/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Multiple-Concept Example 4 revie... [FREE SOLUTION] | 91Ó°ÊÓ

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Multiple-Concept Example 4 reviews the concepts that are involved in this problem. A ruler is accurate when the temperature is \(25^{\circ} \mathrm{C}\). When the temperature drops to \(-14^{\circ} \mathrm{C}\), the ruler shrinks and no longer measures distances accurately. However, the ruler can be made to read correctly if a force of magnitude \(1.2 \times 10^{3} \mathrm{~N}\) is applied to each end so as to stretch it back to its original length. The ruler has a cross-sectional area of \(1.6 \times 10^{-5} \mathrm{~m}^{2},\) and it is made from a material whose coefficient of linear expansion is \(2.5 \times 10^{-5}\left(\mathrm{C}^{0}\right)^{-1}\). What is Young's modulus for the material from which the ruler is made?

Short Answer

Expert verified
Young's modulus for the ruler material is approximately \(1.92 \times 10^{11}\, \text{N/m}^2\).

Step by step solution

01

Recognize the Variables Needed

First, identify the known quantities: temperature change from \(25^{\circ} C\) to \(-14^{\circ} C\), force applied \(F = 1.2 \times 10^3\, \text{N}\), cross-sectional area \(A = 1.6 \times 10^{-5}\, \text{m}^2\), and coefficient of linear expansion \(\alpha = 2.5 \times 10^{-5}\, \text{C}^{-1}\). The goal is to find Young's modulus \(Y\).
02

Calculate the Thermal Expansion

Using the formula for linear expansion \(\Delta L = \alpha L_0 \Delta T\), calculate the change in length \(\Delta L\). \(\Delta T = -14 - 25 = -39^{\circ} C\). So, \(\Delta L = \alpha L_0 \Delta T = 2.5 \times 10^{-5} \times L_0 \times (-39)\).
03

Apply Hooke's Law

Hooke's law for stretching is given by \(F = Y \frac{\Delta L}{L_0} A\). Rearrange this formula to solve for Young's modulus: \(Y = \frac{F}{A \frac{\Delta L}{L_0}}\).
04

Substitute and Solve for Young's Modulus

Substitute the given values into the equation for Young's modulus: \[Y = \frac{1.2 \times 10^3}{1.6 \times 10^{-5} \times \left( \frac{2.5 \times 10^{-5} \times (-39)}{1} \right)}\]. Simplify the values to find \(Y\).
05

Compute Final Value for Young's Modulus

After substituting and simplifying, calculate the value of Young's modulus, which will provide the property of the material that stretches back the ruler to its original length.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Expansion
Thermal expansion is a fascinating concept explaining how materials change size with temperature. When a material, like metal or plastic, is heated, its molecules vibrate more and take up more space. This is why most substances grow larger when they get hotter. Conversely, when they cool down, they shrink. The physical property that describes this behavior is called the **coefficient of linear expansion**.

For example, in the case of the ruler from the problem, the initial temperature is 25°C, well within the normal room temperature range. When the temperature drops to -14°C, the ruler's length decreases. This can be calculated using the formula for linear expansion:
  • **\( \Delta L = \alpha L_0 \Delta T \)**
Here,
  • **\( \alpha \)** is the coefficient of linear expansion**
  • **\( L_0 \)** is the original length**
  • **\( \Delta T \)** is the change in temperature.**
In this scenario, as we substitute the given values, we can compute how much the ruler shrinks due to this change in temperature.
Hooke's Law
Hooke's law is a principle of physics that states the force needed to extend or compress a spring is proportional to the distance you stretch it. This concept is not just limited to springs but applies to most elastic materials to some extent. The law is usually formulated as:
  • **\( F = k \Delta x \)**
Where
  • **\( F \)** is the force applied**
  • **\( \Delta x \)** is the extension (or compression)**
  • **\( k \)** is the spring constant or stiffness**
However, for our ruler example, Hooke's law is adapted for stretching using Young's modulus, which characterizes how a material deforms under stress. The adapted formula becomes:
  • **\( F = Y \frac{\Delta L}{L_0} A \)**
Where
  • **\( Y \)** is Young's modulus which measures material stiffness**
  • **\( \Delta L \)** is the change in length**
  • **\( L_0 \)** is the original length**
  • **\( A \)** is the cross-sectional area**
By knowing how much force (\( 1.2 \times 10^3 \) N) is applied to counteract the shrinking, we can solve for Young's modulus to understand the elastic nature of the material.
Linear Expansion
Linear expansion occurs when the length of an object changes due to temperature shifts. While thermal expansion can happen three-dimensionally— affecting length, width, and height—linear expansion simplifies the concept to just one dimension along the length.

