In this unit we will discuss properties of waves and types of waves. Moreover, we will try to explain situations which can not be explained with light properties of matter. Disturbance of the shape of the elastic matters are transported from one end to other by the particles of that matter, we call this process wave. Be careful, during the transportation, no matter is transported.
Waves are classified in different ways with their properties. For example, mechanical waves and electromagnetic waves are classified according to the medium they transport energy. Water waves and sound waves are examples of mechanical wave on the contrary, light waves, radio waves are examples of electromagnetic waves. Electromagnetic waves can propagate in vacuum but mechanical waves need a medium to transport energy.
Waves can propagate in 1D, 2D and 3D. Spring waves are examples of 1D waves, water waves are examples of 2D waves and light and sound waves are examples of 3D waves.
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Wednesday, February 9, 2011
Magnetism
In ancient times Greece people found a rock that attracts iron, nickel and cobalt. They call them as “magnet “and magnetism comes from here. These rocks were used later by Chinese people to make compasses. Later scientists found that, magnets have always two poles different from electricity. Magnets have two ends or faces called “poles” where the magnetic effect is highest. In last unit we saw that there is again two polarities in electricity, “-“charges and “+” charges. Electricity can exist as monopole but magnetism exists always in dipoles North Pole (represented by N) and South Pole (represented by S). If you break the rock into pieces you get small magnets and each magnet also has two poles N and S.
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ELECTRIC CURRENT
In the last unit we have learned static electric, or charge at rest. However, in this unit we will deal with the charges in motion. We see the electric current everywhere in daily life. Most of the electrical devices work with electric current. In this unit we will try to explain direction of the flow of current, ohm’s law, and resistance of the electric circuit, resistors, measuring the current, and current density
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Electrostatics with Examples
ELECTROSTATICS
Scientist found that if you rub an ebonite rod into silk you observe that rod pulls the paper pieces. Or in winter when you put off your pullover, your hair will be charged and move. We first examine the structure of atom to understand electricity better.
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Scientist found that if you rub an ebonite rod into silk you observe that rod pulls the paper pieces. Or in winter when you put off your pullover, your hair will be charged and move. We first examine the structure of atom to understand electricity better.
HEAT TEMPERATURE AND EXPANSION
In this unit we will learn some concepts like heat, temperature, thermal expansion, thermal energy and phases of matter. Moreoversome misconceptions about heat and temperature will be explained. Since they make confusions in many students’ mind, we give more importance on this subject. In daily life sometimes we use them interchangeably however, in physics they are totally different concepts.
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In this unit we will learn some concepts like heat, temperature, thermal expansion, thermal energy and phases of matter. Moreoversome misconceptions about heat and temperature will be explained. Since they make confusions in many students’ mind, we give more importance on this subject. In daily life sometimes we use them interchangeably however, in physics they are totally different concepts.
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HEAT TEMPERATURE AND EXPANSION
In this unit we will learn some concepts like heat, temperature, thermal expansion, thermal energy and phases of matter. Moreoversome misconceptions about heat and temperature will be explained. Since they make confusions in many students’ mind, we give more importance on this subject. In daily life sometimes we use them interchangeably however, in physics they are totally different concepts.
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Tuesday, February 8, 2011
PROPERTIES OF MATTER
Matter
Everything around us has mass and volume and they occupy space, and we called them as matter. It can be in four sate, like solid, liquid, gas and plasma. We will talk about main properties of matter in this unit like, mass, volume, density, elasticity, inertia...Etc. You can classify matters with their physical or observable properties and chemical or unobservable properties, for example their smells, colors, shapes give you an idea about it. On the contrary unobservable properties like conductivity of the matter cannot be understood from appearance or smells of the matter.
OPTICS
Optic is one of the branch of physics which deals with the light and properties of it. We know that light shows both the particle and wave characteristics. However, in this unit we will learn the particle characteristics of the light. Some of the topics will be covered in this unit are; reflection and refraction of light, plane mirrors, concave and convex mirrors, reflection of light from the mirrors, prisms and behavior of light in different mediums.
ROTATIONAL MOTION
Rotational motion or we can say circular motion can be analyzed in the same way of linear motion. In this unit we will examine the motion of the objects having circular motion. For example, we will find the velocity, acceleration and other concepts related to the circular motion in this section. Uniform circular motion is one of the example of this subject. In uniform circular motion speed of the object is always constant and direc
tion is changing. Thus, velocity of the object is changing and as a result object has acceleration
tion is changing. Thus, velocity of the object is changing and as a result object has accelerationMomentum with Examples
MOMENTUM
Look at the given pictures. If both the car and the truck have same speed, which one can be stopped first? Of course all you say, it is hard to stop truck relative to car. Well, what is the reason making car stop easier? They have same speed but different masses. Can mass effect the stopping time or distance? The answer is again YES! It is hard to stop heavier objects. What we are talking about so far is momentum.
Energy, Work, Power
In this unit we will deal with the energy, mechanical energy, potential energy, kinetic energy, conservation of energy theorem. Moreover, work and power also discussed in this unit. We will try to give examples related to each topic.
Read moreMechanics
Mechanics is one the most important subject in physics. You will learn many concepts and key points under mechanics title those will also be used in other subjects of physics. Mechanics can be defined as the behavior of the particle under any effects. It can be studied under main two topics which are kinematics concerning the motion of the particle and dynamics concerning the causes of motion.
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vectors
vectors addition of vectors components of vectors with examples
In physics and all science branches quantities are categorized in two ways. Scalars and vectors are used for to define quantities. We can use scalars in just indication of the magnitude, they are only numerical value of that quantity. However, if we talk about the vectors we should consider more than numeric value of the quantities. Vectors are explained in detail below.
Study Of Motion in Two Dimensions
Movement in Two Dimensions: This page focuses on several different scenarios involving motion in two dimensions.
Equilibrium and the Equilibrant:
As was said in the study of forces, an object is in a sate of equilibrium with respect to forces when the net force acting on it equals zero. [F(net) = 0 Newtons, in the metric system, or 0 lbs, in the English system.) When an object is in this state of equilibrium, it is either standing still with respect to a reference point or it moving with constant velocity with respect to a reference point. As we have already seen, an object that is in a state of equilibrium will NOT experience an acceleration.
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Equilibrium and the Equilibrant:
As was said in the study of forces, an object is in a sate of equilibrium with respect to forces when the net force acting on it equals zero. [F(net) = 0 Newtons, in the metric system, or 0 lbs, in the English system.) When an object is in this state of equilibrium, it is either standing still with respect to a reference point or it moving with constant velocity with respect to a reference point. As we have already seen, an object that is in a state of equilibrium will NOT experience an acceleration.
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Gas Laws
1. Kinetic Theory
2. Pressure
3. Boyle's Law
4. Charles' Law
5. Gay-Lussac's Law
6. Combined Gas Law
7. Ideal Gas Law
8. Real Gases
Notes should include:
Kinetic Theory: Whenever you study the properties of material substances, you will get into discussing what matter is made of, and how its structure varies among the different phases of matter, solid, liquid, and gas. The core principle in the study of matter and thus the nature of materials is the kinetic theory. The kinetic theory provides us with an explanation of how the particles that make up a substance behave. This is a valuable concept when studying gases. A number of sources tell you that the kinetic theory is based on three assumptions. They are:
1. Matter, regardless of type or form, is made up of very small particles.
2. These particles are constantly in motion, and generally this motion is random.
3. The particles themselves experience collisions with other particles and often with the
walls of a container, if confined to one. These collisions are considered to be elastic.
The kinetic theory applies to gases. When studying gases, particularly in a first course in physics, you generally consider an ideal gas. An ideal gas is a hypothetical gas made up of particles having mass, but no significant volume. Also the particles in the ideal gas exhibit no amount of attractive force for each other or the sides of containers, if so confined.
