/*! 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 24 Grains of fine California beach ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 24

Grains of fine California beach sand are approximately spheres with an average radius of \(50 \mu \mathrm{m}\) and are made of silicon dioxide, which has a density of \(2600 \mathrm{~kg} / \mathrm{m}^{3} .\) What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube \(1.00 \mathrm{~m}\) on an edge?

Short Answer

Expert verified
The mass of sand grains is approximately 103.6 kg.

Step by step solution

01

Find the surface area of the cube

The cube has edges of length 1 meter. The surface area of a cube is calculated as \(6 \times \text{(side length)}^2\). Substitute the side length 1 m into the formula: \[\text{Surface Area of Cube} = 6 \times (1)^2 = 6 \ m^2.\]
02

Find the surface area of a single sand grain

The surface area of a sphere is given by the formula \(4\pi r^2\), where \(r\) is the radius of the sphere. The average radius of a sand grain is given as \(50 \ \mu m = 50 \times 10^{-6} \ m\). Substitute this into the formula: \[\text{Surface Area of a Sand Grain} = 4\pi (50 \times 10^{-6})^2.\]
03

Calculate the total number of sand grains needed

We need to find how many such sand grains we need for the total surface area to equal 6 m虏. Let \( N \) be the number of sand grains. Then, \[N \times 4\pi (50 \times 10^{-6})^2 = 6.\] Solving for \( N \): \[N = \frac{6}{4\pi (50 \times 10^{-6})^2}.\]
04

Calculate the volume of a single sand grain

The volume of a sphere is given by \(\frac{4}{3}\pi r^3\). Substitute the radius of a sand grain, \(50 \times 10^{-6} \ m\), into the formula: \[\text{Volume of a Sand Grain} = \frac{4}{3}\pi (50 \times 10^{-6})^3.\]
05

Calculate the total mass of the sand grains

The mass of an object can be found by multiplying its volume by its density. Let \(V_\text{grain}\) be the volume of one sand grain and \(\rho\) be its density (2600 kg/m^3). The mass of one grain is \(\rho \times V_\text{grain}\). Use the number of grains \(N\) from Step 3 to find the total mass: \[\text{Total Mass} = N \times 2600 \times V_\text{grain}.\]
06

Final Step: Perform Numerical Calculations

Now we need to perform the calculations using the values from the previous steps: 1. Calculate the number of sand grains \( N \) using the formula in Step 3.2. Calculate the volume of one sand grain from Step 4.3. Multiply by the density to get the mass from Step 5. Find: \( \text{Total Mass} = N \times 2600 \times V_\text{grain} \approx 103.6 \text{ kg}. \)

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.

Understanding Spherical Objects
Spheres are fascinating shapes that are defined by their perfectly round form. Every point on the surface of a sphere is equidistant from its center, which gives it its distinctive shape. Spherical objects, like sand grains mentioned in this exercise, often appear in various forms, such as planets, bubbles, or even sports balls.

To analyze such spherical objects, certain geometric properties are crucial. The surface area and volume are two vital characteristics that describe a sphere. The surface area explains how much space is covered by the outer part of the sphere. In mathematics, it is calculated using the formula \[4 \pi r^2\], where \( r \) is the radius.

Additionally, knowing the volume, which measures how much space is contained within the sphere, is vital for calculating the mass and other properties of spherical objects. The volume of a sphere can be determined using the formula \[\frac{4}{3} \pi r^3\]. These formulas help explore and calculate essential features of spherical objects, valuable for both scientific studies and practical applications.
Calculation of Volume and Mass
Calculating the volume and mass of spherical objects is a common task in mathematics and physics. Whether you are figuring out the mass of sand grains or determining the volume of larger objects, the precise formulas ensure accurate calculations.

For a sphere, start with finding its volume using the formula \(\frac{4}{3} \pi r^3\). In the case of sand grains with a radius of \(50 \mu m\), convert this to meters for consistency in calculations. This translates to \(50 \times 10^{-6} \ m\). Applying this radius into the formula yields the volume for a single grain.

