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Chapter 11: Problem 3

ssm Accomplished silver workers in India can pound silver into incredibly thin sheets, as thin as \(3.00 \times 10^{-7} \mathrm{m}\) (about one-hundredth of the thickness a sheet of paper). Find the area of such a sheet that can be formed from \(1.00 \mathrm{kg}\) of silver.

Short Answer

Expert verified
The area of the silver sheet is approximately \(317 \text{ m}^2\).

Step by step solution

01

Determine the Density of Silver

The density of silver is approximately \(10,500 \text{ kg/m}^3\). This information will help us find the volume of the silver sheet.
02

Use the Mass-Density-Volume Relationship

The volume \(V\) of the silver can be found using the formula \(V = \frac{m}{\rho}\), where \(m\) is the mass and \(\rho\) is the density. Substituting the given values, \(V = \frac{1.00 \text{ kg}}{10,500 \text{ kg/m}^3}\).
03

Calculate the Volume of Silver

Calculating the volume, we get \(V \approx 9.52 \times 10^{-5} \text{ m}^3\).
04

Relate Volume to Sheet Area

The volume of the sheet is also given by the product of its area \(A\) and thickness \(t\), so \(V = A \times t\). We know \(t = 3.00 \times 10^{-7} \text{ m}\).
05

Solve for the Area of the Sheet

Rearrange the equation \(A = \frac{V}{t}\) to solve for \(A\). Substitute \(V = 9.52 \times 10^{-5} \text{ m}^3\) and \(t = 3.00 \times 10^{-7} \text{ m}\) to find \(A \approx 3.17 \times 10^{2} \text{ m}^2\).

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Key Concepts

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

Physics Problem Solving
When approaching physics problems, like calculating the area of a silver sheet, it's crucial to follow a structured approach. Begin by comprehending the problem statement. Understand what is given and identify the unknowns. This helps in organizing your thoughts.
In our problem, we're given:
  • The thickness of the silver sheet
  • The mass of silver
  • The density of silver
These parameters guide us in determining the area of the sheet.
Next, apply relevant physics concepts. Here, the relationship between mass, volume, and density is key. Knowing such relationships can significantly simplify complex problems. Finally, ensure each step logically leads to the next, checking calculations to avoid errors.
Volume Calculation
Volume calculation is a fundamental skill in physics. Knowing how to calculate volume helps interpret many physical scenarios.
For solids, liquids, or gases, the volume can be determined if you know the mass and density. The formula is: \[ V = \frac{m}{\rho} \] where
  • \( V \) is the volume
  • \( m \) is the mass
  • \( \rho \) is the density
In our case, given the mass of the silver as 1 kg and its density as 10,500 kg/m鲁, the volume of the silver is calculated as approximately 9.52 x 10鈦烩伒 cubic meters.
This volume allows us to understand how much space the silver occupies when compressed into a sheet.
Understanding and applying this volume calculation formula enables solving diverse physical problems effectively.
Area of a Sheet
Calculating the area of a sheet involves understanding its physical construction. In this case, the sheet is essentially a thin, flat body with given thickness.
The simple relationship guiding this calculation is: \[ A = \frac{V}{t} \] Here:
  • \( A \) is the area
  • \( V \) is the volume of the sheet
  • \( t \) is the thickness of the sheet
For our silver sheet, we calculated:
  • Volume \( V = 9.52 \times 10^{-5} \) m鲁
  • Thickness \( t = 3.00 \times 10^{-7} \) m
Using these, the area \( A \) is determined as approximately 3.17 x 10虏 m虏.
This area represents how large the sheet will be when spread out, respecting the volume and thickness constraints.

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Most popular questions from this chapter

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