Pressure: The study of gases requires that you find a means of measuring quantities of gaseous material other than those commonly used for measuring quantities of materials in either the liquid or solid state. It isn't convenient to simply lay some gas on a balance to mass it or pore some into a graduate cylinder to find its volume. As a consequence, the measurements of Pressure, Temperature, and Volume are used together to describe the quantity of a gas present in a particular situation. You are more likely to be familiar with temperature and volume than you are with the measurement of pressure, though you may have used a pressure gauge in the past for the purpose of checking the pressure of something such as a bicycle tire. Pressure is the measure of the force exerted by the particles of a gas acting on, pressing against, or colliding with one unit area of measure on the surface of a container in which the gas is confined. For example, Pressure is often expressed in PSI, which stands for pounds per square inch. Notice that the phrase per square inch is one unit area of measurement. The SI unit for pressure is the pascal, abbreviated Pa. A pascal is equal to one newton per square meter. 1 Pa = 1 N/m2.
You will often see pressure expressed in kilopascals (kPa), because one pascal is a rather small measurement.
A barometer is a device used to measure atmospheric pressure. You should always remember that the weight of the atmosphere is pressing down on you and everything else on the earth's surface. This weight is the pressure of the atmosphere. On the average the atmospheric pressure near the surface of the earth is 14.7 psi. As with most measurements, pressure requires that there be some ãstandard reference point or measurement to which (or upon which) all other measurements are compared (or defined). When working with pressure you refer to standard atmospheric pressure. Standard atmospheric pressure is defined as the amount of pressure exerted by the atmosphere that will support a column of mercury that is 760 mm above the pool of mercury in a barometer. This measurement is reported as 760 mm Hg. Expressed in pascals this would be 1.013 x 105 Pa or in kilopascals 101.3 kPa. The aneroid barometer, which often looks like a round pressure gauge, works by monitoring the effect atmospheric pressure has on a small chamber from which most, if not all, of the air has been removed. A mechanical linkage that monitors the change in volume of this chamber as the atmospheric pressure changes moves the needle that sits on the numbered scale, thus allowing you to read the atmospheric pressure right off the face of this type of barometer.
Boyles' Law: Back in the 17th century, a British scientist by the name of Robert Boyle discovered the relationship between the pressure and the volume of a gas. This relationship has become well known to chemistry and physics students as Boyle's Law. His law says that the volume of a gas varies inversely with the pressure of the gas, if the temperature of the gas remains constant. The equation for this law can be written as Pi Vi = Pf Vf. (i = initial measurement and f = final measurement) The relationship can also be written as PV = C, where C represents a numerical constant. The letter k could also be used to represent the constant. If you were to graph the data that supports this law, you would get a first quadrant hyperbola.
Charles' Law: Near the end of the 18th century a scientist by the name of Jacques Charles discovered the relationship between temperature and the volume of a gas. This relationship has also become well known to chemistry and physics students. It is referred to as Charles' Law. This law says that the volume of a gas varies directly with the temperature of the gas, if the pressure of the gas remains constant. The equation for this law is Vi / Vf = Ti / Tf. If you were to graph data that supports this law, you would get a straight line.
Gay-Lussac's Law: In the early 19th century a scientist by the name of Joseph Gay-Lussac discovered the relationship between temperature and pressure. This law says that the pressure of a gas varies directly with the pressure of the gas, if the volume of the gas remains constant. The equation for this law is Pi / Pf = Ti / Tf. If you were to graph data that supports this law, you would get a straight line.
A noteworthy comment: Boyle's, Charles', and Gay-Lussac's laws are not really laws in the sense that we use the term law today. Rather, they are reasonable approximations that are considered correct for real gases as long as they are not too close to their condensation points and their pressures and densities are not very high. However, because calling these statements laws has a strong historical basis, we will continue to call them laws. Remember, for the purpose of learning relationships, we are tending to ignore the effect of attractive forces on the behavior of gases. We are treating them ideally, with certain limitations placed on the conditions under which we are studying them. One of those conditions involves completely elastic collisions.
Combined Gas Law: Since it is not always likely that one of the three variables of temperature, pressure, and volume will remain constant a relationship that allows for all three to change simultaneously is needed. This relationship is called the combined gas law. The equation for this law can be written as Pi Vi / Ti = Pf Vf / Tf. If you were to state this relationship in words, you could say that the product of the pressure and volume of a gas varies directly with its temperature.
ãIdeal Gas Law: Decreasing the volume and / or increasing the temperature of a gas are not the only means of affecting the pressure of a gas. Up until now we have ignored the effect that the number of particles has on the pressure of a gas. For example, consider the effect of adding more air (molecules) to a bicycle or automobile tire. As you do so, the pressure goes up. This happens because when the temperature and volume of a gas are held constant but the number of particles is increased the density of the gas increases causing more collisions in the same amount of space including collisions with the surface are of the wall of the container in which the gas is being held. This means that the pressure has increased, as the number of collisions against the wall per unit area represent a way of expressing the force pushing on that wall.
It can also be said that, if the temperature of a gas remains constant, the product of its pressure and volume varies directly with the number of molecules of gas present. A mathematician would write this relationship as PV a n , where n is the number of particles (molecules) present. The symbol "a" means "varies directly with". The number of particles are usually expressed using a unit called the mole. A mole of a substance is equal to 6.02 x 1023 particles or molecules. This rather large number is called Avogadro's number. For example, 0.5 moles of oxygen molecules equals 3.01 x 1023 molecules of oxygen. Likewise, 2.0 moles of carbon dioxide molecules equals 1.204 x 1024 molecules of carbon dioxide molecules.
[In the 19th century Amedeo Avogadro, an Italian scientist came up with the following hypothesis. He said "equal volumes of gases at the same pressure and temperature contain equal numbers of molecules". The term mole itself is defined as the amount of substance that contains as many atoms or molecules as there are in 0.012 kg of carbon - 12. Avogadro was not able to determine the actual number of particles in the mole during his lifetime. It required 20th century technology to do so. However, the idea is his, so to honor him, we name the number after him.]
Since the product PV is also varies directly with temperature, a more expansive relationship may be expressed that says the product of the pressure and volume for a gas varies directly with the product of its temperature and the amount of gas present. Mathematically this can be written as PV µ nT. From this statement the mathematical relationship can be written in the form of a ratio that is equal to a numerical constant. This relationship can be expressed as PV / nT = k , where k is a numerical constant called the Boltzman's constant. The Boltzman's constant requires that the value for n be expressed in terms of the actual number of molecules of gas present. The relationship can also be expressed as PV / nT = R. The letter k is replaced with the letter R. In this situation the value for n is expressed in terms of the number of moles of gas present. The letter R is sometimes referred to as the universal (gas) constant. This equation is often rearranged and written as PV = nRT. This is the form of the equation is most often written to represent the ideal gas law. It allows you to work with four variables simultaneously.
[The value for R will depend upon what units you use for P, V, n, T. In high school chemistry books, R is often given as 0.082 L Atm / mol K. This requires that P be expressed in atmospheres, V in Liters, n in number of moles, and T in kelvins. In SI units, R is expressed as 8.315 J / mol K. If you are trying to figure out how the product of P and V could get you a value of 8.315 J, you need to remember that in SI units Pressure is expressed in N / m2 and Volume is expressed in m3. When you multiply the two together you get the N m. The N m is the same as the joule. (Work = Force x distance)]
[To find the value for Boltzman's constant, k, you would need to take the value for R expressed in SI units, which is a value expressed per mole and divide it by Avogadro's number to get the value ãper individual molecule. Thus k is equal to 1.38 x 10-23.]
Real Gases: At low temperatures and high pressures real gases deviate from the laws we are in the process of studying. As long as the intermolecular forces are having little impact on the motion of the molecules and the collisions are relatively elastic, real gases behave pretty much the way that these relationships suggest. The problem occurs as gases cool down approaching their condensation temperatures and their low kinetic energies allow for intermolecular forces to have a significant effect and when high pressures (high densities) force the molecules very close together and allow for intermolecular forces to also have a significant effect.
