/*! 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 59 A piece of wood is \(0.600 \math... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 59

A piece of wood is \(0.600 \mathrm{~m}\) long, \(0.250 \mathrm{~m}\) wide, and \(0.080 \mathrm{~m}\) thick. Its density is \(600 \mathrm{~kg} / \mathrm{m}^{3}\). What volume of lead must be fastened underneath it to sink the wood in calm water so that its top is just even with the water level? What is the mass of this volume of lead?

Short Answer

Expert verified
The required volume of lead is \(0.000423 \text{ m}^3\) and its mass is \(4.8 \text{ kg}\).

Step by step solution

01

Calculate the Volume of Wood

The volume of the piece of wood can be found using the formula for the volume of a rectangular prism: \( V = ext{length} \times ext{width} \times ext{thickness} \). Therefore, we calculate: \( V_{ ext{wood}} = 0.600 \, \text{m} \times 0.250 \, \text{m} \times 0.080 \, \text{m} = 0.012 \, \text{m}^3 \).
02

Calculate the Mass of Wood

The mass of the wood is calculated using its volume and density: \( m = \rho \times V \). The density is \( 600 \, \text{kg/m}^3 \), so the mass of the wood is: \( m_{ ext{wood}} = 600 \, \text{kg/m}^3 \times 0.012 \, \text{m}^3 = 7.2 \, \text{kg} \).
03

Determine the Buoyant Force Exerted by Water

To float just at the water surface, the buoyant force on the wood plus lead must equal the total weight of the wood and lead. The buoyant force is equal to the weight of the water displaced. The volume of water displaced is equal to the volume of the wood, so the buoyant force is \( F_b = V_{ ext{wood}} \times \rho_{ ext{water}} \times g = 0.012 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 = 117.72 \, \text{N} \).
04

Find the Required Additional Mass of Lead

The buoyant force must equal the total weight of wood plus lead to keep the wood just even with the water level. Since \( F_b = 117.72 \, \text{N} \) and the weight of the wood is \( m_{ ext{wood}} \times g = 7.2 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 70.632 \, \text{N} \), the weight of the lead must be \( 117.72 \, \text{N} - 70.632 \, \text{N} = 47.088 \, \text{N} \).
05

Calculate the Mass of the Lead

The mass of the lead is calculated from its weight \( W = m \times g \). Thus, \( m_{ ext{lead}} = \frac{47.088 \, \text{N}}{9.81 \, \text{m/s}^2} = 4.8 \, \text{kg} \).
06

Find the Volume of Lead Needed

Use the density of lead \( \rho_{\text{lead}} = 11340 \, \text{kg/m}^3 \) to find the volume of lead: \( V = \frac{m}{\rho} \). Therefore, \( V_{\text{lead}} = \frac{4.8 \, \text{kg}}{11340 \, \text{kg/m}^3} = 0.000423 \text{ m}^3 \).

