This document contains the chapter outline and answers to questions for a physics and measurement chapter. The chapter outline lists the main topics covered, including standards of length, mass and time, dimensional analysis, and significant figures. The answers to questions section provides worked solutions to sample problems related to these topics, such as density calculations, unit conversions, and dimensional analysis questions.
This document contains a chapter outline and solutions to problems for a physics textbook. The chapter outline lists topics such as standards of length, mass and time, matter and model-building, density and atomic mass. The solutions provide worked examples and calculations for problems related to these topics, such as calculating density using mass and volume, dimensional analysis, and unit conversions.
This document contains a chapter outline and answers to questions about physics and measurement. The chapter outline lists topics like standards of length, mass and time, density and atomic mass, and dimensional analysis. The answers to questions section provides explanations and calculations in response to multiple choice and free response questions about these topics. For example, it explains why atomic clocks and pulsars can serve as highly accurate time standards, and it calculates densities, masses, numbers of atoms, and rates of change using the relevant physics equations and units.
This document contains physics questions and answers related to mechanics. Some key points:
- Question 11 discusses calculating the percent uncertainty in the volume of a spherical beach ball.
- Question 19 expresses a sum with the correct number of significant figures.
- Question 51 calculates the volume of the Moon and determines how many Moons would be needed to equal Earth's volume.
- Other questions cover topics like calculating components of velocity, displacement and time calculations for motion problems, projectile motion, and tensions in hanging objects. The document provides worked examples of physics equations and calculations.
This document discusses elasticity and harmonic motion. It analyzes how forces affect the elastic properties of materials and the nature of oscillatory motion. Simple harmonic motion is described as motion where the restoring force is proportional to displacement from equilibrium. The period of oscillation depends on the mass and spring constant. Examples are given of using harmonic motion to measure an astronaut's mass in orbit by having them sit in a spring-mounted chair and observing the period of oscillation.
This document contains the chapter outline and answers to questions for a physics and measurement chapter. The chapter outline lists the main topics covered, including standards of length, mass and time, dimensional analysis, and significant figures. The answers to questions section provides worked solutions to sample problems related to these topics, such as density calculations, unit conversions, and dimensional analysis questions.
This document contains answers to multiple choice and calculation questions about physics concepts. Key details include:
- Density varies with temperature and pressure and requires accurate measurement of mass and volume.
- Atomic clocks use electromagnetic waves emitted by atoms, and pulsars are also highly precise astronomical clocks.
- Different crystal structures result from variations in electron configurations of different elements.
- Vector displacement can equal zero while distance is a non-zero scalar quantity.
- Conversions between units include miles, feet, gallons, liters, meters, centimeters, nanoseconds, days, and years.
The document provides data and equations to solve several physics problems related to fluid dynamics and material properties. It gives the dimensions of a room and calculates the mass of air in the room. It also gives properties of nitrogen gas and calculates the mass of nitrogen in a storage tank. Finally, it provides equations and properties to calculate the maximum speed of a small aluminum sphere falling through air, and the time to reach 95% of this speed.
The document provides data and equations to solve several physics problems related to fluid dynamics and material properties. It gives the dimensions of a room and calculates the mass of air in the room. It also gives properties of nitrogen gas and calculates the mass of nitrogen in a storage tank. Finally, it provides equations and properties to calculate the maximum speed of a small aluminum sphere falling through air, and the time to reach 95% of this maximum speed.
This document contains a chapter outline and solutions to problems for a physics textbook. The chapter outline lists topics such as standards of length, mass and time, matter and model-building, density and atomic mass. The solutions provide worked examples and calculations for problems related to these topics, such as calculating density using mass and volume, dimensional analysis, and unit conversions.
This document contains a chapter outline and answers to questions about physics and measurement. The chapter outline lists topics like standards of length, mass and time, density and atomic mass, and dimensional analysis. The answers to questions section provides explanations and calculations in response to multiple choice and free response questions about these topics. For example, it explains why atomic clocks and pulsars can serve as highly accurate time standards, and it calculates densities, masses, numbers of atoms, and rates of change using the relevant physics equations and units.
This document contains physics questions and answers related to mechanics. Some key points:
- Question 11 discusses calculating the percent uncertainty in the volume of a spherical beach ball.
- Question 19 expresses a sum with the correct number of significant figures.
- Question 51 calculates the volume of the Moon and determines how many Moons would be needed to equal Earth's volume.
- Other questions cover topics like calculating components of velocity, displacement and time calculations for motion problems, projectile motion, and tensions in hanging objects. The document provides worked examples of physics equations and calculations.
This document discusses elasticity and harmonic motion. It analyzes how forces affect the elastic properties of materials and the nature of oscillatory motion. Simple harmonic motion is described as motion where the restoring force is proportional to displacement from equilibrium. The period of oscillation depends on the mass and spring constant. Examples are given of using harmonic motion to measure an astronaut's mass in orbit by having them sit in a spring-mounted chair and observing the period of oscillation.
This document contains the chapter outline and answers to questions for a physics and measurement chapter. The chapter outline lists the main topics covered, including standards of length, mass and time, dimensional analysis, and significant figures. The answers to questions section provides worked solutions to sample problems related to these topics, such as density calculations, unit conversions, and dimensional analysis questions.
This document contains answers to multiple choice and calculation questions about physics concepts. Key details include:
- Density varies with temperature and pressure and requires accurate measurement of mass and volume.
- Atomic clocks use electromagnetic waves emitted by atoms, and pulsars are also highly precise astronomical clocks.
- Different crystal structures result from variations in electron configurations of different elements.
- Vector displacement can equal zero while distance is a non-zero scalar quantity.
- Conversions between units include miles, feet, gallons, liters, meters, centimeters, nanoseconds, days, and years.
The document provides data and equations to solve several physics problems related to fluid dynamics and material properties. It gives the dimensions of a room and calculates the mass of air in the room. It also gives properties of nitrogen gas and calculates the mass of nitrogen in a storage tank. Finally, it provides equations and properties to calculate the maximum speed of a small aluminum sphere falling through air, and the time to reach 95% of this speed.
The document provides data and equations to solve several physics problems related to fluid dynamics and material properties. It gives the dimensions of a room and calculates the mass of air in the room. It also gives properties of nitrogen gas and calculates the mass of nitrogen in a storage tank. Finally, it provides equations and properties to calculate the maximum speed of a small aluminum sphere falling through air, and the time to reach 95% of this maximum speed.
This document contains examples and calculations related to statistics, physics, and engineering. It includes:
1) Calculations of distances, speeds, volumes, densities, and other physical quantities.
2) Examples of statistical analysis such as calculating means, standard deviations, and control limits from data sets.
3) Physics problems involving concepts like force, weight, pressure, and fluid dynamics.
This chapter discusses various physics concepts including:
1) Conversions between different units of time, distance, and speed.
2) Calculating the number of steps from Earth to a nearby star and the number of reports needed to describe the distance to the moon.
3) Determining the number of planes needed based on fuel consumption rates and crude oil production.
4) Calculating forces, weights, and densities in various physics problems.
1) The experiment investigated the relationship between the period of a simple harmonic oscillation and the mass of the oscillating system using a hacksaw blade.
2) Mass was added to the end of the blade in increments and the period was measured from graphs of the oscillation.
3) The results showed a linear relationship between the period squared and the mass, following the predicted relationship but with a y-intercept due to the non-zero mass of the blade.
This document provides an overview of key topics in physics and physical measurements. It discusses:
- The goals and scope of physics as the study of natural phenomena and how nature works.
- The importance of measurements in physics for making quantitative comparisons between laws and the natural world, and the International System of Units (SI) for measurements.
- Common physical quantities like length, mass, time, temperature and their typical orders of magnitude.
Numerical problem pile capacity (usefulsearch.org) (useful search)Make Mannan
A 10m long concrete pile with a square cross section of 450x450mm driven into clay with an undrained cohesion of 35 kPa has an ultimate load capacity of 600 kN. If the cross section is reduced to 250x250mm and length increased to 20m, the ultimate load capacity will be 614.6 kN.
A group of 16 piles 10m long and 0.8m in diameter in a 12m thick clay layer has a base resistance of 452.16 kN for a single pile and a group side resistance of 14,469.12 kN assuming 100% efficiency.
The load carrying capacity of a concrete pile driven into sand using a 4 ton hammer with
The document presents a study that uses the non-linear Gauss-Newton technique to interpret spherical gravity anomaly data. It generates synthetic gravity data for a spherical ore body and adds 10% noise. It then applies the Gauss-Newton method to extract parameters of the ore body, like depth and radius, from the synthetic data. For a field gravity dataset over a salt dome, it initially estimates the depth and radius using direct interpretation methods. It then applies the Gauss-Newton method to the field data, improving the estimates of the depth and radius. The study examines the performance of the Gauss-Newton method on both synthetic and field gravity data for a spherical source body.
1) The document provides conceptual problems and their solutions related to measurement and vectors in physics. It covers topics like SI base units, significant figures, and vector operations.
2) Multiple choice questions test the understanding of concepts like unit conversions, vector directions, and approximations using dimensional analysis. Diagrammatic representations are provided for vector addition problems.