[The critical temperature of a substance is the temperature to which a gaseous substance must be cooled before it can be liquified by pressure. The critical pressure of a substance is the pressure needed to liquify a gaseous substance that is at its critical temperature. At temperatures above the critical temperature a substance is usually called a gas. At temperatures below the critical temperature it is usually called a vapor, assuming it has not in liquid or solid state. The triple point of a substance is an exact combination of a single temperature and pressure measurement where all three phases of the substance can exist simultaneously in a state of physical equilibrium.]
Vocabulary: gas, temperature, standard temperature, volume, standard volume, pressure, atmospheric pressure, standard atmospheric pressure, STP, mole, density, Pascal (the unit), Atmosphere (the unit), Liter, barometer, manometer, elastic collision, Boyle’s Law, Charles’ Law, interpolation, extrapolation, Combined gas law, Ideal Gas Law, universal gas constant, Boltzmann constant, real gas, Avogadro’s number
Skills to be learned:
Solve Pressure Conversion Problems
Solve Boyle's Law Problems
Solve Charles' Law Problems
Solve Gay-Lussac's Law Problems
Solve Combined Gas Law Problems
Solve Ideal Gas Law Problems
Assignments:
Textbook: NA, Sources will be announced
WB Exercises: PS#13-1, PS#13-2, PS#13-3 and PS#13-4
Activities: TBA
Resourses:
This Handout and the Overhead and Board Notes discussed in class
WB Lessons and Problem Sets
www.physicsphenomena.com - “Gas Laws”
Study Of Waves, Energy, Light, and Sound
Waves and Energy: Energy can be transferred by waves.
Categories of Waves:
1. Mechanical Waves: This type of wave requires a medium (material) to travel through. Examples of such waves would be waves on a jump rope, a slinky, surface waves on water, and sound waves.
2. Electromagnetic Waves: This type of wave requires no medium (substance) to travel through. Visible Light and X-rays are examples of two kinds of electromagnetic waves.
Three General Types of Waves:
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Categories of Waves:
1. Mechanical Waves: This type of wave requires a medium (material) to travel through. Examples of such waves would be waves on a jump rope, a slinky, surface waves on water, and sound waves.
2. Electromagnetic Waves: This type of wave requires no medium (substance) to travel through. Visible Light and X-rays are examples of two kinds of electromagnetic waves.
Three General Types of Waves:
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Study Of Motion in Two Dimensions
Movement in Two Dimensions: This page focuses on several different scenarios involving motion in two dimensions.
Equilibrium and the Equilibrant:
As was said in the study of forces, an object is in a sate of equilibrium with respect to forces when the net force acting on it equals zero. [F(net) = 0 Newtons, in the metric system, or 0 lbs, in the English system.) When an object is in this state of equilibrium, it is either standing still with respect to a reference point or it moving with constant velocity with respect to a reference point. As we have already seen, an object that is in a state of equilibrium will NOT experience an acceleration.
Consider the diagram below. If Vector A and Vector B are added the sum is the resultant. The Equilibrant is the vector that is equal in magnitude to the Resultant, but opposite in direction. When the three vectors, Vector A, B and C are added the sum is 0 Newtons and the object is in a state of equilibrium with respect to the forces acting on it. The object is either at rest or moving with constant velocity. According to Newton's laws of motion and object in a state of equilibrium would not experience any acceleration since the F(net) would be zero newtons.
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Equilibrium and the Equilibrant:
As was said in the study of forces, an object is in a sate of equilibrium with respect to forces when the net force acting on it equals zero. [F(net) = 0 Newtons, in the metric system, or 0 lbs, in the English system.) When an object is in this state of equilibrium, it is either standing still with respect to a reference point or it moving with constant velocity with respect to a reference point. As we have already seen, an object that is in a state of equilibrium will NOT experience an acceleration.
Consider the diagram below. If Vector A and Vector B are added the sum is the resultant. The Equilibrant is the vector that is equal in magnitude to the Resultant, but opposite in direction. When the three vectors, Vector A, B and C are added the sum is 0 Newtons and the object is in a state of equilibrium with respect to the forces acting on it. The object is either at rest or moving with constant velocity. According to Newton's laws of motion and object in a state of equilibrium would not experience any acceleration since the F(net) would be zero newtons.
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Light, An Electromagnetic Wave Phenomenon:
1. Light Travels in a Straight Line
2. The Speed of Light
3. Transmission and Absorption of Light
4. Illumination by a Point Source
5. Light - An Electromagnetic Wave
6. Light and Color Vision
7. Polarization of Light
8. Interference in Thin Films
Notes should include:
Light Travels in a Straight Line: Light travels in a straight line. Have you ever noticed light rays being made visible by moisture, dust, or smoke in the air? You can see the rays and the fact that light does travel in a straight line. If it were not for diffraction and the scattering of light, the effects of light traveling in straight lines would be more obvious to us. For example, if light rays were not diffracted around corners and scattered as they reflect off of objects, imagine how dark areas that were in shadow would actually be. The straight lines that are used to represent light waves are called rays. Diagrams that are drawn using rays to represent light waves in order to represent the behavior of light are called ray diagrams and the use of ray diagrams to study light is called ray optics.
The Speed of Light: Light is so fast that people once thought it traveled from one point to another instantaneously. Around 1676, an astronomer by the name of Olaf Roemer observed eclipses of one of Jupiter's moons as it disappeared behind the planet and reappeared a while later. He made a number of careful time measurements of this phenomenon. He made observations over several months and realized that he was getting an increasing error between his measurements nightly and a table he had initially constructed which provided a projection of the data that he expected to collect over time. He concluded that, though the planet Jupiter moved only a little in its large orbit, the Earth had moved significantly further along its path, which took it further and further from Jupiter each day that passed during the observations. This meant that the light reflected off of the moon being observed had to travel a greater distance to get to Earth as each day passed. From his observations, Olaf concluded that light must have some fixed speed.
As the nineteenth century ended and the twentieth began a scientist by the name of Albert A. Michelson came up with a means of getting a reasonable measurement of the speed of light. He set out to measure the time for light to complete a round trip between two mountains that were 35 kilometers apart. That is a total distance of 70 kilometers. Remember that, because light travels quite fast, a relatively long distance is needed to measure its speed. On one mountain he set up a rotating octagonal mirror. He aimed a beam of light towards this rotating mirror on the top of this first mountain. On the second mountain were two mirrors side by side both partially facing one another and both ãpartially facing the first mountain, such that a beam of light coming off of the rotating mirror would bounce (reflect) off one mirror onto the other and then bounce (reflect) off the second mirror right back towards the rotating octagonal mirror on the first mountain. An observer was placed in line with the source of the light beam and the octagonal mirror. The beam of light was only visible to the observer when the mirror was rotated at the proper speed. Any other speed rather than the correct one would result in the beam being reflected in some direction other than the position of the observer. The motor turning the mirror was started from rest and sped up until the light beam was seen by the observer. From the rate of rotation of the mirror the speed of light was computed.
Transmission and Absorption of Light: Materials that allow light to pass through them are said to transmit light. If you can see through these materials they are called transparent. If they allow light to pass through them, but you cannot see objects through them they are called translucent. A luminous object gives off (emits) light waves. If a body reflects light it is said to be an illuminated object. The brightness of a light source is expressed using a measurement called luminous intensity. The unit of measure for luminous intensity is the candela. This is the international unit for luminous intensity measurement. One candela is the brightness of a light source consisting of one square centimeter of a white thorium oxide powder heated to a temperature of 2046 kelvins. The rate at which light is emitted from a source is called the luminous flux. Luminous flux is measured using the unit called the lumen (a unit of power). One lumen is the light energy being emitted from a point source of one candela intensity and flowing outwards through a one square meter area on a transparent sphere surrounding the point source having a diameter of one meter. Light can be absorbed as well as emitted. The rate at which light energy falls on one unit area of surface is called illuminance. Illuminance is measured in the units of lumens per square meter. Because light from a light source spreads out as it travels outwards from a light source the light dims, that is, with distance the illumination decreases. This means that the illuminance of a surface varies inversely as the square of the distance between the light source and the surface. To increase the illumination of a surface, you must either increase the intensity of the light source or decrease the distance between it and the surface it is illuminating.