To better understand, think about how a railroad track or a metal bridge expands in the heat. Shouldn't they crack in warmer weather? Luckily, engineers design them with tiny gaps allowing room to expand and contract. The concept involves relying on the coefficient of linear expansion which differs between materials. It determines how much a material expands per degree change in temperature.
  • **Formula: \( \Delta L = \alpha L_0 \Delta T \)**
Using our previous parameters:
  • \( \alpha \)** is the coefficient of linear expansion measured in per °C.**
  • \( \Delta T \)** stands for temperature change**
  • \( L_0 \)** is the initial length**
In the exercise, the problem tasks us with calculating the length change of a ruler exposed to different temperatures. The value is pivotal as it provides insight into how much the material contracts or expands, impacting the calculation of Young's modulus and ensuring the correct measurement from a structural perspective.

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Most popular questions from this chapter

During an all-night cram session, a student heats up a one-half liter \(\left(0.50 \times 10^{-3} \mathrm{~m}^{3}\right)\) glass (Pyrex) beaker of cold coffee. Initially, the temperature is \(18^{\circ} \mathrm{C}\), and the beaker is filled to the brim. A short time later when the student retums, the temperature has risen to \(92^{\circ} \mathrm{C}\). The coefficient of volume expansion of coffee is the same as that of water. How much coffee (in cubic meters) has spilled out of the beaker?

A steel ruler is calibrated to read true at \(20.0^{\circ} \mathrm{C}\). A draftsman uses the ruler at \(40.0^{\circ} \mathrm{C}\) to draw a line on a \(40.0^{\circ} \mathrm{C}\) copper plate. As indicated on the warm ruler, the length of the line is \(0.50 \mathrm{~m}\). To what temperature should the plate be cooled, such that the length of the line truly becomes \(0.50 \mathrm{~m}\) ?

A locomotive wheel is \(1.00 \mathrm{~m}\) in diameter. A \(25.0-\mathrm{kg}\) steel band has a temperature of \(20.0^{\circ} \mathrm{C}\) and a diameter that is \(6.00 \times 10^{-4} \mathrm{~m}\) less than that of the wheel. What is the smallest mass of water vapor at \(100^{\circ} \mathrm{C}\) that can be condensed on the steel band to heat it, so that it will fit onto the wheel? Do not ignore the water that results from the condensation.

Interactive Solution \(\underline{12.61} 12.61\) at provides a model for solving problems such as this. A \(42-\mathrm{kg}\) block of ice at \(0{ }^{\circ} \mathrm{C}\) is sliding on a horizontal surface. The initial speed of the ice is \(7.3 \mathrm{~m} / \mathrm{s}\) and the final speed is \(3.5 \mathrm{~m} / \mathrm{s}\). Assume that the part of the block that melts has a very small mass and that all the heat generated by kinetic friction goes into the block of ice, and determine the mass of ice that melts into water at \(0^{\circ} \mathrm{C}\).

When you take a bath, how many kilograms of hot water \(\left(49.0^{\circ} \mathrm{C}\right)\) must you mix with cold water \(\left(13.0^{\circ} \mathrm{C}\right)\) so that the temperature of the bath is \(36.0^{\circ} \mathrm{C} ?\) The total mass of water (hot plus cold) is \(191 \mathrm{~kg}\). Ignore any heat flow between the water and its external surroundings.

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