3) An example environmental debate question is included that involves estimating the number of disposable diapers used in the US annually based on population data, then calculating the resulting landfill volume and area requirements.
The document uses equations to calculate the thickness of a layer of seawater gaining heat at a rate of 1.5 x 10-7Ks-1. It determines the layer is 2.5 meters thick. It then calculates the temperature change of the top 1mm of ocean gaining 4 x 10-3 Wm-2 of heat over 30 days, finding the temperature change to be 0.025K.
This document summarizes the derivation of the 2-dimensional wave equation to model the propagation of a disturbance through a thin stretched membrane. It shows that the wave equation can be derived from the Lagrangian and Hamilton's principle. Initial and boundary conditions are prescribed, and the wave equation is solved using separation of variables to obtain a double Fourier series solution representing the membrane displacement as a function of space and time.
Modern physics paul a. tipler 6ª edição solutio manualIzabela Ferreira
This document is the preface to an instructor solutions manual for the problems in the textbook "Modern Physics, Sixth Edition" by Paul A. Tipler and Ralph A. Llewellyn. It contains solutions to every problem in the textbook and is intended for instructors, not distribution to students. It was prepared by Mark J. Llewellyn and includes an introduction and table of contents organizing the solutions by chapter. The preface explains the purpose and contents of the manual and provides contact information for the author.
Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solut...KirkMcdowells
Full download : https://alibabadownload.com/product/fox-and-mcdonalds-introduction-to-fluid-mechanics-9th-edition-pritchard-solutions-manual/ Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solutions Manual
1) The document describes a numerical simulation of the spherical collapse of dark matter perturbations in an expanding universe. It simulates both cold dark matter (CDM) and warm dark matter (WDM) cases.
2) For CDM, the simulation shows collapse times depend on the initial overdensity but are approximately symmetric around the turnaround time. Regions of all masses collapse as long as the initial overdensity exceeds a threshold.
3) For WDM, the simulation adds a pressure term to account for the thermal velocity of WDM in the early universe. This term slows collapse compared to CDM and can potentially prevent collapse of low-mass regions.
This document summarizes a study that uses a non-linear Gauss-Newton technique to interpret spherical gravity anomalies. It generates synthetic gravity data for a spherical ore body and applies the Gauss-Newton method to extract parameters like depth and dimension. When applied to noise-free data, the method accurately resolves the parameters in 22 iterations with low error. Adding 10% noise increases the error by 20% but parameters are still estimated reasonably well. The method is also applied to interpret a published field gravity dataset from Texas.
Serway, raymond a physics for scientists and engineers (6e) solutionsTatiani Andressa
This document contains a chapter outline and sample questions and solutions for a physics and measurement chapter. The chapter outline lists topics like standards of length, mass and time, density and atomic mass, and dimensional analysis. The questions and solutions provide examples of calculations involving converting between units, determining densities, and applying dimensional analysis.
This document contains a chapter outline and answers to questions about physics and measurement. The chapter outline lists topics like standards of length, mass and time, matter and model-building, density and atomic mass, and dimensional analysis. The answers to questions section provides explanations and calculations in response to questions about these topics. For example, it explains that atomic clocks are based on electromagnetic waves emitted by atoms, and that density varies with temperature and pressure.
1. The document provides examples of converting between different units of measurement using metric prefixes like micro, pico, and nano. Conversions include kilometers to microns, centimeters to microns, and yards to microns.
2. Additional examples convert between inches and picas, points, and other units like grys and almudes which are used to measure volumes of grains.
3. Conversion factors are also provided for seconds, days, weeks, centuries and other units of time. Examples show converting between these different units of time.
1. The document provides examples of unit conversions using metric prefixes and other conversion factors. Calculations are shown for distances, areas, volumes, and time conversions between various units like kilometers, meters, centimeters, microseconds, seconds and years.
2. Geometric formulas are used to calculate properties of the Earth like circumference, surface area and volume. Examples of unit conversions involving length, area, volume, time and other units are also provided.
3. Examples involve calculating distances, areas, volumes, rotations, uncertainties and time differences between various measurement systems including metric, English, and other historical units. Careful application of conversion factors and significant figures is emphasized.
1. The document provides examples of converting between different units of measurement, such as miles to kilometers, gallons to liters, seconds to years, etc. Conversions involve setting up proportional relationships between units and calculating using conversion factors.
2. Examples show calculating uncertainties in measurements based on the precision of the measuring instrument. Uncertainty is estimated as a percentage error relative to the measured quantity.
3. To determine average values and uncertainties, the document uses the maximum and minimum possible values based on the precision of the original measurements. The uncertainty is taken as half the range between the maximum and minimum average values.
1. The document provides examples of converting between different units of measurement, such as miles to kilometers, gallons to liters, seconds to years, etc. Conversions involve setting up proportional relationships between units and calculating using conversion factors.
2. Examples show calculating uncertainties in measurements based on the precision of the measuring instrument. Uncertainty is estimated as a percentage error relative to the measured quantity.
3. To determine average values and uncertainties, the document uses the maximum and minimum possible values based on the precision of the original measurements. The uncertainty is taken as half the range between the maximum and minimum average values.
1. This document contains conceptual problems and their solutions related to measurement and vectors in physics. It covers topics like SI base units, significant figures, dimensions of physical quantities, and vector operations.
2. Conceptual questions are answered by determining the relevant concepts, relationships between quantities, and applying definitions. Numerical problems are solved by substituting values into appropriate equations.
3. Estimation problems involve approximating quantities using given values and reasonable assumptions. Diagrams are used to illustrate vector concepts and solutions.
This document contains examples and calculations related to statistics, physics, and engineering. It includes:
1) Calculations of distances, speeds, volumes, densities, and other physical quantities.
2) Examples of statistical analysis such as calculating means, standard deviations, and control limits from data sets.
3) Physics problems involving concepts like force, weight, pressure, and fluid dynamics.
This chapter discusses various physics concepts including:
1) Conversions between different units of time, distance, and speed.
2) Calculating the number of steps from Earth to a nearby star and the number of reports needed to describe the distance to the moon.
3) Determining the number of planes needed based on fuel consumption rates and crude oil production.
4) Calculating forces, weights, and densities in various physics problems.
1) The experiment investigated the relationship between the period of a simple harmonic oscillation and the mass of the oscillating system using a hacksaw blade.
2) Mass was added to the end of the blade in increments and the period was measured from graphs of the oscillation.
3) The results showed a linear relationship between the period squared and the mass, following the predicted relationship but with a y-intercept due to the non-zero mass of the blade.
This document provides an overview of key topics in physics and physical measurements. It discusses:
- The goals and scope of physics as the study of natural phenomena and how nature works.
- The importance of measurements in physics for making quantitative comparisons between laws and the natural world, and the International System of Units (SI) for measurements.
- Common physical quantities like length, mass, time, temperature and their typical orders of magnitude.
Numerical problem pile capacity (usefulsearch.org) (useful search)Make Mannan
A 10m long concrete pile with a square cross section of 450x450mm driven into clay with an undrained cohesion of 35 kPa has an ultimate load capacity of 600 kN. If the cross section is reduced to 250x250mm and length increased to 20m, the ultimate load capacity will be 614.6 kN.
A group of 16 piles 10m long and 0.8m in diameter in a 12m thick clay layer has a base resistance of 452.16 kN for a single pile and a group side resistance of 14,469.12 kN assuming 100% efficiency.
The load carrying capacity of a concrete pile driven into sand using a 4 ton hammer with
The document presents a study that uses the non-linear Gauss-Newton technique to interpret spherical gravity anomaly data. It generates synthetic gravity data for a spherical ore body and adds 10% noise. It then applies the Gauss-Newton method to extract parameters of the ore body, like depth and radius, from the synthetic data. For a field gravity dataset over a salt dome, it initially estimates the depth and radius using direct interpretation methods. It then applies the Gauss-Newton method to the field data, improving the estimates of the depth and radius. The study examines the performance of the Gauss-Newton method on both synthetic and field gravity data for a spherical source body.
1) The document provides conceptual problems and their solutions related to measurement and vectors in physics. It covers topics like SI base units, significant figures, and vector operations.
2) Multiple choice questions test the understanding of concepts like unit conversions, vector directions, and approximations using dimensional analysis. Diagrammatic representations are provided for vector addition problems.
3) An example environmental debate question is included that involves estimating the number of disposable diapers used in the US annually based on population data, then calculating the resulting landfill volume and area requirements.
The document uses equations to calculate the thickness of a layer of seawater gaining heat at a rate of 1.5 x 10-7Ks-1. It determines the layer is 2.5 meters thick. It then calculates the temperature change of the top 1mm of ocean gaining 4 x 10-3 Wm-2 of heat over 30 days, finding the temperature change to be 0.025K.
This document summarizes the derivation of the 2-dimensional wave equation to model the propagation of a disturbance through a thin stretched membrane. It shows that the wave equation can be derived from the Lagrangian and Hamilton's principle. Initial and boundary conditions are prescribed, and the wave equation is solved using separation of variables to obtain a double Fourier series solution representing the membrane displacement as a function of space and time.