Light, An Electromagnetic Wave: Visible Light is one kind of a large category of electromagnetic waves. All properties of waves apply to light waves. A major characteristic of electromagnet waves is that unlike mechanical waves, they require no medium through which to travel. They can travel through a vacuum, such as in space. The whole continuous spectrum of light (electromagnetic) waves includes gamma rays, x-rays, ultra violet, visible, infra red, micro, and radio. These waves are listed here from the highest frequency and shortest wavelength electromagnetic waves to the lowest frequency and the longest wavelength electromagnetic waves.
Light and Color Vision: Issac Newton may be among the first to have seen that a glass prism will break white light into its constituent colors of the rainbow (ROY G BIV). This arrangement of colors is called the visible spectrum. White light is a mixture of the colors of the rainbow. If you were to mix the colors back together you would get white light ãagain. What happens when you see color in an object in the presence of white light? What you are seeing is the color that is not absorbed by the material upon which the white light is shining. The color you see is the wavelength(s) of light reflected back to your eye. For example, red cloth contains dyes, which contain chemical pigments that absorb all colors of visible light except red. The red color is seen because the red is the wavelength of visible light not being absorbed. In a pure blue light source, what color would you see when you looked at the cloth that looked red in white light?
Polarization of Light: When a wave is passed along a medium such as a rope, you can imagine that the waves could pass through an imaginary slot if the wave is oriented in the same direction as the slot. Otherwise the rope wave would be caught or stopped by the slot and not allowed to continue beyond the slot. The concept of polarization of light is similar. Imagine that ordinary light rays are like a bunch of ropes generating waves with all sorts of different orientations. Now imagine a polarizing filter to be like a slot. The only waves that will get through are those that are oriented correctly. They must match the slot or in the case of light waves the orientation of the polarizing material, such as is found in sunglasses having polarizing lenses. Instead of calling a polarizing filter a slot it is called a filter and the orientation of the waves that pass through this filter are said to lie along a particular plane. These waves are said to be plane polarized, because they have the proper orientation. If you wear polarized sunglasses, you will probably notice that the light passing through the filtering lenses is dimmer, because on the average only about half of the rays get through.
Interference in Thin Films: If you have every noticed that soap film and films of oily materials often appear to have colors of the rainbow spread across their surface. This occurs because the thin film varies in thickness and the thickness determines which color of light will reflect back to you eyes. For simplicity imagine a very, very thin prism. As you move from the narrow tip downwards towards the base the prism increases in thickness. Anywhere that the thickness of the prism is odd number multiples of one quarter of a wave length of a specific color (1 x 1/4l, 3 x 1/4l, 5 x 1/4l, etc.) you will see a line or band of that color only. This is caused by a combination of two different reflected rays. When a light ray strikes the surface of a transparent material a portion of it is reflected and if the medium is denser than the air through which the light ray was traveling before it reached the material the reflected wave is inverted. A portion of the light ray will enter the material and be reflected off of the other side and return essentially along its original path. As this internally reflected light ray emerges from the material it will experience interference with the externally reflected portion of the original light ray.
If the two reflected waves are in phase they would cancel at this point because one is inverted. However, if the internally reflected wave is one half wavelength out of phase with the externally reflected wave they will reinforce each other and you will see the color having that particular wavelength and not those colors which were canceled by interference. The positions along the surface of the material where a specific color appears as a line or band are the points where the thicknesses are the odd number multiples of one fourth of the wavelength for that color. Since very thin prisms or films ãvary in thickness, you will see many different lines or bands of color on the surface, because there is likely to be at least one position with the correct thickness for each color somewhere on the film. It is also highly likely that you would see many repeating lines or bands of color, because there are likely to be multiple positions of the same thickness in the film. If the film is actually moving, you are likely to see the lines or bands of colors swirling around as you do in a soap or oil film.
The reason that odd number multiples of one quarter of a color's wavelength is required for the thickness of the film for lines or bands of color to be produced has to do with the length of the internally reflected ray's path. For the interference between the two reflected waves to produce a reinforced wave rather than a diminished or canceled wave the internally reflected wave must be one half of a wavelength out of phase with the externally reflected wave. This requires that the path length of the internally reflected wave be 1/2 of a wavelength more than some multiple of the whole wavelength so the two waves will be out of phase by one half of a wavelength. Because the wave travels first forward through the medium and then backwards through the medium after reflection, the material need only be 1/4 of a wavelength thicker than some multiple of a wavelength for the wave to have traveled the extra 1/2 wavelength. To help you understand this process always remember that, if the thickness of the film were to be simple multiples of a wavelength, the internally reflected wave would be in phase with the externally reflected and inverted wave. This would result in the complete canceling of the two reflected waves because crests and troughs of equal magnitude but opposite in direction would be meeting and canceling one another.
Vocabulary: light, ray model, luminous, illuminated, luminous flux, lumen, illuminance, lux, candela, luminous intensity, transparent, translucent, opaque, spectrum, primary color, secondary color, complementary color, dye, pigment, primary pigment, secondary pigment, thin-film interference, polarized
Skills to be learned:
Solve problems involving the speed of light.
Solve problems involving luminous intensity and illuminance.
Assignments:
Textbook: Read / Study / Learn Chapter 16 about light
WB Exercise(s):
Activities: TBA
References:
This Handout and Overhead and Board Notes discussed in class
Textbook Chapter 16
WB Lessons and Problem Sets
© www.physicsphenomena.com / Light, An Electromagnetic Phenomena
Study Of Waves, Energy, Light, and Soun
Waves and Energy: Energy can be transferred by waves.
Categories of Waves:
1. Mechanical Waves: This type of wave requires a medium (material) to travel through. Examples of such waves would be waves on a jump rope, a slinky, surface waves on water, and sound waves.
2. Electromagnetic Waves: This type of wave requires no medium (substance) to travel through. Visible Light and X-rays are examples of two kinds of electromagnetic waves.
Three General Types of Waves:
1. Transverse Wave: This type of wave causes the particles in a medium to vibrate perpendicular to the direction in which the the wave moves. Two examples are that of waves sent along a jump rope or along a spring such as a slinky.
2. Longitudinal Wave: This type of wave causes the particles in a medium to vibrate parallel to the direction in which the wave moves. Two examples are a compression wave moving along a spring (produced by compressing and then releasing a few coils of a spring) and a sound wave.
3. Surface Wave: This type of wave has characteristice of both the transverse wave and the longitudinal wave. Surface waves cause particles on the surface of a medium to move both horizontally and verticlly in sort of a rolling, cicular motion as the wave moves forward. An example are the surface waves moving across a body of water.
More Facts about waves: A medium vibrates in response to a wave moving along through the material, but the medium itselve does not move with the wave. A pulse is a single didturbance in the medium. A single ripple spreading out across a pond would be an example of a single pulse. A wave train is a series of pulses at regular intervals. A series of waves rolling up on a shore is an example of a wave train.
Common Terminology: A wave crest is the highest point on a transverse wave or the point of maximum compression on a longitudinal wave. A wave trough is the lowest point on a transverse wave or the point of least compression on a longitudinal wave. The wavelength of a wave is the distance between corresponding points on consecutive waves. The wave height of a wave is the maximum displacement of the wave from the rest or equilibrium position when no wave is present.
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Categories of Waves:
1. Mechanical Waves: This type of wave requires a medium (material) to travel through. Examples of such waves would be waves on a jump rope, a slinky, surface waves on water, and sound waves.
2. Electromagnetic Waves: This type of wave requires no medium (substance) to travel through. Visible Light and X-rays are examples of two kinds of electromagnetic waves.
Three General Types of Waves:
1. Transverse Wave: This type of wave causes the particles in a medium to vibrate perpendicular to the direction in which the the wave moves. Two examples are that of waves sent along a jump rope or along a spring such as a slinky.
