/*! 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 5 Accomplished silver workers in I... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 5

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.

Short Answer

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

Step by step solution

01

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.
02

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:\[V = \frac{m}{\rho} = \frac{1.00 \text{ kg}}{10,500 \text{ kg/m}^3} = 9.52 \times 10^{-5} \text{ m}^3.\]
03

Calculate the Area of the Sheet

The area \(A\) of the silver sheet is found by dividing the volume by the thickness \(t = 3.00 \times 10^{-7} \text{ m}\):\[A = \frac{V}{t} = \frac{9.52 \times 10^{-5} \text{ m}^3}{3.00 \times 10^{-7} \text{ m}} = 3.17 \times 10^{2} \text{ m}^2.\]
04

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

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/m鲁.
  • 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鈥檚 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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Mariana trench is located in the floor of the Pacific Ocean at a depth of about 11000 \(\mathrm{m}\) below the surface of the water. The density of seawater is \(1025 \mathrm{~kg} / \mathrm{m}^{3}\). (a) If an underwater vehicle were to explore such a depth, what force would the water exert on the vehicle's observation window (radius \(=0.10 \mathrm{~m}\) )? (b) For comparison, determine the weight of a jetliner whose mass is \(1.2 \times 10^{5} \mathrm{~kg}\).

A suitcase (mass \(m=16 \mathrm{~kg}\) ) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures \(0.50 \mathrm{~m}\) by \(0.15 \mathrm{~m} .\) The elevator is moving upward, the magnitude of its acceleration being \(1.5 \mathrm{~m} / \mathrm{s}^{2}\). What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?

Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is closed. Both containers are filled initially to the same height of \(1.00 \mathrm{~m}\), one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.

One way to administer an inoculation is with a "gun" that shoots the vaccine through a narrow opening. No needle is necessary, for the vaccine emerges with sufficient speed to pass directly into the tissue beneath the skin. The speed is high, because the vaccine \(\left(\rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\right)\) is held in a reservoir where a high pressure pushes it out. The pressure on the surface of the vaccine in one gun is \(4.1 \times 10^{6} \mathrm{~Pa}\) above the atmospheric pressure outside the narrow opening. The dosage is small enough that the vaccine's surface in the reservoir is nearly stationary during an inoculation. The vertical height between the vaccine's surface in the reservoir and the opening can be ignored. Find the speed at which the vaccine emerges.

A room has a volume of \(120 \mathrm{~m}^{3}\). An air-conditioning system is to replace the air in this room every twenty minutes, using ducts that have a square cross section. Assuming that air can be treated as an incompressible fluid, find the length of a side of the square if the air speed within the ducts is (a) \(3.0 \mathrm{~m} / \mathrm{s}\) and \((\mathrm{b}) 5.0 \mathrm{~m} / \mathrm{s}\).
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Modern Physics

Read Explanation

Waves Physics

Read Explanation

Measurements

Read Explanation

Mechanics and Materials

Read Explanation

Electricity and Magnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.