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 Calculation
Density is a fundamental concept in physics that describes how much mass is contained within a given volume. It is calculated using the formula: \( \rho = \frac{m}{V} \), where \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume. In our exercise, we calculated the density of wood, which was given as \( 600 \, \text{kg/m}^3 \). This tells us how tightly packed the wood's molecules are compared to the amount of space they occupy. Understanding density helps in comparing how solid substances are in different objects. Higher density means more mass in a smaller space, which affects how materials behave in different conditions like floating or sinking.
Buoyant Force
Buoyant force is a force exerted by a fluid on an object placed in it. This force acts in the upward direction, counteracting the weight of the object and enabling it to float. According to Archimedes' Principle, the buoyant force on an object is equal to the weight of the fluid it displaces. The formula is straightforward: \( F_b = \rho_{\text{water}} \times V_{\text{displaced}} \times g \). In our wood and lead problem, the buoyant force needed to submerge the wood just to the water surface was calculated as \( 117.72 \, \text{N} \). This force must balance the total weight of the wood and the lead. Remembering this principle is critical in solving problems related to objects submerged in liquids.
Mass and Volume Relationship
The relationship between mass and volume is directly linked to density. When you know two of these values (mass, volume, or density), you can easily find the third. For example, you can calculate the mass by multiplying density (\( \rho \)) by volume (\( V \)): \[ m = \rho \times V \] This equation helps in determining the amount of matter present within a certain space.In the given exercise, we used this relationship to find the mass of the wood to be \( 7.2 \, \text{kg} \) by using its density and volume. We also used it to determine the mass of lead needed to support the wood's weight in water. Understanding this relationship helps in identifying how changes in the composition or structure of an object might affect its behavior in fluid dynamics.
Floating and Sinking Principles
The principles of floating and sinking are essential in understanding why objects behave differently when placed in fluid. An object floats if its density is less than the fluid it's in. Conversely, it sinks if its density is more. This is often guided by comparing the object's density to the fluid's density. For the wood initially floating and later sinking when lead is added, we balance densities using mass and volume calculations. When trying to sink the wood, the density of the wood and the lead together must overcome the buoyant force by displacing enough water. In practical terms, the key to floating or sinking lies in understanding how to manipulate these variables to achieve the desired result. This exercise demonstrates that altering the mass (by adding lead) changes the overall density to sink the wood into water.

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 piston of a hydraulic automobile lift is \(0.30 \mathrm{~m}\) in diameter. What gauge pressure, in pascals, is required to lift a car with a mass of \(1200 \mathrm{~kg}\) ? Now express this pressure in atmospheres.

Water is flowing in a cylindrical pipe of varying circular crosssectional area, and at all points the water completely fills the pipe. (a) At one point in the pipe, the radius is \(0.150 \mathrm{~m}\). What is the speed of the water at this point if the volume flow rate in the pipe is \(1.20 \mathrm{~m}^{3} / \mathrm{s} ?\) (b) At a second point in the pipe, the water speed is \(3.80 \mathrm{~m} / \mathrm{s}\). What is the radius of the pipe at this point?

Elephants under pressure. An elephant can swim or walk with its chest several meters underwater while the animal breathes through its trunk, which remains above the water surface and acts like a snorkel. The elephant's tissues are at an increased pressure due to the surrounding water, but the lungs are at atmospheric pressure because they are connected to the air through the trunk. Figure 13.47 shows the gauge pressures in an elephant's lungs and abdomen when the elephant's chest is submerged to a particular depth in a lake. In this situation, the elephant's diaphragm, which separates the lungs from the abdomen, must maintain the difference in pressure between the lungs and the abdomen. The diaphragm of an elephant is typically \(3.0 \mathrm{~cm}\) thick and \(120 \mathrm{~cm}\) in diameter. If the elephant were to snorkel in saltwater, which is more dense than freshwater, would the maximum depth at which it could snorkel be different from the depth in freshwater? A. Yes; that depth would increase because there is less pressure at a given depth in saltwater than in freshwater. B. Yes; that depth would decrease because there is greater pressure at a given depth in saltwater than in freshwater. C. No, because pressure differences within the submerged elephant depend only on the density of air, not on the density of the water. D. No, because the buoyant force on the elephant would be the same in both cases.

Glaucoma. Under normal circumstances, the vitreous humor, a jelly-like substance in the main part of the eye, exerts a pressure of up to \(24 \mathrm{~mm}\) of mercury that maintains the shape of the eye. If blockage of the drainage duct for aqueous humor causes this pressure to increase to about \(50 \mathrm{~mm}\) of mercury, the condition is called glaucoma. What is the increase in the total force (in newtons) on the walls of the eye if the pressure increases from \(24 \mathrm{~mm}\) to \(50 \mathrm{~mm}\) of mercury? We can quite accurately model the eye as a sphere \(2.5 \mathrm{~cm}\) in diameter.

A block of wood has a density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\). It is placed in a fluid in which it floats with two-thirds of its volume submerged. What is the density of the fluid?
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Wave Optics

Read Explanation

Rotational Dynamics

Read Explanation

Fields in Physics

Read Explanation

Space Physics

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.