Modern physics paul a. tipler 6ª edição solutio manualIzabela Ferreira
This document is the preface to an instructor solutions manual for the problems in the textbook "Modern Physics, Sixth Edition" by Paul A. Tipler and Ralph A. Llewellyn. It contains solutions to every problem in the textbook and is intended for instructors, not distribution to students. It was prepared by Mark J. Llewellyn and includes an introduction and table of contents organizing the solutions by chapter. The preface explains the purpose and contents of the manual and provides contact information for the author.
Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solut...KirkMcdowells
Full download : https://alibabadownload.com/product/fox-and-mcdonalds-introduction-to-fluid-mechanics-9th-edition-pritchard-solutions-manual/ Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solutions Manual
1) The document describes a numerical simulation of the spherical collapse of dark matter perturbations in an expanding universe. It simulates both cold dark matter (CDM) and warm dark matter (WDM) cases.
2) For CDM, the simulation shows collapse times depend on the initial overdensity but are approximately symmetric around the turnaround time. Regions of all masses collapse as long as the initial overdensity exceeds a threshold.
3) For WDM, the simulation adds a pressure term to account for the thermal velocity of WDM in the early universe. This term slows collapse compared to CDM and can potentially prevent collapse of low-mass regions.
This document summarizes a study that uses a non-linear Gauss-Newton technique to interpret spherical gravity anomalies. It generates synthetic gravity data for a spherical ore body and applies the Gauss-Newton method to extract parameters like depth and dimension. When applied to noise-free data, the method accurately resolves the parameters in 22 iterations with low error. Adding 10% noise increases the error by 20% but parameters are still estimated reasonably well. The method is also applied to interpret a published field gravity dataset from Texas.
Serway, raymond a physics for scientists and engineers (6e) solutionsTatiani Andressa
This document contains a chapter outline and sample questions and solutions for a physics and measurement chapter. The chapter outline lists topics like standards of length, mass and time, density and atomic mass, and dimensional analysis. The questions and solutions provide examples of calculations involving converting between units, determining densities, and applying dimensional analysis.
This document contains a chapter outline and answers to questions about physics and measurement. The chapter outline lists topics like standards of length, mass and time, matter and model-building, density and atomic mass, and dimensional analysis. The answers to questions section provides explanations and calculations in response to questions about these topics. For example, it explains that atomic clocks are based on electromagnetic waves emitted by atoms, and that density varies with temperature and pressure.
1. The document provides examples of converting between different units of measurement using metric prefixes like micro, pico, and nano. Conversions include kilometers to microns, centimeters to microns, and yards to microns.
2. Additional examples convert between inches and picas, points, and other units like grys and almudes which are used to measure volumes of grains.
3. Conversion factors are also provided for seconds, days, weeks, centuries and other units of time. Examples show converting between these different units of time.
1. The document provides examples of unit conversions using metric prefixes and other conversion factors. Calculations are shown for distances, areas, volumes, and time conversions between various units like kilometers, meters, centimeters, microseconds, seconds and years.
2. Geometric formulas are used to calculate properties of the Earth like circumference, surface area and volume. Examples of unit conversions involving length, area, volume, time and other units are also provided.
3. Examples involve calculating distances, areas, volumes, rotations, uncertainties and time differences between various measurement systems including metric, English, and other historical units. Careful application of conversion factors and significant figures is emphasized.
1. The document provides examples of converting between different units of measurement, such as miles to kilometers, gallons to liters, seconds to years, etc. Conversions involve setting up proportional relationships between units and calculating using conversion factors.
2. Examples show calculating uncertainties in measurements based on the precision of the measuring instrument. Uncertainty is estimated as a percentage error relative to the measured quantity.
3. To determine average values and uncertainties, the document uses the maximum and minimum possible values based on the precision of the original measurements. The uncertainty is taken as half the range between the maximum and minimum average values.
1. The document provides examples of converting between different units of measurement, such as miles to kilometers, gallons to liters, seconds to years, etc. Conversions involve setting up proportional relationships between units and calculating using conversion factors.
2. Examples show calculating uncertainties in measurements based on the precision of the measuring instrument. Uncertainty is estimated as a percentage error relative to the measured quantity.
3. To determine average values and uncertainties, the document uses the maximum and minimum possible values based on the precision of the original measurements. The uncertainty is taken as half the range between the maximum and minimum average values.
1. This document contains conceptual problems and their solutions related to measurement and vectors in physics. It covers topics like SI base units, significant figures, dimensions of physical quantities, and vector operations.
2. Conceptual questions are answered by determining the relevant concepts, relationships between quantities, and applying definitions. Numerical problems are solved by substituting values into appropriate equations.
3. Estimation problems involve approximating quantities using given values and reasonable assumptions. Diagrams are used to illustrate vector concepts and solutions.
This document appears to be the solutions manual for the fourth edition of the textbook "Mathematics for Physical Chemistry" by Robert G. Mortimer. It provides solutions to nearly all of the exercises and problems from that textbook. The solutions manual contains 16 chapters that align with the textbook's chapter structure. Each chapter solution includes step-by-step workings for multiple sample problems drawn from the corresponding textbook chapter. The preface acknowledges that some errors may exist in the solutions and invites readers to report any errors found.
1. The document provides examples of unit conversions between various systems of measurement including centimeters, meters, yards, miles, inches, and picas.
2. Size comparisons are made between everyday objects like grapes and stars, as well as microscopic objects like atoms, cell nuclei, and protons.
3. Calculations are shown for time intervals, rates, and conversions between seconds, minutes, hours, days, weeks, and years.
1. This document summarizes key concepts from a chapter on basic considerations in thermodynamics and fluid mechanics. It defines concepts such as density, pressure, power, energy, mass flux, and flow rate using mathematical equations involving mass, length, time, and other fundamental units.
2. The document also discusses dimensional analysis and demonstrates how to determine the dimensions of physical quantities like force, viscosity, heat flux, and specific heat. Dimensional equations are derived and simplified to fundamental units.
3. Examples are provided for performing unit conversions between the English and SI systems. Problems are also presented and solved for determining values like density, pressure, and flow rates using the defined equations and given data.
This document provides answers to questions about electric current and resistance from Chapter 27. It includes calculations of current, resistance, resistivity, and other electrical properties. Key points addressed include:
- How geometry and temperature affect resistance
- Models for electrical conduction and how resistance changes with temperature
- Properties of superconductors and electrical power calculations
- Questions cover topics like current density, resistivity, Ohm's law, and resistances of different materials.
This document summarizes key concepts and calculations from Chapter 1 of a chemistry textbook. It contains:
1) Definitions and examples of closed and open systems, mixtures, and phases.
2) Practice problems calculating quantities like moles, mass, volume, and molarity using conversion factors and the ideal gas law.
3) Explanations and examples of how pressure, volume, temperature, and amount of substance are related based on gas laws like Boyle's law, Charles' law, Avogadro's law, and the combined gas law.
This document provides a summary of Chapter 1 from an introductory physics textbook. It includes 43 conceptual and calculation problems covering topics like:
- Standards of length, mass and time in the SI system
- The building blocks of matter like atoms, protons, neutrons
- Dimensional analysis and units
- Uncertainty in measurement and significant figures
- Conversion between different units
- Order of magnitude estimates
- Coordinate systems
- Basic trigonometry
The problems address foundational concepts taught in a first chapter and are meant to test the reader's understanding of definitions and ability to set up and solve straightforward calculations.
The document provides solutions to several exercises related to slurry transport. For Exercise 4.1, the solution analyzes shear stress and shear rate data for a phosphate slurry and determines it follows a power-law relationship with a flow index of 0.15 and consistency index of 23.4 Ns0.15/m2. Exercise 4.2 verifies an equation for pressure drop in pipe flow of a power-law fluid. Exercise 4.3 similarly verifies an equation incorporating a yield stress. Subsequent exercises provide solutions for pressure drop, slurry concentration, and rheological properties calculations using data given.
The Sydney Opera House was designed by Danish architect Jørn Utzon between 1955 and 1973. Its distinctive shell design represents a ship under full sail, with each concrete shell being a segment of a sphere with a radius of 75.2 meters. The shells are covered with over 1 million white and cream tiles. The Sydney Opera House sits on 1.3 hectares of land and is 183 meters long and 120 meters wide at its widest point.
Mecánica de Fluidos_Merle C. Potter, David C. Wiggert_3ed SolucionarioEdgar Ortiz Sánchez
The document provides information about free engineering solucionarios (solution manuals) and textbooks that can be downloaded. It states that the solution manuals contain fully solved and clearly explained solutions to all the problems in the textbooks. Visitors are invited to download the files for free. The document then lists the contents of Chapter 1 from an engineering textbook, which includes concepts like mass, density, pressure, power, energy, viscosity, and dimensional analysis.
1) The document discusses units and vectors, which are fundamental tools in physics. It introduces the International System of Units (SI units) including common units like meters, seconds, kilograms.
2) Vectors are quantities that have both magnitude and direction, and can be represented graphically or through components. The document covers how to add and multiply vectors through their components and graphically.
3) It also discusses scalar and vector products of vectors, as well as changing units and dimensional analysis, which are important concepts for solving physics problems.
