Problem 5 - Physics

Question

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

Step-by-Step Solution

Answer

The area of the silver sheet is \(3.17 \times 10^{2} \text{ m}^2\).

Step 1: Understand the Problem

We need to find the area of a silver sheet that can be formed from 1.00 kg of silver, given the thickness of the sheet is \(3.00 \times 10^{-7} \text{ m}\). To do this, we will find the volume of silver and then calculate the area using the volume and thickness.

Step 2: Find the Volume of Silver

We know the mass \(m = 1.00 \text{ kg}\) and the density of silver \( \rho = 10,500 \text{ kg/m}^3\). The volume \(V\) is given by:\[V = \frac{m}{\rho} = \frac{1.00 \text{ kg}}{10,500 \text{ kg/m}^3} = 9.52 \times 10^{-5} \text{ m}^3.\]

Step 3: Calculate the Area of the Sheet

The area \(A\) of the silver sheet is found by dividing the volume by the thickness \(t = 3.00 \times 10^{-7} \text{ m}\):\[A = \frac{V}{t} = \frac{9.52 \times 10^{-5} \text{ m}^3}{3.00 \times 10^{-7} \text{ m}} = 3.17 \times 10^{2} \text{ m}^2.\]

Step 4: Verify and Conclude

We have calculated that the area of the silver sheet is \(3.17 \times 10^{2} \text{ m}^2\). Double-check calculations to ensure accuracy and verify that the steps align with the problem's requirements.

Key Concepts

Density of Silver

Density is a significant physical property of materials which helps to determine how much mass is contained in a unit volume of a substance. The density of silver is an important value in this calculation because it allows us to find out how much space a given mass of silver occupies.

• The density of silver is approximately 10,500 kg/m³.
• This means that for every cubic meter of silver, the mass is 10,500 kilograms.

In our exercise, knowing the density enables us to calculate how much volume 1 kg of silver will take up. This foundational step is crucial for understanding how we can transform the mass of silver into its corresponding volume before shaping it into a sheet.

Thickness

Thickness plays a crucial role when calculating the area of an object from its volume. The thickness of a sheet refers to how tall or deep it is when measured through its smallest dimension.

• In this problem, the thickness of the silver sheet is given as \(3.00 \times 10^{-7} \text{ m}\).
• This is extremely thin, similar to one-hundredth the thickness of a regular piece of paper.

By knowing the thickness, we can link the volume of the silver with how large a sheet can be spread out while maintaining this specific thinness. It acts as the bridge to convert from volume to a more two-dimensional area.

Volume Calculation

Understanding how to calculate volume is fundamental in physics and materials science. Volume represents the amount of three-dimensional space occupied by an object or substance.

• Given: mass \(m = 1.00 \text{ kg}\), density \(\rho = 10,500 \text{ kg/m}^3\).
• The formula to find volume \(V\) is \(V = \frac{m}{\rho}\).

With the provided density and mass, the calculation becomes:\[V = \frac{1.00 \text{ kg}}{10,500 \text{ kg/m}^3} = 9.52 \times 10^{-5} \text{ m}^3.\] This computation allows us to determine how much space the silver takes up before transforming it into any particular shape, like a sheet.

Mass of Silver

The mass is the amount of matter in an object, and when dealing with materials like silver, it’s important to understand how mass relates to other physical properties.

• In this exercise, we start with a mass of 1.00 kg of silver.
• This mass, combined with density, will help in finding the volume.

Once the volume is found, we can calculate the area of the thin sheet of silver. The reasons for knowing the mass can include determining how much silver is needed for production or understanding the constraints in creating different shapes from a particular amount of silver.