2. Longitudinal Wave: This type of wave causes the particles in a medium to vibrate parallel to the direction in which the wave moves. Two examples are a compression wave moving along a spring (produced by compressing and then releasing a few coils of a spring) and a sound wave.
3. Surface Wave: This type of wave has characteristice of both the transverse wave and the longitudinal wave. Surface waves cause particles on the surface of a medium to move both horizontally and verticlly in sort of a rolling, cicular motion as the wave moves forward. An example are the surface waves moving across a body of water.
More Facts about waves: A medium vibrates in response to a wave moving along through the material, but the medium itselve does not move with the wave. A pulse is a single didturbance in the medium. A single ripple spreading out across a pond would be an example of a single pulse. A wave train is a series of pulses at regular intervals. A series of waves rolling up on a shore is an example of a wave train.
Common Terminology: A wave crest is the highest point on a transverse wave or the point of maximum compression on a longitudinal wave. A wave trough is the lowest point on a transverse wave or the point of least compression on a longitudinal wave. The wavelength of a wave is the distance between corresponding points on consecutive waves. The wave height of a wave is the maximum displacement of the wave from the rest or equilibrium position when no wave is present.
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Study Of Motion
A. Describing, Measuring, and Displaying Motion Scenarios: In everyday conversation, as well as in physics, math, and engineering, the motion of objects are discussed. The most common discussions focus on either constant speed or acceleration scenarios. In everyday conversation usually a particular object is being discussed ranging from a skate board to an automobile or some other form of motorized transportation. In the sciences and in engineering expressing information about constant speed, velocity (not exactly the same as speed), and acceleration are concepts that are quantified and need to be measured, recorded, and displayed. Two common methods of displaying such observations (generally called data) is through the use of ticker tape measurements and vectors. You may click on either of the two links below to obtain more information about these methods of expressing data about motion.
1. Motion described using "Ticker Tape"
2. Motion described using "Vectors"
B. "Motion in One Dimension" – Notes on motion and solving problems involving motion.
1. Displaying Problem Solving Information (including variable symbols)
Displaying Problem Solving Information1. All problem solving solutions use a five step problem solving format consisting of a(n):
Given statement, G:
Find statement, F:
Equation statement, E:
Substitution statement, S:
Answer statement, A:
2. A bar over a variable symbol is used to indicate an average value.
3. A subscript of “aver”, or ave enclosed in parentheses, (ave), following a variable symbol could also be used.
4. A scalar quantity variable is printed in physics text books in non bold type.
5. A vector quantity variable is printed in physics text books in bold type, sometimes with an arrow above the symbol.
2. Constant Speed and Velocity
Constant Speed is the measure of the rate of "how fast" an object is moving without concern for the direction in which the object is traveling.
Speed belongs to a category of measurements called scalar quantities. A scalar quantity only answers the quantitative question "how much?".
Constant Velocity is a measure of both an object's speed and the direction in which the object is traveling.
Velocity belongs to a category of measurement called vector quantities. A vector quantity answers two questions, which are "how much?" and "in what direction?". Direction is most often expressed as an angle measured from some reference point.
--> Example of Average Speed
Calculation of Average SpeedA bicycle rider travels 22 km in 3.25 hours. What is the average speed of the bicycle rider? The magnitude of v(ave) is average speed
G: distance (d) = 22 km; time interval (t) = 3.25 hours
F: Average speed, v(ave)
E: v(ave) = d / t
S: v(ave) = 22 km / 3.25 h
A: v(ave) = 6.8 km / h
--> Example of calculating Displacement during Constant Velocity
Displacement is the distance (a scalar measurement) an object has moved and the direction the object has moved. It is a vector measurement. Displacement is also called "change in position".
Calculation of Displacement during Constant VelocityA vehicle moves along at 60 km / h for a time interval of 30 minutes due North (directly North). What is the displacement of the vehicle?
G: v(ave) = 60 Km / h, t = 30 minutes; Note 30 minutes is 1/2 hour or 0.5 h
F: Displacement, d
E: d = v(ave) t
S: d = [60 km / h] [0.5 h]
A: d = 30 km
3. Uniform (constant) Acceleration
Acceleration is also a vector quantity and represents a measure of the rate of change in the velocity of an object. Variable symbols representing vector quantity measurements can be written with an arrow placed over the variable symbol or in a typed font printed in bold type.
--> Example of Average Velocity
Calculation of Average VelocityA high speed train takes 2.00 hours to go from Paris 454 km due south to Lyons. What is the average velocity of the train?
G: displacement (d) = 454 km, S; time interval (t) = 2.00 hr
F: Average velocity, v
E: v = d / t
S: v = 454 km, S / 2.00 hr
A: v = 227 km / hr, S
--> Example of Acceleration
Acceleration is the change in velocity divided by the interval of time in which the change occurs. The unit of acceleration is m/s^2 or km/h^2, etc.
Calculation of AccelerationThe velocity of an automobile increases from 0 to 14 m/s, E, in 3.5 s. What is its acceleration?
G: initial velocity v(i) = 0 m/s ; final velocity v(f) = 14 m/s, E ; time interval (t) = 3.5 s
F: Acceleration (a)
E: a = [v(f) - v(i)] / t
S: a = (14 m/s – 0 m/s) / 3.5 s
A: a = 4.0 m/s^2, E
--> Example of Negative Acceleration
Calculation of Negative AccelerationAn automobile slows from 14.0 m/s, E, to 7.0 m/s, E, in 20 s. Find its acceleration?
G: intial velocity v(i) = 14.0 m/s, E ; final velocity v(f) = 7.0 m/s, E ; time interval (t) = 2.0 s
F: Acceleration (a)
E: a = [v(f) - v(i)] / t
S: a = (7.0 m/s – 14 m/s) / 2.0 s
A: a = -3.5 m/s^2, E
4. Other variations of Uniform Acceleration Problems
Final Velocity can be found from the initial velocity, acceleration, and time interval.
--> Example of calculating Final Velocity after uniform acceleration from initial velocity, acceleration, and time
Final Velocity after Uniform AccelerationA ball rolling down an incline for 5.0 s undergoes a uniform acceleration of 4.2 m/s^2. If the ball has an initial velocity of 2.0 m/s when it starts down the incline, what is its final velocity?
G: time (t) = 5.0 s ; acceleration (a) = 4.2 m/s^2 ; initial velocity v(i) = 2.0 m/s forward down the incline
F: final velocity v(f)
E: v(f) = v(i) + at
S: v(f) = 2.0 m/s + 4.2 m/s^2 (5.0 s)
A: v(f) = 23 m/s, forward down the incline
The average velocity of a uniformly accelerating object is the middle velocity as in v(ave) = [v(f)+ v(i)] / 2
Remember that Displacement is the distance (a scalar measurement) an object has moved and the direction the object has moved. It is a vector measurement. Displacement is also called "change in position".
The displacement of an object can be found from its initial and final velocities and the time interval over which the change occurs.
--> Example of calculating Displacement during Uniform Acceleration from final velocity, initial velocity, and time
Displacement during Uniform AccelerationWhat is the displacement of a train as it is accelerated uniformly from 22 m/s to 44 m/s, west in a 20.0 s time interval?
G: initial velocity v(i) = 22 m/s, W ; final velocity v(f) = 44 m/s, W ; time interval (t) = 20.0 s
F: displacement, d
E: d = [ ( v(f) + v(i) ) / 2] t
S: d = [ (44 m/s + 22 m/s) / 2 ] (20.0 s)
A: d = 660 m, W
The displacement can also be found from the initial velocity, acceleration, and time interval
--> Example of calculating Displacement during Uniform Acceleration from initial Velocity, Acceleration, and Time
Displacement during Uniform AccelerationA car starting from rest is accelerated at 6.1 m/s^2, S. What is the car’s displacement during the first 7.0 s of acceleration?