The use of Nauplii and metanauplii artemia in aquaculture (brine shrimp).pptxMAGOTI ERNEST
Although Artemia has been known to man for centuries, its use as a food for the culture of larval organisms apparently began only in the 1930s, when several investigators found that it made an excellent food for newly hatched fish larvae (Litvinenko et al., 2023). As aquaculture developed in the 1960s and ‘70s, the use of Artemia also became more widespread, due both to its convenience and to its nutritional value for larval organisms (Arenas-Pardo et al., 2024). The fact that Artemia dormant cysts can be stored for long periods in cans, and then used as an off-the-shelf food requiring only 24 h of incubation makes them the most convenient, least labor-intensive, live food available for aquaculture (Sorgeloos & Roubach, 2021). The nutritional value of Artemia, especially for marine organisms, is not constant, but varies both geographically and temporally. During the last decade, however, both the causes of Artemia nutritional variability and methods to improve poorquality Artemia have been identified (Loufi et al., 2024).
Brine shrimp (Artemia spp.) are used in marine aquaculture worldwide. Annually, more than 2,000 metric tons of dry cysts are used for cultivation of fish, crustacean, and shellfish larva. Brine shrimp are important to aquaculture because newly hatched brine shrimp nauplii (larvae) provide a food source for many fish fry (Mozanzadeh et al., 2021). Culture and harvesting of brine shrimp eggs represents another aspect of the aquaculture industry. Nauplii and metanauplii of Artemia, commonly known as brine shrimp, play a crucial role in aquaculture due to their nutritional value and suitability as live feed for many aquatic species, particularly in larval stages (Sorgeloos & Roubach, 2021).
This presentation explores a brief idea about the structural and functional attributes of nucleotides, the structure and function of genetic materials along with the impact of UV rays and pH upon them.
Comparing Evolved Extractive Text Summary Scores of Bidirectional Encoder Rep...University of Maribor
Slides from:
11th International Conference on Electrical, Electronics and Computer Engineering (IcETRAN), Niš, 3-6 June 2024
Track: Artificial Intelligence
https://www.etran.rs/2024/en/home-english/
Nucleophilic Addition of carbonyl compounds.pptxSSR02
Nucleophilic addition is the most important reaction of carbonyls. Not just aldehydes and ketones, but also carboxylic acid derivatives in general.
Carbonyls undergo addition reactions with a large range of nucleophiles.
Comparing the relative basicity of the nucleophile and the product is extremely helpful in determining how reversible the addition reaction is. Reactions with Grignards and hydrides are irreversible. Reactions with weak bases like halides and carboxylates generally don’t happen.
Electronic effects (inductive effects, electron donation) have a large impact on reactivity.
Large groups adjacent to the carbonyl will slow the rate of reaction.
Neutral nucleophiles can also add to carbonyls, although their additions are generally slower and more reversible. Acid catalysis is sometimes employed to increase the rate of addition.
What is greenhouse gasses and how many gasses are there to affect the Earth.moosaasad1975
What are greenhouse gasses how they affect the earth and its environment what is the future of the environment and earth how the weather and the climate effects.
EWOCS-I: The catalog of X-ray sources in Westerlund 1 from the Extended Weste...Sérgio Sacani
Context. With a mass exceeding several 104 M⊙ and a rich and dense population of massive stars, supermassive young star clusters
represent the most massive star-forming environment that is dominated by the feedback from massive stars and gravitational interactions
among stars.
Aims. In this paper we present the Extended Westerlund 1 and 2 Open Clusters Survey (EWOCS) project, which aims to investigate
the influence of the starburst environment on the formation of stars and planets, and on the evolution of both low and high mass stars.
The primary targets of this project are Westerlund 1 and 2, the closest supermassive star clusters to the Sun.
Methods. The project is based primarily on recent observations conducted with the Chandra and JWST observatories. Specifically,
the Chandra survey of Westerlund 1 consists of 36 new ACIS-I observations, nearly co-pointed, for a total exposure time of 1 Msec.
Additionally, we included 8 archival Chandra/ACIS-S observations. This paper presents the resulting catalog of X-ray sources within
and around Westerlund 1. Sources were detected by combining various existing methods, and photon extraction and source validation
were carried out using the ACIS-Extract software.
Results. The EWOCS X-ray catalog comprises 5963 validated sources out of the 9420 initially provided to ACIS-Extract, reaching a
photon flux threshold of approximately 2 × 10−8 photons cm−2
s
−1
. The X-ray sources exhibit a highly concentrated spatial distribution,
with 1075 sources located within the central 1 arcmin. We have successfully detected X-ray emissions from 126 out of the 166 known
massive stars of the cluster, and we have collected over 71 000 photons from the magnetar CXO J164710.20-455217.
Phenomics assisted breeding in crop improvementIshaGoswami9
As the population is increasing and will reach about 9 billion upto 2050. Also due to climate change, it is difficult to meet the food requirement of such a large population. Facing the challenges presented by resource shortages, climate
change, and increasing global population, crop yield and quality need to be improved in a sustainable way over the coming decades. Genetic improvement by breeding is the best way to increase crop productivity. With the rapid progression of functional
genomics, an increasing number of crop genomes have been sequenced and dozens of genes influencing key agronomic traits have been identified. However, current genome sequence information has not been adequately exploited for understanding
the complex characteristics of multiple gene, owing to a lack of crop phenotypic data. Efficient, automatic, and accurate technologies and platforms that can capture phenotypic data that can
be linked to genomics information for crop improvement at all growth stages have become as important as genotyping. Thus,
high-throughput phenotyping has become the major bottleneck restricting crop breeding. Plant phenomics has been defined as the high-throughput, accurate acquisition and analysis of multi-dimensional phenotypes
during crop growing stages at the organism level, including the cell, tissue, organ, individual plant, plot, and field levels. With the rapid development of novel sensors, imaging technology,
and analysis methods, numerous infrastructure platforms have been developed for phenotyping.
Remote Sensing and Computational, Evolutionary, Supercomputing, and Intellige...University of Maribor
Slides from talk:
Aleš Zamuda: Remote Sensing and Computational, Evolutionary, Supercomputing, and Intelligent Systems.
11th International Conference on Electrical, Electronics and Computer Engineering (IcETRAN), Niš, 3-6 June 2024
Inter-Society Networking Panel GRSS/MTT-S/CIS Panel Session: Promoting Connection and Cooperation
https://www.etran.rs/2024/en/home-english/
hematic appreciation test is a psychological assessment tool used to measure an individual's appreciation and understanding of specific themes or topics. This test helps to evaluate an individual's ability to connect different ideas and concepts within a given theme, as well as their overall comprehension and interpretation skills. The results of the test can provide valuable insights into an individual's cognitive abilities, creativity, and critical thinking skills
ANAMOLOUS SECONDARY GROWTH IN DICOT ROOTS.pptxRASHMI M G
Abnormal or anomalous secondary growth in plants. It defines secondary growth as an increase in plant girth due to vascular cambium or cork cambium. Anomalous secondary growth does not follow the normal pattern of a single vascular cambium producing xylem internally and phloem externally.
1. 1
Physics and Measurement
CHAPTER OUTLINE
1.1 Standards of Length, Mass, and
Time
1.2 Matter and Model-Building
1.3 Dimensional Analysis
1.4 Conversion of Units
1.5 Estimates and Order-of-
Magnitude Calculations
1.6 Significant Figures
ANSWERS TO QUESTIONS
* An asterisk indicates an item new to this edition.
Q1.1 Density varies with temperature and pressure. It would
be necessary to measure both mass and volume very
accurately in order to use the density of water as a
standard.
Q1.2 (a) 0.3 millimeters (b) 50 microseconds
(c) 7.2 kilograms
*Q1.3 In the base unit we have (a) 0.032 kg (b) 0.015 kg
(c) 0.270 kg (d) 0.041 kg (e) 0.27 kg. Then the ranking is
c = e > d > a > b
Q1.4 No: A dimensionally correct equation need not be true.
Example: 1 chimpanzee = 2 chimpanzee is dimension-
ally correct.
Yes: If an equation is not dimensionally correct, it cannot
be correct.
*Q1.5 The answer is yes for (a), (c), and (f). You cannot add or subtract a number of apples and a
number of jokes. The answer is no for (b), (d), and (e). Consider the gauge of a sausage, 4 kgր2 m,
or the volume of a cube, (2 m)3
. Thus we have (a) yes (b) no (c) yes (d) no (e) no (f) yes
*Q1.6 41 € ≈ 41 € (1 Lր1.3 €)(1 qtր1 L)(1 galր4 qt) ≈ (10ր1.3) gal ≈ 8 gallons, answer (c)
*Q1.7 The meterstick measurement, (a), and (b) can all be 4.31 cm. The meterstick measurement and
(c) can both be 4.24 cm. Only (d) does not overlap. Thus (a) (b) and (c) all agree with the
meterstick measurement.
*Q1.8 0.02(1.365) = 0.03. The result is (1.37 ± 0.03) × 107
kg. So (d) 3 digits are significant.
