/*! 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 4 If an object has a constant dens... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 4

If an object has a constant density \(\delta\) and a volume \(V\), what is its mass?

Short Answer

Expert verified
Mass is calculated by \( m = \delta \times V \).

Step by step solution

01

Understand the Given Information

We know that density, denoted here as \( \delta \), is a measure of how much mass there is per unit volume for an object. The object has a given volume \( V \). Our task is to find the mass given these two values.
02

Use the Formula for Density

The formula relating density \( \delta \), mass \( m \), and volume \( V \) is \( \delta = \frac{m}{V} \). This equation states that density is equal to mass divided by volume.
03

Rearrange the Formula to Find Mass

To find the mass \( m \), rearrange the formula to solve for \( m \). Multiply both sides of \( \delta = \frac{m}{V} \) by \( V \) to isolate \( m \): \( m = \delta \times V \).
04

Substitute the Values

Substitute the given values of density \( \delta \) and volume \( V \) into the formula \( m = \delta \times V \) to calculate the mass. Ensure all units are consistent before performing the calculation.

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
Density is a fundamental concept in understanding how matter is packed into a given space. It's like figuring out how tightly packed a crowd is in a room. When we talk about density, we refer to the relationship between mass and volume. In more straightforward terms, density tells us how much stuff is crammed into a certain amount of space. A denser object has more mass in the same volume compared to a less dense one. For example:
  • A brick and a sponge of the same size - the brick is denser because it is heavier for its size.
  • Oil and water - Oil is less dense than water; that鈥檚 why it floats on top.
Understanding density is essential in various fields, from physics to engineering, as it helps in predicting how substances will behave under different conditions.
Volume
Volume refers to the amount of space an object takes up. Imagine filling a container with water and knowing how much water it holds 鈥� that's volume. It's measured in units like liters or cubic meters. The concept of volume is everywhere around us. Think about:
  • The amount of liquid your bottle can contain.
  • The space inside a box.
  • The area your bed occupies in your room.
To calculate the volume of regular shapes, like cubes or spheres, we use simple formulas. For example: - For a cube: - Volume = side length 脳 side length 脳 side length, or - Volume = length鲁. Understanding volume helps in real-life as it contributes to planning activities like packing, construction, or even cooking.
Formula for Density
The formula that connects mass, density, and volume provides a clear pathway to solving many practical problems in physics and engineering. The core formula is written as \( \delta = \frac{m}{V} \), where:
  • \( \delta \) is the density (mass per unit volume),
  • \( m \) is the mass, and
  • \( V \) is the volume.
If you need to find the mass, the formula can be rearranged to \( m = \delta \times V \). This rearrangement is essential when you know the density and volume but need the mass.By plugging the known values into the formula, you can easily compute the unknown variable. This process is especially useful in experiments, quality control in manufacturing, and in designing materials with specific properties.

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

Find the center of mass of the lamina described by the region \(R\) in the plane and its density function \(\delta(x, y)\) Note: these are the same lamina as in Exercises \(11-18\). \(R\) is the triangle with corners \((-1,0),(1,0),\) and (0,1) \(\delta(x, y)=2 \mathrm{lb} / \mathrm{in}^{2}\)

A solid is described along with its density function. Find the mass of the solid using cylindrical coordinates. Bounded by the cylinders \(x^{2}+y^{2}=4\) and \(x^{2}+y^{2}=9,\) between the planes \(z=0\) and \(z=10\) with density function \(\delta(x, y, z)=z\).

Set up the triple integrals that give the volume of \(D\) in all 6 orders of integration, and find the volume of \(D\) by evaluating the indicated triple integral. \(D\) is bounded by the planes \(y=0, y=2, x=1, z=0\) and \(z=(3-x) / 2\) Evaluate the triple integral with order \(d x d y d z\).

Two surfaces \(f_{1}(x, y)\) and \(f_{2}(x, y)\) and a region \(R\) in the \(x, y\) plane are given. Set up and evaluate the double integral that finds the volume between these surfaces over \(R\). \(f_{1}(x, y)=x^{2}+y^{2}, f_{2}(x, y)=-x^{2}-y^{2}\); \(R\) is the square with corners (0,0) and (2,3) .

Give an informal interpretation of what \(" \iiint_{D} d V^{\prime \prime}\) means.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

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.