) is the original length of the wire
**Cross-Sectional Area Calculation:**
Convert diameter to meters: \(2.5 \, \text{cm} = 0.025 \, \text{m}\).
Calculate the radius: \(r = 0.0125 \, \text{m}\).
Area, \(A = \pi r^2 = \pi (0.0125)^2\).
Substituting the values will give us the force required to stretch the wire by the specified amount.
**Conclusion:**
This calculation helps understand the application of material properties in practical scenarios, like determining the force required to cause a specified deformation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff015db4-0da8-42fc-9d60-01a672218c35%2F27084624-acb0-4a44-aaae-a3ee249d3068%2Fjcqxn5i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title:** Calculating the Force Required to Stretch a Steel Wire
**Introduction:**
In this problem, we explore the concept of elasticity and how force is required to stretch a material. A student is conducting an experiment involving a wire. Here's the given scenario:
**Problem Statement:**
A student is testing a wire for his project. The wire has:
- A length of 10.0 meters
- A diameter of 2.5 centimeters
The wire is made of steel. We need to calculate how much force is required to stretch this wire by 1 centimeter. For this calculation, assume Young's modulus (\(Y\)) for steel is \(200 \times 10^9 \, \text{N/m}^2\).
**Solution Approach:**
To solve this, we can use Hooke’s Law in terms of Young’s modulus. The formula is:
\[
F = \frac{Y \cdot A \cdot \Delta L}{L}
\]
Where:
- \(F\) is the force required
- \(Y\) is Young's modulus
- \(A\) is the cross-sectional area of the wire
- \(\Delta L\) is the change in length (1 cm = 0.01 m)
- \(L\) is the original length of the wire
**Cross-Sectional Area Calculation:**
Convert diameter to meters: \(2.5 \, \text{cm} = 0.025 \, \text{m}\).
Calculate the radius: \(r = 0.0125 \, \text{m}\).
Area, \(A = \pi r^2 = \pi (0.0125)^2\).
Substituting the values will give us the force required to stretch the wire by the specified amount.
**Conclusion:**
This calculation helps understand the application of material properties in practical scenarios, like determining the force required to cause a specified deformation.
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