1
SOLUTIONS TO PROBLEMS
Section 1.1 Standards of Length, Mass, and Time
P1.1 Modeling the Earth as a sphere, we find its volume as
4
3
4
3
6 37 10 1 08 103 6 3 21 3
π πr = ×( ) = ×. .m m .
Its density is then ρ = =
×
×
= ×
m
V
5 98 10
1 08 10
5 52 10
24
21 3
3 3.
.
.
kg
m
kg m . This value is intermediate
between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000
to 3 000 kgրm3
. The average density of the Earth is significantly higher, so higher-density material
must be down below the surface.
ISMV1_5103_01.indd 1ISMV1_5103_01.indd 1 10/27/06 4:33:21 PM10/27/06 4:33:21 PM
2. P1.2 With V = ( )( )base area height V r h= ( )π 2
and ρ =
m
V
, we have
ρ
π π
= =
( ) ( )
m
r h2 2
9
1
19 5 39 0
10
1
kg
mm mm
mm
m
3
3
. .
⎛⎛
⎝⎜
⎞
⎠⎟
= ×ρ 2 15 104 3
. .kg m
P1.3 Let V represent the volume of the model, the same in ρ =
m
V
for both. Then ρiron kg= 9 35. V
and ρgold
gold
=
m
V
. Next,
ρ
ρ
gold
iron
gold
kg
=
m
9 35.
and mgold
3 3
3
kg
19.3 10 kg/m
kg/m
=
×
×
⎛
9 35
7 86 103
.
.⎝⎝⎜
⎞
⎠⎟ = 23 0. kg .
*P1.4 ρ = m V/ and V r d d= = =( / ) ( / ) ( / ) /4 3 4 3 2 63 3 3
π π π where d is the diameter.
Then ρ π
π
= =
×
×
= ×
−
−
6
6 1 67 10
2 4 10
2 33
27
15 3
m d/
( . )
( . )
.
kg
m
11017 3
kg/m
2.3 10 kg/m /(11.3 10 kg/m )=17 3 3 3
× × it is 20 × 1012
times the density of lead .
P1.5 For either sphere the volume is V r=
4
3
3
π and the mass is m V r= =ρ ρ π
4
3
3
. We divide
this equation for the larger sphere by the same equation for the smaller:
m
m
r
r
r
rs s s
ᐉ ᐉ ᐉ
= = =
ρ π
ρ π
4 3
4 3
5
3
3
3
3
.
Then r rsᐉ = = ( ) =5 4 50 1 71 7 693
. . .cm cm .
Section 1.2 Matter and Model-Building
P1.6 From the figure, we may see that the spacing between diagonal planes is half the distance
between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from
the Pythagorean theorem, L L Ldiag = +2 2
. Thus, since the atoms are separated by a distance
L = 0 200. nm, the diagonal planes are separated by
1
2
0 1412 2
L L+ = . nm .
Section 1.3 Dimensional Analysis
P1.7 (a) This is incorrect since the units of ax[ ] are m s2 2
, while the units of v[ ] are m s .
(b) This is correct since the units of y[ ] are m, and cos kx( ) is dimensionless if k[ ] is in m−1
.
P1.8 (a) Circumference has dimensions of L.
(b) Volume has dimensions of L3
.
(c) Area has dimensions of L2
.
Expression (i) has dimension L L L2 1 2 2
( ) =
/
, so this must be area (c).
Expression (ii) has dimension L, so it is (a).
Expression (iii) has dimension L L L2 3
( )= , so it is (b). Thus, (a) ii; (b) iii;(c) i= = = .
2 Chapter 1
ISMV1_5103_01.indd 2ISMV1_5103_01.indd 2 10/27/06 12:27:01 PM10/27/06 12:27:01 PM
3. Physics and Measurement 3
P1.9 Inserting the proper units for everything except G,
kg m
s
kg
m2
⎡
⎣⎢
⎤
⎦⎥ =
[ ]
[ ]
G
2
2
.
Multiply both sides by m[ ]2
and divide by kg[ ]2
; the units of G are m
kg s
3
2
⋅
.
Section 1.4 Conversion of Units
P1.10 Apply the following conversion factors:
1 2 54in cm= . , 1 86 400d s= , 100 1cm m= , and 10 19
nm m=
1
32
2 54 10 102 9
in day
cm in m cm nm m
8
⎛
⎝
⎞
⎠
( )( )( )−
.
66 400 s day
nm s= 9 19. .
This means the proteins are assembled at a rate of many layers of atoms each second!
P1.11 Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we
should expect the area to be about A ≈( )( )=30 50 1500m m m2
.
Categorize: We model the lot as a perfect rectangle to use Area = Length × Width. Use the
conversion:1 m 3.281 ft= .
Analyze: A LW= = ( )⎛
⎝
⎞
⎠
( )100
1
3 281
150
1
3 28
ft
m
ft
ft
m
. . 11
1 39 103
ft
= 1 390 m m2 2⎛
⎝
⎞
⎠
= ×. .
Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper
units of m2
. Unit conversion is a common technique that is applied to many problems.
P1.12 (a) V = ( )( )( ) = ×40.0 m 20.0 m 12.0 m . m3
9 60 103
V = × ( ) = ×9 60 10 3 39 103 5 3
. m 3.28 ft 1 m ft3 3
.
(b) The mass of the air is
m V= = ( ) ×( )= ×ρair
3 3
kg m 9.60 10 m . k1 20 1 15 103 4
. gg.
The student must look up weight in the index to find
F mgg = = ×( )( )= ×1.15 10 kg 9.80 m s 1.13 10 N4 2 5
.
Converting to pounds,
Fg = ×( )( ) = ×1 13 10 2 54 105 4
. N 1 lb 4.45 N lb. .
*P1.13 The area of the four walls is (3.6 + 3.8 + 3.6 + 3.8)m (2.5 m) = 37 m2
. Each sheet in the book
has area (0.21 m) (0.28 m) = 0.059 m2
. The number of sheets required for wallpaper is
37 m2
ր0.059 m2
= 629 sheets = 629 sheets(2 pagesր1 sheet) = 1260 pages.
The pages from volume one are inadequate, but the full version has enough pages.
ISMV1_5103_01.indd 3ISMV1_5103_01.indd 3 10/27/06 4:18:09 PM10/27/06 4:18:09 PM
4. 4 Chapter 1
P1.14 (a) Seven minutes is 420 seconds, so the rate is
r = = × −30 0
420
7 14 10 2.
.
gal
s
gal s .
(b) Converting gallons first to liters, then to m3
,
r = ×( )⎛
⎝⎜
⎞
⎠⎟
−
−
7 14 10
3 786 102
3
.
.
gal s
L
1 gal
m3
11 L
m s3
⎛
⎝⎜
⎞
⎠⎟
= × −
r 2 70 10 4
. .
(c) At that rate, to fill a 1-m3
tank would take
t =
×
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟ =−
1
2 70 10
1
1 034
m
m s
h
3 600
3
3
.
. h .
P1.15 From Table 14.1, the density of lead is 1 13 104
. kg m3
× , so we should expect our calculated
value to be close to this number. This density value tells us that lead is about 11 times denser than
water, which agrees with our experience that lead sinks.
Density is defined as mass per volume, in ρ =
m
V
. We must convert to SI units in the calculation.
ρ =
⎛
⎝⎜
⎞
⎠⎟
⎛23 94
2 10
1
1000
100
13
.
.
g
cm
kg
g
cm
m⎝⎝⎜
⎞
⎠⎟ =
⎛
⎝⎜
⎞
⎠⎟
3
3
23 94
2 10
1
1000
1 00.
.
g
cm
kg
g
00
1
1 14 104000 cm
m
. kg m
3
3
3
⎛
⎝
⎜
⎞
⎠
⎟ = ×
At one step in the calculation, we note that one million cubic centimeters make one cubic meter.
Our result is indeed close to the expected value. Since the last reported significant digit is not
certain, the difference in the two values is probably due to measurement uncertainty and should
not be a concern. One important common-sense check on density values is that objects which sink
in water must have a density greater than 1 g cm3
, and objects that float must be less dense than
water.
P1.16 The weight flow rate is 1 200
2 000 1 1ton
h
lb
ton
h
60 min
min
60
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠ s
lb s⎛
⎝
⎞
⎠
= 667 .
P1.17 (a) 8 10 1 112
×⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
$
1 000 $ s
h
3 600 s
day
24 h
yr
365 days
years⎛
⎝
⎞
⎠
⎛
⎝⎜
⎞
⎠⎟ =
1
250
(b) The circumference of the Earth at the equator is 2 6 378 10 4 01 103 7
π . .×( )= ×m m. The
length of one dollar bill is 0.155 m so that the length of 8 trillion bills is 1 24 1012
. × m.
Thus, the 8 trillion dollars would encircle the Earth
1 24 10
1
3 09 10
12
4.
. .
×
×
= ×
m
4.01 0 m
times7
P1.18 V Bh= =
( )( )⎡⎣ ⎤⎦1
3
13 0 43 560
3
481
. acres ft acre2
fft
ft3
( )
= ×9 08 107
. ,
or
V = ×( ) ×⎛
⎝⎜
⎞
⎠⎟
=
−
9 08 10
2 83 10
1
2 5
7
2
.