G: intial velocity (vi) = 0 m/s ; acceleration (a) = 6.1 m/s^2 ; time interval (t) = 7.0 s
F: displacement (d)
E: d = d(i) + v(i) t + 0.5 a t^2
S: d = 0 m + 0 m/s (7.0 s) + 0.5 (6.1 m/s^2) (7.0 s)^2
A: d = 150 m, S
Final velocity can also be found from the initial velocity, acceleration, and distance traveled. And Acceleration can be found from the initial velocity, final velocity, and the distance traveled.
--> Example of calculating Acceleration during Uniform Acceleration from final velocity, intial velocity, and displacement
Calculating AccelerationAn airplane must achieve a velocity of +71 m/s for takeoff. If the runway is 1.0 x 10^3 m long, what must the acceleration be?
G: initial velocity (vi) = 0 m/s ; final velocity (vf) = +71 m/s ; displacement (d) = 1,000 m
F: acceleration (a)
E: a = [ v(f)^2 – v(i)^2 ] / [ 2 d ]
S: a = [ (71 m/s)^2 – (0 m/s)^2 ] / [ 2 x 1000 m ]
A: +2.5 m/s^2
The displacement is the same if the initial and final velocities are exchanged and acceleration is changed to deceleration
--> Example of finding Displacement when Velocities and Times are known
Calculating DisplacementDetermine the displacement during constant (uniform) acceleration of an airplane that is accelerated from +66 m/s to +88 m/s in 12.0 s.
G: initial velocity v(i) = +66.0 m/s ; final velocity v(f) = +88.0 m/s ; time (t) = 12.0 s.
F: displacement, d
E: d = [ ( v(f) + v(i) ) / 2 ] t
S: d = [ (88.0 m/s + 66.0 m/s) / 2 ] [ 12 s ]
A: d = +924 m
Acceleration due to gravity problems requires the use of the same basic equations, but the acceleration symbol a is changed to a g for the acceleration due to gravity.
--> Example of a falling body (object) problem
Calculations pertaining to Falling BodiesA 20 kg object is suspended 100 m in the air by a large construction crane. The mass is released and allowed to fall to the ground.
a. How long will it take for the object to reach the ground?
b. What is the final velocity of the object just before it strikes the ground?
Solution to part a:
The equation d = v(i) t + 0.5 g t^2 allows you to solve for the time. Since the initial velocity for an object that is dropped is zero, the vi t in the equation will drop out. The d = 0.5 g t^2 portion of the equation is used to solve for the time.
E: t = (2 d / g )^0.5
S: t = [ 2 (100 m) / 9.8 m/s^2 ]^0.5
A: t = 4.52 s
Solution to part b:
Note: The acceleration downwards is the acceleration due to gravity, so a = g.
E: v(f) = v(i) + a t
S: v(f) = 0 m/s + 9.8 m/s^2 (4.52 s)
A: v(f) = 44.3 m/s, downwards
It would be proper to express the answer as a -44.3 m/s, where the negative in front of the number implies downwards. The use of the negative for the downwards direction would be valid in situations defined by you or others where up was defined as being the positive direction and down was defined as being the downward direction.
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1. Motion described using "Ticker Tape"
2. Motion described using "Vectors"
B. "Motion in One Dimension" – Notes on motion and solving problems involving motion.
1. Displaying Problem Solving Information (including variable symbols)
Displaying Problem Solving Information1. All problem solving solutions use a five step problem solving format consisting of a(n):
Given statement, G:
Find statement, F:
Equation statement, E:
Substitution statement, S:
Answer statement, A:
2. A bar over a variable symbol is used to indicate an average value.
3. A subscript of “aver”, or ave enclosed in parentheses, (ave), following a variable symbol could also be used.
4. A scalar quantity variable is printed in physics text books in non bold type.
5. A vector quantity variable is printed in physics text books in bold type, sometimes with an arrow above the symbol.
2. Constant Speed and Velocity
Constant Speed is the measure of the rate of "how fast" an object is moving without concern for the direction in which the object is traveling.
Speed belongs to a category of measurements called scalar quantities. A scalar quantity only answers the quantitative question "how much?".
Constant Velocity is a measure of both an object's speed and the direction in which the object is traveling.
Velocity belongs to a category of measurement called vector quantities. A vector quantity answers two questions, which are "how much?" and "in what direction?". Direction is most often expressed as an angle measured from some reference point.
--> Example of Average Speed
Calculation of Average SpeedA bicycle rider travels 22 km in 3.25 hours. What is the average speed of the bicycle rider? The magnitude of v(ave) is average speed
G: distance (d) = 22 km; time interval (t) = 3.25 hours
F: Average speed, v(ave)
E: v(ave) = d / t
S: v(ave) = 22 km / 3.25 h
A: v(ave) = 6.8 km / h
--> Example of calculating Displacement during Constant Velocity
Displacement is the distance (a scalar measurement) an object has moved and the direction the object has moved. It is a vector measurement. Displacement is also called "change in position".
Calculation of Displacement during Constant VelocityA vehicle moves along at 60 km / h for a time interval of 30 minutes due North (directly North). What is the displacement of the vehicle?
G: v(ave) = 60 Km / h, t = 30 minutes; Note 30 minutes is 1/2 hour or 0.5 h
F: Displacement, d
E: d = v(ave) t
S: d = [60 km / h] [0.5 h]
A: d = 30 km
3. Uniform (constant) Acceleration
Acceleration is also a vector quantity and represents a measure of the rate of change in the velocity of an object. Variable symbols representing vector quantity measurements can be written with an arrow placed over the variable symbol or in a typed font printed in bold type.
--> Example of Average Velocity
Calculation of Average VelocityA high speed train takes 2.00 hours to go from Paris 454 km due south to Lyons. What is the average velocity of the train?
G: displacement (d) = 454 km, S; time interval (t) = 2.00 hr
F: Average velocity, v
E: v = d / t
S: v = 454 km, S / 2.00 hr
A: v = 227 km / hr, S
--> Example of Acceleration
Acceleration is the change in velocity divided by the interval of time in which the change occurs. The unit of acceleration is m/s^2 or km/h^2, etc.
Calculation of AccelerationThe velocity of an automobile increases from 0 to 14 m/s, E, in 3.5 s. What is its acceleration?
G: initial velocity v(i) = 0 m/s ; final velocity v(f) = 14 m/s, E ; time interval (t) = 3.5 s
F: Acceleration (a)
E: a = [v(f) - v(i)] / t
S: a = (14 m/s – 0 m/s) / 3.5 s
A: a = 4.0 m/s^2, E
--> Example of Negative Acceleration
Calculation of Negative AccelerationAn automobile slows from 14.0 m/s, E, to 7.0 m/s, E, in 20 s. Find its acceleration?
G: intial velocity v(i) = 14.0 m/s, E ; final velocity v(f) = 7.0 m/s, E ; time interval (t) = 2.0 s
F: Acceleration (a)
E: a = [v(f) - v(i)] / t
S: a = (7.0 m/s – 14 m/s) / 2.0 s
A: a = -3.5 m/s^2, E
4. Other variations of Uniform Acceleration Problems
Final Velocity can be found from the initial velocity, acceleration, and time interval.
--> Example of calculating Final Velocity after uniform acceleration from initial velocity, acceleration, and time
Final Velocity after Uniform AccelerationA ball rolling down an incline for 5.0 s undergoes a uniform acceleration of 4.2 m/s^2. If the ball has an initial velocity of 2.0 m/s when it starts down the incline, what is its final velocity?
G: time (t) = 5.0 s ; acceleration (a) = 4.2 m/s^2 ; initial velocity v(i) = 2.0 m/s forward down the incline
F: final velocity v(f)
E: v(f) = v(i) + at
S: v(f) = 2.0 m/s + 4.2 m/s^2 (5.0 s)
A: v(f) = 23 m/s, forward down the incline
The average velocity of a uniformly accelerating object is the middle velocity as in v(ave) = [v(f)+ v(i)] / 2
Remember that Displacement is the distance (a scalar measurement) an object has moved and the direction the object has moved. It is a vector measurement. Displacement is also called "change in position".