Once the volume is determined, calculating the mass becomes straightforward. Multiply the volume by the density of the material. Silicon dioxide, in this example, has a density of \(2600 \ kg/m^3\). For a collective mass, as required by the exercise, multiply the mass of one grain by the total number of grains to find the overall mass.
Exploring Silicon Dioxide Density
Silicon dioxide, or \(SiO_2\), is a common compound that contributes to the structure of sand grains. Its density is an important factor in determining the mass of spherical objects composed of this material. A substance's density is a measure of its mass per unit volume, typically expressed in \(kg/m^3\).

For silicon dioxide, the density is approximately \(2600 \ kg/m^3\). This high density is indicative of silicon dioxide's solid, compact nature. Understanding the density allows for accurate mass calculations when dealing with objects like sand grains, because the mass depends on both the volume and the density.

In practical applications, knowledge of silicon dioxide density aids in estimating the weight of sand or other \(SiO_2\)-based materials. This is essential not only in academic exercises but also in industries ranging from construction to technology, where precise calculations of material quantities are required for successful project execution.

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

As a contrast between the old and the modern and between the large and the small, consider the following: In old rural England 1 hide (between 100 and 120 acres) was the area of land needed to sustain one family with a single plough for one year. (An area of 1 acre is equal to \(4047 \mathrm{~m}^{2} .\) ) Also, 1 wapentake was the area of land needed by 100 such families. In quantum physics, the cross-sectional area of a nucleus (defined in terms of the chance of a particle hitting and being absorbed by it ) is measured in units of barns, where 1 barn is \(1 \times 10^{-28} \mathrm{~m}^{2}\). (In nuclear physics jargon, if a nucleus is "large," then shooting a particle at it is like shooting a bullet at a barn door, which can hardly be missed.) What is the ratio of 25 wapentakes to 11 barns?

Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height \(H=1.70 \mathrm{~m},\) and stop the watch when the top of the Sun again disappears. If the elapsed time is \(t=11.1 \mathrm{~s},\) what is the radius \(r\) of Earth?

A ton is a measure of volume frequently used in shipping, but that use requires some care because there are at least three types of tons: A displacement ton is equal to 7 barrels bulk, a freight ton is equal to 8 barrels bulk, and a register ton is equal to 20 barrels bulk. A barrel bulk is another measure of volume: 1 barrel bulk \(=0.1415 \mathrm{~m}^{3} .\) Suppose you spot a shipping order for "73 tons" of M\&M candies, and you are certain that the client who sent the order intended "ton" to refer to volume (instead of weight or mass, as discussed in Chapter 5 ). If the client actually meant displacement tons, how many extra U.S. bushels of the candies will you erroneously ship if you interpret the order as (a) 73 freight tons and (b) 73 register tons? \(\left(1 \mathrm{~m}^{3}=28.378\right.\) U.S. bushels. \()\)

The cubit is an ancient unit of length based on the distance between the elbow and the tip of the middle finger of the measurer. Assume that the distance ranged from 43 to \(53 \mathrm{~cm},\) and suppose that ancient drawings indicate that a cylindrical pillar was to have a length of 9 cubits and a diameter of 2 cubits. For the stated range, what are the lower value and the upper value, respectively, for (a) the cylinder's length in meters, (b) the cylinder's length in millimeters, and (c) the cylinder's volume in cubic meters?

Gold, which has a density of \(19.32 \mathrm{~g} / \mathrm{cm}^{3},\) is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If a sample of gold, with a mass of \(27.63 \mathrm{~g},\) is pressed into a leaf of \(1.000 \mu \mathrm{m}\) thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius \(2.500 \mu \mathrm{m},\) what is the length of the fiber?
See all solutions

Recommended explanations on Physics Textbooks

Quantum Physics

Read Explanation

Electrostatics

Read Explanation

Fundamentals of Physics

Read Explanation

Nuclear Physics

Read Explanation

Astrophysics

Read Explanation

Modern Physics

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.