.
.
ft
m
ft
3
3
3
77 106
× m3
FIG. P1.18
B
h
B
h
ISMV1_5103_01.indd 4ISMV1_5103_01.indd 4 10/27/06 12:27:03 PM10/27/06 12:27:03 PM
5. Physics and Measurement 5
P1.19 Fg = ( ) ×( )2 50 2 00 10 2 0006
. .tons block blocks lb tton lbs( ) = ×1 00 1010
.
P1.20 (a) d d
d
nucleus, scale nucleus, real
atom, scale
=
ddatom, real
m
ft
1.06
⎛
⎝
⎜
⎞
⎠
⎟ = ×( )−
2 40 10
30015
.
××
⎛
⎝
⎞
⎠
= ×−
−
10 m
ft10
3
6 79 10. , or
dnucleus, scale ft mm 1 ft= ×( )(−
6 79 10 304 83
. . )) = 2 07. mm
(b) V
V
r
r
ratom
nucleus
atom
nucleus
atom
= =
4 3
4 3
3
3
π
π
/
/ rr
d
dnucleus
atom
nucleus
⎛
⎝⎜
⎞
⎠⎟ =
⎛
⎝⎜
⎞
⎠⎟ =
×
3 3
1 06. 110
2 40 10
8 62 10
10
15
3
13
−
−
×
⎛
⎝⎜
⎞
⎠⎟
= ×
m
m
times
.
. as large
P1.21 V At= so t
V
A
= =
×
= ×
−
−3 78 10
25 0
1 51 10 151
3
4.
.
.
m
m
m or
3
2
mµ( )
P1.22 (a) A
A
r
r
r
r
Earth
Moon
Earth
Moon
2
Earth
Moon
= =
⎛
⎝
4
4
2
π
π ⎜⎜
⎞
⎠⎟ =
×( )( )
×
⎛
⎝
⎜
⎞
2 6
8
6 37 10 100
1 74 10
.
.
m cm m
cm ⎠⎠
⎟ =
2
13 4.
(b) V
V
r
r
r
r
Earth
Moon
Earth
Moon
Earth
Mo
= =
4 3
4 3
3
3
π
π
/
/ oon
3
m cm m
cm
⎛
⎝⎜
⎞
⎠⎟ =
×( )( )
×
6 37 10 100
1 74 10
6
8
.
.
⎛⎛
⎝
⎜
⎞
⎠
⎟ =
3
49 1.
P1.23 To balance, m mFe Al= or ρ ρFe Fe Al AlV V=
ρ π ρ π
ρ
ρ
Fe Fe Al Al
Al Fe
Fe
A
4
3
4
3
3 3⎛
⎝⎜
⎞
⎠⎟ = ⎛
⎝
⎞
⎠
=
r r
r r
ll
cm c
⎛
⎝⎜
⎞
⎠⎟ = ( )⎛
⎝
⎞
⎠
=
1 3 1 3
2 00
7 86
2 70
2 86
/ /
.
.
.
. mm .
P1.24 The mass of each sphere is
m V
r
Al Al Al
Al Al
= =ρ
π ρ4
3
3
and
m V
r
Fe Fe Fe
Fe Fe
= =ρ
π ρ4
3
3
.
Setting these masses equal,
4
3
4
3
3 3
π ρ π ρAl Al Fe Fer r
= and r rAl Fe
Fe
Al
=
ρ
ρ
3 .
The resulting expression shows that the radius of the aluminum sphere is directly proportional to
the radius of the balancing iron sphere. The sphere of lower density has larger radius. The
fraction
ρ
ρ
Fe
Al
is the factor of change between the densities, a number greater than 1. Its cube root
is a number much closer to 1. The relatively small change in radius implies a change in volume
sufficient to compensate for the change in density.
ISMV1_5103_01.indd 5ISMV1_5103_01.indd 5 10/27/06 12:27:04 PM10/27/06 12:27:04 PM
6. 6 Chapter 1
Section 1.5 Estimates and Order-of-Magnitude Calculations
P1.25 Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ball as a
sphere of diameter 0.038 m. The volume of the room is 4 4 3 48× × = m3
, while the volume of one ball is
4
3
0 038
2 87 10
3
5π .
.
m
2
m3⎛
⎝
⎞
⎠
= × −
.
Therefore, one can fit about 48
2 87 10
105
6
.
~
× −
ping-pong balls in the room.
As an aside, the actual number is smaller than this because there will be a lot of space in the
room that cannot be covered by balls. In fact, even in the best arrangement, the so-called “best
packing fraction” is 1
6
2 0 74π = . so that at least 26% of the space will be empty. Therefore, the
above estimate reduces to 1 67 10 0 740 106 6
. . ~× × .
P1.26 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft.
Thus, the tire would make 50 000 5280 1 3 107
mi ft mi rev 8 ft rev ~( )( )( ) = × 110 rev7
.
P1.27 Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then
V = ( )( )( )( ) =0 5 1 3 0 5 0 3 0 10. . . . .m m m m3
.
The mass of this volume of water is
m Vwater water
3 3
kg m m kg= = ( )( )=ρ 1 000 0 10 100. ~ 1102
kg .
Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub.
The mass of copper required is
m Vcopper copper
3 3
kg m m k= = ( )( )=ρ 8 920 0 10 892. gg kg~ 103
.
*P1.28 The time required for the task is
10
1 1 1
1
9
$
s
1 $
h
3600 s
working day⎛
⎝⎜
⎞
⎠⎟
⎛
⎝
⎞
⎠ 66
1
58
h
bad yr
300 working days
y⎛
⎝
⎞
⎠
⎛
⎝⎜
⎞
⎠⎟ = rr
Since you are already around 20 years old, you would have a miserable life and likely die
before accomplishing the task. You have better things to do. Say no.
P1.29 Assume: Total population = 107
; one out of every 100 people has a piano; one tuner can serve about
1 000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year). Therefore,
# tuners ~
1 tuner
1 000 pianos
1 piano
10
⎛
⎝⎜
⎞
⎠⎟ 00 people
people tuners
⎛
⎝⎜
⎞
⎠⎟ =( )10 1007
.
Section 1.6 Significant Figures
P1.30 METHOD ONE
We treat the best value with its uncertainty as a binomial 21 3 0 2 9 8 0 1. . . .±( ) ±( )cm cm,
A = ( )± ( )± ( )± ( )( )[ ]21 3 9 8 21 3 0 1 0 2 9 8 0 2 0 1. . . . . . . . cm2
.
The first term gives the best value of the area. The cross terms add together to give the
uncertainty and the fourth term is negligible.
A = ±209 42 2
cm cm .
METHOD TWO
We add the fractional uncertainties in the data.
A = ( )( )± +
⎛
⎝
⎞
⎠
=21 3 9 8
0 2
21 3
0 1
9 8
209. .
.
.
.
.
cm cm ccm cm cm2 2 2
2 209 4± = ±%
ISMV1_5103_01.indd 6ISMV1_5103_01.indd 6 10/28/06 2:41:27 AM10/28/06 2:41:27 AM
7. Physics and Measurement 7
P1.31 (a) 3 (b) 4 (c) 3 (d) 2
P1.32 r
m
= ±( ) = ±( )×
= ±
−
6 50 0 20 6 50 0 20 10
1 85 0
2
. . cm . . m
. .. kg02
4
3
3
( )
=
( )
ρ
π
m
r
also, δ ρ
ρ
δ δ
= +
m
m
r
r
3 .
In other words, the percentages of uncertainty are cumulative. Therefore,
δ ρ
ρ
= +
( )
=
0 02
1 85
3 0 20
6 50
0 103
.
.
.
.
. ,
ρ
π
=
( ) ×( )
= ×−
1 85
6 5 10
1 61 10
4
3
2 3
3 3.
.
.
m
kg m
and
ρ δ ρ± = ±( )× = ±( )×1 61 0 17 10 1 6 0 2 103 3
. . . .kg m kg m3 33
.
P1.33 (a) 756.??
37.2?
0.83
+ 2.5?
796./ /5 3 = 797
(b) 0 003 2 356 3 1 140 16. 2 s.f. . 4 s.f. . 2 s.f.( )× ( ) = = (( ) 1 1.
(c) 5.620 4 s.f. >4 s.f. 17.656= 4 s.f. 17( )× ( ) = ( )π ..66
P1.34 We work to nine significant digits:
1 1
365 242199 24 6
yr yr
d
1 yr
h
1 d
=
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝
⎞
⎠
. 00 60
31556 926 0
min
1 h
s
1 min
s
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
= . .
*P1.35 The tax amount is $1.36 − $1.25 = $0.11. The tax rate is $0.11ր$1.25 = 0.088 0 = 8.80%
*P1.36 (a) We read from the graph a vertical separation of 0.3 spaces = 0.015 g .
(b) Horizontally, 0.6 spaces = 30 cm2
.
(c) Because the graph line goes through the origin, the same percentage describes the vertical
and the horizontal scatter: 30 cm2
ր380 cm2
= 8% .