The displacement of an object can be found from its initial and final velocities and the time interval over which the change occurs.
--> Example of calculating Displacement during Uniform Acceleration from final velocity, initial velocity, and time
Displacement during Uniform AccelerationWhat is the displacement of a train as it is accelerated uniformly from 22 m/s to 44 m/s, west in a 20.0 s time interval?
G: initial velocity v(i) = 22 m/s, W ; final velocity v(f) = 44 m/s, W ; time interval (t) = 20.0 s
F: displacement, d
E: d = [ ( v(f) + v(i) ) / 2] t
S: d = [ (44 m/s + 22 m/s) / 2 ] (20.0 s)
A: d = 660 m, W
The displacement can also be found from the initial velocity, acceleration, and time interval
--> Example of calculating Displacement during Uniform Acceleration from initial Velocity, Acceleration, and Time
Displacement during Uniform AccelerationA car starting from rest is accelerated at 6.1 m/s^2, S. What is the car’s displacement during the first 7.0 s of acceleration?
G: intial velocity (vi) = 0 m/s ; acceleration (a) = 6.1 m/s^2 ; time interval (t) = 7.0 s
F: displacement (d)
E: d = d(i) + v(i) t + 0.5 a t^2
S: d = 0 m + 0 m/s (7.0 s) + 0.5 (6.1 m/s^2) (7.0 s)^2
A: d = 150 m, S
Final velocity can also be found from the initial velocity, acceleration, and distance traveled. And Acceleration can be found from the initial velocity, final velocity, and the distance traveled.
--> Example of calculating Acceleration during Uniform Acceleration from final velocity, intial velocity, and displacement
Calculating AccelerationAn airplane must achieve a velocity of +71 m/s for takeoff. If the runway is 1.0 x 10^3 m long, what must the acceleration be?
G: initial velocity (vi) = 0 m/s ; final velocity (vf) = +71 m/s ; displacement (d) = 1,000 m
F: acceleration (a)
E: a = [ v(f)^2 – v(i)^2 ] / [ 2 d ]
S: a = [ (71 m/s)^2 – (0 m/s)^2 ] / [ 2 x 1000 m ]
A: +2.5 m/s^2
The displacement is the same if the initial and final velocities are exchanged and acceleration is changed to deceleration
--> Example of finding Displacement when Velocities and Times are known
Calculating DisplacementDetermine the displacement during constant (uniform) acceleration of an airplane that is accelerated from +66 m/s to +88 m/s in 12.0 s.
G: initial velocity v(i) = +66.0 m/s ; final velocity v(f) = +88.0 m/s ; time (t) = 12.0 s.
F: displacement, d
E: d = [ ( v(f) + v(i) ) / 2 ] t
S: d = [ (88.0 m/s + 66.0 m/s) / 2 ] [ 12 s ]
A: d = +924 m
Acceleration due to gravity problems requires the use of the same basic equations, but the acceleration symbol a is changed to a g for the acceleration due to gravity.
--> Example of a falling body (object) problem
Calculations pertaining to Falling BodiesA 20 kg object is suspended 100 m in the air by a large construction crane. The mass is released and allowed to fall to the ground.
a. How long will it take for the object to reach the ground?
b. What is the final velocity of the object just before it strikes the ground?
Solution to part a:
The equation d = v(i) t + 0.5 g t^2 allows you to solve for the time. Since the initial velocity for an object that is dropped is zero, the vi t in the equation will drop out. The d = 0.5 g t^2 portion of the equation is used to solve for the time.
E: t = (2 d / g )^0.5
S: t = [ 2 (100 m) / 9.8 m/s^2 ]^0.5
A: t = 4.52 s
Solution to part b:
Note: The acceleration downwards is the acceleration due to gravity, so a = g.
E: v(f) = v(i) + a t
S: v(f) = 0 m/s + 9.8 m/s^2 (4.52 s)
A: v(f) = 44.3 m/s, downwards
It would be proper to express the answer as a -44.3 m/s, where the negative in front of the number implies downwards. The use of the negative for the downwards direction would be valid in situations defined by you or others where up was defined as being the positive direction and down was defined as being the downward direction.
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About this Site
My web site Physics Net Class has been created to assist me in my work as a teacher of physics as well as share information related to physics with the public at large. The web site is a living thing having been born January 21, 2011 on my home computer. Slowly, but steadily, the site evolves. It seems very small now, but at the rate it swallows up knowledge it may grow very large indeed.
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Welcome to Physics Net Class
This Web Site is dedicated to all of us who are students of Physics or just naturally curious about how our world and our universe works. This Web Site is for students in private, public, and home schools who have a genuine interest or curiosity in science as it relates to physics.
"Physics is fun" says the author, because Physics is about toys, machines, gizmos, and all sorts of wonderful things.
For those interested in Science education and Physics education this Web Site addresses simple Physics lessons and Physics problem solving skills that can be used at the elementary, junior high, high school, and college level education.
And finally this web site is for my students to use as a gateway to the internet and the world wide web in their study of physics and astronomy.
"Physics is fun" says the author, because Physics is about toys, machines, gizmos, and all sorts of wonderful things.
For those interested in Science education and Physics education this Web Site addresses simple Physics lessons and Physics problem solving skills that can be used at the elementary, junior high, high school, and college level education.
And finally this web site is for my students to use as a gateway to the internet and the world wide web in their study of physics and astronomy.
Physics Questions
Do heavier objects fall more slowly than lighter objects?
See solution
Physics Questions — Problem 2
Why do objects float in liquids denser than themselves?
See solution
Physics Questions — Problem 3
A particle is moving around in a circle and its position is given in polar coordinates as x = Rcosθ, and y = Rsinθ, where R is the radius of the circle, and θ is in radians. From these equations derive the equation for centripetal acceleration.
See solution
Physics Questions — Problem 4
How come in free fall you feel weightless even though gravity is pulling down on you? (ignore air resistance when answering this question).
See solution
Physics Questions — Problem 5
What is the difference between centripetal acceleration and centrifugal force?
See solution
Physics Questions — Problem 6
What is the difference between energy and power?
See solution
Physics Questions — Problem 7
Two identical cars collide head on. Each car is traveling at 100 km/h. The impact force on each car is the same as hitting a solid wall at:
(a) 100 km/h
(b) 200 km/h
(c) 150 km/h
(d) 50 km/h
See solution
Physics Questions — Problem 8
Why is it possible to drive a nail into a piece of wood with a hammer, but it is not possible to push a nail in by hand?
See solution
Physics Questions — Problem 9
An archer pulls back 0.75 m on a bow which has a stiffness of 200 N/m. The arrow weighs 50 g. What is the velocity of the arrow immediately after release?
See solution
Physics Questions — Problem 10
When a moving car encounters a patch of ice the brakes are applied. Why is it desirable to keep the wheels rolling on the ice without locking up?
See solution
Solutions For High School Physics Questions
Solution For Problem # 1No. If an object is heavier the force of gravity is greater, but since it has greater mass the acceleration is the same, so it moves at the same speed (if we neglect air resistance). If we look at Newton's second law, F = ma. The force of gravity is F = mg, where m is the mass of the object and g is the acceleration due to gravity.
Equating, we have mg = ma. Therefore, a = g.
If there was no air resistance, a feather would fall at the same speed as an apple.
Solution For Problem # 2
If an object were completely immersed in a liquid denser than it, the resulting buoyant force would exceed the weight of the object. This is because the weight of the liquid displaced by the object is greater than the weight of the object (since the liquid is denser). As a result, the object cannot remain completely submerged and it floats. The scientific name for this phenomenon is Archimedes Principle.
Solution For Problem # 3
Without loss of generality, we only need to look at the equation for the x-position, since we know that centripetal acceleration points towards the center of the circle. Thus, when θ = 0, the second derivative of x with respect to time must be the centripetal acceleration.