(d) Choose a grid point on the line far from the origin: slope = 0.31 gր600 cm2
= 0.000 52 gրcm2
=
(0.000 52 gրcm2
)(10 000 cm2
ր1 m2
) = 5.2 g/m2
.
(e) For any and all shapes cut from this copy paper, the mass of the cutout is proportional to
its area. The proportionality constant is 5.2 g/m2
± 8%, where the uncertainty is estimated.
(f) This result should be expected if the paper has thickness and density that are uniform
within the experimental uncertainty. The slope is the areal density of the paper, its mass
per unit area.
ISMV1_5103_01.indd 7ISMV1_5103_01.indd 7 10/28/06 2:41:45 AM10/28/06 2:41:45 AM
8. 8 Chapter 1
*P1.37 15 players = 15 players (1 shiftր1.667 player) = 9 shifts
*P1.38 Let o represent the number of ordinary cars and s the number of trucks. We have
o = s + 0.947s = 1.947s, and o = s + 18. We eliminate o by substitution: s + 18 = 1.947s
0.947s = 18 and s = 18ր0.947 = 19 .
*P1.39 Let s represent the number of sparrows and m the number of more interesting birds.
We have sրm = 2.25 and s + m = 91. We eliminate m by substitution: m = sր2.25
s + sր2.25 = 91 1.444s = 91 s = 91ր1.444 = 63 .
*P1.40 For those who are not familiar with solving equations numerically, we provide a detailed solution.
It goes beyond proving that the suggested answer works.
The equation 2 3 5 70 04 3
x x x− + − = is quartic, so we do not attempt to solve it with algebra. To
find how many real solutions the equation has and to estimate them, we graph the expression:
x −3 −2 −1 0 1 2 3 4
y x x x= − + −2 3 5 704 3
158 −24 −70 −70 −66 −52 26 270
We see that the equation y = 0 has two roots, one around x = −2 2.
and the other near x = +2 7. . To home in on the first of these solutions
we compute in sequence: When x = −2 2. , y = −2 20. . The root must be
between x = −2 2. and x = −3. When x = −2 3. , y = 11 0. . The root is
between x = −2 2. and x = −2 3. . When x = −2 23. , y = 1 58. . The root
is between x = −2 20. and x = −2 23. . When x = −2 22. , y = 0 301. . The
root is between x = −2 20. and −2.22. When x = −2 215. , y = −0 331. .
The root is between x = −2 215. and −2.22. We could next try
x = −2 218. , but we already know to three-digit precision that the root
is x = −2 22. .
*P1.41 We require sin cosθ θ= −3 , or
sin
cos
θ
θ
= −3, or tanθ = −3.
For tan a tan−
−( ) = −( )1
3 3rc , your calculator may return −71.6°,
but this angle is not between 0° and 360° as the problem
requires. The tangent function is negative in the second quad-
rant (between 90° and 180°) and in the fourth quadrant (from
270° to 360°). The solutions to the equation are then
360 71 6 288° ° °− =. and 180 71 6 108° ° °− =. .
y
x
FIG. P1.40
360°0
tanθ
θ
FIG. P1.41
ISMV1_5103_01.indd 8ISMV1_5103_01.indd 8 10/27/06 12:27:07 PM10/27/06 12:27:07 PM
9. Physics and Measurement 9
*P1.42 We draw the radius to the initial point and the radius to the final point.
The angle θ between these two radii has its sides perpendicular, right side
to right side and left side to left side, to the 35° angle between the original
and final tangential directions of travel. A most useful theorem from
geometry then identifies these angles as equal: θ = °35 . The whole
circumference of a 360° circle of the same radius is 2π R. By proportion,
then 2
360
840π R
°
=
°
m
35
.
R =
°
°
= = ×
360
2
840 840
1 38 103
π
m
35
m
0.611
m.
We could equally well say that the measure of the angle in radians is
θ
π
= ° = °
°
⎛
⎝
⎞
⎠
= =35 35
2
0 611
840radians
360
rad
m
.
RR
.
Solving yields R = 1 38. km.
*P1.43 Mass is proportional to cube of length: m = kᐉ3
mƒ
րmi
= (ᐉf
րᐉi
)3
.
Length changes by 15.8%: ᐉf
= ᐉi
+ 0.158 ᐉi
= 1.158 ᐉi
.
Mass increase: mf
= mi
+ 17.3 kg.
Eliminate by substitution:
m
m
f
f −
= =
17 3
1 158 1 5533
.
. .
kg
mf
= 1.553 mf
− 26.9 kg 26.9 kg = 0.553 mf
mf
= 26.9 kgր0.553 = 48. 6 kg .
*P1.44 We use substitution, as the most generally applicable method for solving simultaneous equations.
We substitute p q= 3 into each of the other two equations to eliminate p:
3
1
2
3
1
2
1
2
2 2 2
qr qs
qr qs qt
=
+ =
⎧
⎨
⎪
⎩⎪
.
These simplify to
3
3 2 2 2
r s
r s t
=
+ =
⎧
⎨
⎩
. We substitute to eliminate s:
3 3
12
2 2 2
2 2
r r t
r t
+( ) =
=
. We solve for the
combination t
r
:
t
r
2
2
12= .
t
r
= −either or3 46 3 46. .
*P1.45 Solve the given equation for ∆t: ∆t = 4QLրkπd2
(Th
− Tc
) = [4QLրkπ (Th
− Tc
)] [1ր d2
].
(a) Making d three times larger with d2
in the bottom of the fraction makes
∆t nine times smaller .
(b) ∆t is inversely proportional to the square of d.
(c) Plot ∆t on the vertical axis and 1/d2
on the horizontal axis.
(d) From the last version of the equation, the slope is 4QL/kπ(Th
− Tc
) . Note that this quantity
is constant as both ∆t and d vary.
FIG. P1.42
R
θ
N
i
f
S
EW
35.0°
ISMV1_5103_01.indd 9ISMV1_5103_01.indd 9 10/28/06 2:42:03 AM10/28/06 2:42:03 AM
10. 10 Chapter 1
Additional Problems
P1.46 It is desired to find the distance x such that
x
x100
1 000
m
m
=
(i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x). Thus, it is seen that
x2 5
100 1 000 1 00 10= ( )( ) = ×m m m2
.
and therefore
x = × =1 00 10 3165
. m m2
.
*P1.47 (a) The mass is equal to the mass of a sphere of radius 2.6 cm and density 4.7 gրcm3
, minus the
mass of a sphere of radius a and density 4.7 gրcm3
plus the mass of a sphere of radius a and
density 1.23 gրcm3
.
m = ρ1
4πr3
ր3 − ρ1
4πa3
ր3 + ρ2
4πa3
ր3
= (4.7 gրcm3
)4π(2.6 cm)3
ր3 − (4.7 gրcm3
)4π(a)3
ր3 + (1.23 gրcm3
)4π(a)3
ր3
m = 346 g − (14.5 g/cm3
)a3
(b) For a = 0 the mass is a maximum, (c) 346 g . (d) Yes . This is the mass of the
uniform sphere we considered in the first term of the calculation.
(e) For a = 2.60 cm the mass is a minimum, (f) 346 − 14.5(2.6)3
= 90.6 g . (g) Yes . This
is the mass of a uniform sphere of density 1.23 gրcm3
.
(h) (346 g + 90.6 g)ր2 = 218 g (i) No . The result of part (a) gives 346 g − (14.5 gրcm3
)
(1.3 cm)3
= 314 g, not the same as 218 g.
(j) We should expect agreement in parts b-c-d, because those parts are about a uniform
sphere of density 4.7 g/cm3
. We should expect agreement in parts e-f-g, because those
parts are about a uniform liquid drop of density 1.23 g/cm3
. The function m(a) is not a
linear function, so a halfway between 0 and 2.6 cm does not give a value for m halfway
between the minimum and maximum values. The graph of m versus a starts at a = 0 with
a horizontal tangent. Then it curves down more and more steeply as a increases. The
liquid drop of radius 1.30 cm has only one eighth the volume of the whole sphere, so its
presence brings down the mass by only a small amount, from 346 g to 314 g.
(k) No change, so long as the wall of the shell is unbroken.
*P1.48 (a) We have B + C(0) = 2.70 gրcm3
and B + C(14 cm) = 19.3 gրcm3
. We know B = 2.70 g/cm3
and we solve for C by subtracting: C(14 cm) = 16.6 gրcm3
so C = 1.19 g/cm4
.
(b) m x dx= +∫ (2.70 g/cm g/cm cm3 4 2
cm
1 19 9
0
14
. )( )
= 24.3 g/cm g/cm2
0
14 cm c
dx xdx+∫ 10 7
0
14
.
mm
= (24.3 g/cm)(14 cm – 0) + (10.7 g/c
∫
mm cm)
= 340 g + 1046 g = 1.
2 2
)[( ] /14 0 2−
339 kg
ISMV1_5103_01.indd 10ISMV1_5103_01.indd 10 10/27/06 12:27:09 PM10/27/06 12:27:09 PM
11. Physics and Measurement 11
P1.49 The scale factor used in the “dinner plate” model is
S =
×
= × −0 25
1 0 10
2 105
6.
.
m
lightyears
.5 m lightyeears.
The distance to Andromeda in the scale model will be
D D Sscale actual
6
2.0 10 lightyears 2.5 10= = ×( ) × −66
m lightyears m( )= 5 0. .
*P1.50 The rate of volume increase is
dV
dt
d
dt
r r
dr
dt
r
dr
dt
= = =
4
3
4
3
3 43 2 2
π π π .
(a) dVրdt = 4 π(6.5 cm)2
(0.9 cmրs) = 478 cm3
/s
(b)
dr
dt
dV dt
r
= = =
/ /
(
.
4
478
4 13
0 2252
π π
cm s
cm)
cm
3
2
33
s/
(c) When the balloon radius is twice as large, its surface area is four times larger. The new
volume added in one second in the inflation process is equal to this larger area times an
extra radial thickness that is one-fourth as large as it was when the balloon was smaller.
P1.51 One month is
1 30 24 3 600 2 592 106
mo day h day s h s= ( )( )( ) = ×. .
Applying units to the equation,
V t t= ( ) +( )1 50 0 008 00 2
. .Mft mo Mft mo3 3 2
.
Since1 106
Mft ft3 3
= ,
V t t= ×( ) + ×( )1 50 10 0 008 00 106 6 2
. .ft mo ft mo3 3 2
.
Converting months to seconds,
V t=
×
×
+
×1 50 10 0 008 00 106 6
. .ft mo
2.592 10 s mo
3
6
fft mo
2.592 10 s mo
3 2
6
×( )2
2
t .
Thus, V t tft ft s ft s3 3 3 2
[ ] . .= ( ) + ×( )−
0 579 1 19 10 9 2
.
*P1.52
′α (deg) α(rad) tan α( ) sin α( ) difference between α and tanα
15.0 0.262 0.268 0.259 2.30%
20.0 0.349 0.364 0.342 4.09%
30.0 0.524 0.577 0.500 9.32%
33.0 0.576 0.649 0.545 11.3%
31.0 0.541 0.601 0.515 9.95%
31.1 0.543 0.603 0.516 10.02%
We see that ␣ in radians, tan(␣) and sin(␣) start out together from zero and diverge only slightly
in value for small angles. Thus 31 0. º is the largest angle for which tan
tan
.
α α
α
−
< 0 1.
ISMV1_5103_01.indd 11ISMV1_5103_01.indd 11 10/27/06 12:27:09 PM10/27/06 12:27:09 PM
12. 12 Chapter 1
P1.53 2 15 0
2 39
2 39
π r
r
h
r
h
=
=
= °
= ( )
. m
. m
tan 55.0
. m tan(( . ) m55 0 3 41° = .
P1.54 Let d represent the diameter of the coin and h its thickness.
The mass of the gold is
m V At
d
dh t= = = +
⎛
⎝⎜
⎞
⎠⎟ρ ρ ρ
π
π
2
4
2
where t is the thickness of the plating.
m =
( )
+ ( )( )
⎡
⎣
⎢
⎤
⎦
⎥ ×19 3 2
2 41
4
2 41 0 178 0 18 1
2
.
.
. . .π π 00
0 003 64
0 003 64 10
4−
( )
=
= ×
.
. $
grams
cost grams grram cents= =$ . .0 036 4 3 64
This is negligible compared to $4.98.
P1.55 The actual number of seconds in a year is
86 400 s day 365.25 day yr 31557 600 s yr( )( ) = .
The percent error in the approximation is
π ×( )−( )
×
10 31557 600
31557 600
100
7
s yr s yr
s yr
% == 0 449. % .
P1.56 v =
⎛
⎝⎜
⎞
⎠⎟5 00
220
1
.
furlongs
fortnight
yd
furlongg
m
yd
fortnight
day
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟
0 914 4
1
1
14
.
ss
day
hrs
hr
s
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝
⎞
⎠
⎛
⎝⎜
⎞
⎠⎟ =
1
24
1
3 600
8 3. 22 10 4
× −
m s
This speed is almost 1 mmրs; so we might guess the creature was a snail, or perhaps a sloth.
P1.57 (a) The speed of rise may be found from
v =
( )
=
Vol rate of flow
(Area:
cm3
π D2
4
16 5
/ )
. ss
cm
cm s
π 6 30 4
0 5292
.
.
( )
=
/
.
(b) Likewise, at a 1.35 cm diameter,
v =
( )
=
16 5
1 35 4
11 52
.
.
.
cm s
cm
cm s
3
π /
.
FIG. P1.53
55°55°
h
rr
h
ISMV1_5103_01.indd 12ISMV1_5103_01.indd 12 10/27/06 12:27:10 PM10/27/06 12:27:10 PM
13. Physics and Measurement 13
P1.58 The density of each material is ρ
π π
= = =
m
V
m
r h
m
D h2 2
4
.
Al:
g
cm cm
g
cm3
ρ
π
=
( )
( ) ( )
=
4 51 5
2 52 3 75
2 752
.
. .
. TThe tabulated value
g
cm
is sma3
2 70 2. %⎛
⎝
⎞
⎠
lller.
Cu:
g
.23 cm .06 cm
g
cm3
ρ
π
=
( )
( ) ( )
=
4 56 3
1 5
9 362
.
. TThe tabulated value
g
cm
is sma3
8 92 5. %⎛
⎝
⎞
⎠
lller.
Brass:
4.4 g
.54 cm .69 cm
g
ρ
π
=
( )
( ) ( )
=
4 9
1 5
8 912 .
ccm3
Sn:
g
.75 cm .74 cm
g
cm3
ρ
π
=
( )
( ) ( )
=
4 69 1
1 3
7 682
.
.
Fe:
16.1 g
.89 cm .77 cm
g
cm
ρ
π
=
( )
( ) ( )
=
4 2
1 9
7 882 . 33 3
The tabulated value
g
cm
is7 86 0 3. . %⎛
⎝
⎞
⎠
ssmaller.
P1.59 V20 mpg
cars mi yr
mi gal
5.0 10= = ×
( )( )10 10
20
8 4
110
gal yr
V25 mpg
cars mi yr
mi gal
4.0 10= = ×
( )( )10 10
25
8 4
110
gal yr
Fuel saved gal yr25 mpg 20 mpg= − = ×V V 1 0 1010
.
P1.60 The volume of the galaxy is
π πr t2 21 2 19 61
10 10 10= ( ) ( )m m m3
~ .
If the distance between stars is 4 × 1016
m, then there is one star in a volume on the order of
4 10 1016 3 50
×( )m m3
~ .
The number of stars is about 10
10
10
61
50
11m
m star
stars
3
3
~ .
ANSWERS TO EVEN-NUMBERED PROBLEMS
P1.2 2.15 × 104
kgրm3
P1.4 2.3 × 1017
kgրm3
is twenty trillion times larger than the density of lead.
P1.6 0.141 nm
P1.8 (a) ii (b) iii (c) i
P1.10 9.19 nmրs
P1.12 (a) 3.39 × 105
ft3
(b) 2.54 × 104
lb
P1.14 (a) 0.071 4 galրs (b) 2.70 × 10−4
m3
րs (c) 1.03 h
ISMV1_5103_01.indd 13ISMV1_5103_01.indd 13 10/27/06 12:27:11 PM10/27/06 12:27:11 PM
14. 14 Chapter 1
P1.16 667 lbրs
P1.18 2 57 106 3
. × m
P1.20 (a) 2.07 mm (b) 8.57 × 1013
times as large
P1.22 (a) 13.4; (b) 49.1
P1.24 r rAl Fe
Fe
Al
=
⎛
⎝⎜
⎞
⎠⎟
ρ
ρ
1 3
P1.26 ~10 rev7
P1.28 No. There is a strong possibility that you would die before finishing the task, and you have much
more productive things to do.
P1.30 209 4 2
±( ) cm
P1.32 (1.61 ± 0.17) × 103
kgրm3
P1.34 31 556 926.0 s
P1.36 (a) 0.015 g (b) 30 cm2
(c) 8% (d) 5.2 gրm2
(e) For any and all shapes cut from this copy paper,
the mass of the cutout is proportional to its area. The proportionality constant is 5.2 gրm2
± 8%,
where the uncertainty is estimated. (f) This result is to be expected if the paper has thickness
and density that are uniform within the experimental uncertainty. The slope is the areal density of
the paper, its mass per unit area.
P1.38 19
P1.40 see the solution
P1.42 1.38 km
P1.44 either 3.46 or −3.46
P1.46 316 m
P1.48 (a) ρ = 2.70 gրcm3
+ 1.19 gրcm4
x (b) 1.39 kg
P1.50 (a) 478 cm3
րs (b) 0.225 cmրs (c) When the balloon radius is twice as large, its surface area is
four times larger. The new volume added in one increment of time in the inflation process is equal
to this larger area times an extra radial thickness that is one-fourth as large as it was when the bal-
loon was smaller.
P1.52 0.542 rad
P1.54 3 64. cents; no
P1.56 8 32 10 4
. × −
m s; a snail
P1.58 see the solution
P1.60 ~1011
stars
ISMV1_5103_01.indd 14ISMV1_5103_01.indd 14 10/27/06 12:27:11 PM10/27/06 12:27:11 PM