The first derivative of x with respect to time t is:
x = —Rsinθ(dθ/dt)
The second derivative of x with respect to time t is:
x = —Rcosθ(dθ/dt)2—Rsinθ(d2θ/dt2)
In both of the above equations the chain rule of Calculus is used and by assumption θ is a function of time. Therefore, θ can be differentiated with respect to time.
Now, evaluate the second derivative at θ = 0.
We have,
x = —R(dθ/dt)2
The term dθ/dt is usually called the angular velocity, which is the rate of change of the angle θ. It has units of radians/second.
For convenience we can set w ≡ dθ/dt.
Therefore,
x = —Rw2
This is the well-known form for the centripetal acceleration equation.
Solution For Problem # 4
The reason you feel weightless is because there is no force pushing against you, since you are not in contact with anything. Gravity is pulling equally on all the particles in your body. This creates a sensation where no forces are acting on you and you feel weightless. It would be the same sensation as if you were floating in space.
Solution For Problem # 5
Centripetal acceleration is the acceleration an object experiences as it travels a certain velocity along an arc. The centripetal acceleration points towards the center of the arc.
Centrifugal force is the imaginary force an unrestrained object experiences as it moves around an arc. This force acts opposite to the direction of centripetal acceleration. For example, if a car makes a sharp right turn the passengers would tend to slide in their seats away from the center of the turn, towards the left (if they are not wearing their seat belts, that is). The passengers would feel as if they are experiencing a force. This is defined as centrifugal force.
Solution For Problem # 6
Power is the rate of energy being generated or consumed. For example, if an engine produces 1000 Watts of power (where Watts is Joules/second), then after an hour the total energy produced by the engine is 1000 Joules/second X 3600 seconds = 3,600,000 Joules.
Solution For Problem # 7
The answer is (a).
Since the collision is head on and each car is identical and traveling at the same speed, the force of impact experienced by each car is equal and opposite. This means that the impact is the same as hitting a solid wall at 100 km/h.
Solution For Problem # 8
When you swing a hammer you increase its kinetic energy, so that by the time it strikes the nail it imparts a large force which drives the nail into the wood.
The hammer is basically an energy reservoir to which you are adding energy during the course of the swing, and which is released all at once upon impact. This results in the impact force greatly exceeding the maximum force you can exert by just pushing on the nail.
Solution For Problem # 9
This can be solved using an energy method.
We can solve this by equating the potential energy of the bow to the kinetic energy of the arrow.
The bow can be treated as a type of spring. The potential energy of a spring is:
(1/2)kx2, where k is the stiffness and x is the amount the spring is stretched, or compressed.
Therefore, the potential energy PE of the bow is:
PE = (1/2)(200)(0.75)2 = 56.25 J
The kinetic energy of a particle is:
(1/2)mv2, where m is the mass and v is the velocity.
The arrow can be treated as a particle since it is not rotating upon release.
Therefore, the kinetic energy KE of the arrow is:
KE = (1/2)(0.05)v2
If we assume energy is conserved, then
PE = KE
Solving for the velocity of the arrow v we get
v = 47.4 m/s
Solution For Problem # 10
Static friction is greater than kinetic friction.
Static friction exists if the wheels keep rolling on the ice without locking up, resulting in maximum braking force. However, if the wheels lock up then kinetic friction takes over since there is relative slipping between wheel and ice. This reduces the braking force and the car takes longer to stop.
Anti-lock braking systems (ABS) on a vehicle prevent the wheels from locking up when the brakes are applied, thus minimizing the amount of time it takes for the vehicle to reach a complete stop. Also, by preventing the wheels from locking up you have greater control of the vehicle.
Equating, we have mg = ma. Therefore, a = g.
If there was no air resistance, a feather would fall at the same speed as an apple.
Solution For Problem # 2
If an object were completely immersed in a liquid denser than it, the resulting buoyant force would exceed the weight of the object. This is because the weight of the liquid displaced by the object is greater than the weight of the object (since the liquid is denser). As a result, the object cannot remain completely submerged and it floats. The scientific name for this phenomenon is Archimedes Principle.
Solution For Problem # 3
Without loss of generality, we only need to look at the equation for the x-position, since we know that centripetal acceleration points towards the center of the circle. Thus, when θ = 0, the second derivative of x with respect to time must be the centripetal acceleration.
The first derivative of x with respect to time t is:
x = —Rsinθ(dθ/dt)
The second derivative of x with respect to time t is:
x = —Rcosθ(dθ/dt)2—Rsinθ(d2θ/dt2)
In both of the above equations the chain rule of Calculus is used and by assumption θ is a function of time. Therefore, θ can be differentiated with respect to time.
Now, evaluate the second derivative at θ = 0.
We have,
x = —R(dθ/dt)2
The term dθ/dt is usually called the angular velocity, which is the rate of change of the angle θ. It has units of radians/second.
For convenience we can set w ≡ dθ/dt.
Therefore,
x = —Rw2
This is the well-known form for the centripetal acceleration equation.
Solution For Problem # 4
The reason you feel weightless is because there is no force pushing against you, since you are not in contact with anything. Gravity is pulling equally on all the particles in your body. This creates a sensation where no forces are acting on you and you feel weightless. It would be the same sensation as if you were floating in space.
Solution For Problem # 5
Centripetal acceleration is the acceleration an object experiences as it travels a certain velocity along an arc. The centripetal acceleration points towards the center of the arc.
Centrifugal force is the imaginary force an unrestrained object experiences as it moves around an arc. This force acts opposite to the direction of centripetal acceleration. For example, if a car makes a sharp right turn the passengers would tend to slide in their seats away from the center of the turn, towards the left (if they are not wearing their seat belts, that is). The passengers would feel as if they are experiencing a force. This is defined as centrifugal force.
Solution For Problem # 6
Power is the rate of energy being generated or consumed. For example, if an engine produces 1000 Watts of power (where Watts is Joules/second), then after an hour the total energy produced by the engine is 1000 Joules/second X 3600 seconds = 3,600,000 Joules.
Solution For Problem # 7
The answer is (a).
Since the collision is head on and each car is identical and traveling at the same speed, the force of impact experienced by each car is equal and opposite. This means that the impact is the same as hitting a solid wall at 100 km/h.
Solution For Problem # 8
When you swing a hammer you increase its kinetic energy, so that by the time it strikes the nail it imparts a large force which drives the nail into the wood.
The hammer is basically an energy reservoir to which you are adding energy during the course of the swing, and which is released all at once upon impact. This results in the impact force greatly exceeding the maximum force you can exert by just pushing on the nail.
Solution For Problem # 9
This can be solved using an energy method.
We can solve this by equating the potential energy of the bow to the kinetic energy of the arrow.
The bow can be treated as a type of spring. The potential energy of a spring is:
(1/2)kx2, where k is the stiffness and x is the amount the spring is stretched, or compressed.
Therefore, the potential energy PE of the bow is:
PE = (1/2)(200)(0.75)2 = 56.25 J
The kinetic energy of a particle is:
(1/2)mv2, where m is the mass and v is the velocity.
The arrow can be treated as a particle since it is not rotating upon release.
Therefore, the kinetic energy KE of the arrow is:
KE = (1/2)(0.05)v2
If we assume energy is conserved, then
PE = KE
Solving for the velocity of the arrow v we get
v = 47.4 m/s
Solution For Problem # 10
Static friction is greater than kinetic friction.
Static friction exists if the wheels keep rolling on the ice without locking up, resulting in maximum braking force. However, if the wheels lock up then kinetic friction takes over since there is relative slipping between wheel and ice. This reduces the braking force and the car takes longer to stop.
Anti-lock braking systems (ABS) on a vehicle prevent the wheels from locking up when the brakes are applied, thus minimizing the amount of time it takes for the vehicle to reach a complete stop. Also, by preventing the wheels from locking up you have greater control of the vehicle.
Physics Questions
I put together a collection of physics questions to help you understand physics better. These set of questions are divided into two parts. The first set of questions is for high school level. These are listed below. The second set of questions is for university and college level. These are listed in the next section.
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