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Light as an Electromagnetic Wave
Quiz by Jeanne Kelly G. Buslay
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1st Weekly Assessment in General Physics 2
3rd Weekly Assessment in General Physics 2
CHARACTERISTICS OF LIFE The world is filled with familiar objects, such as tables, rocks, plants, pets, and automobiles. Which of these objects are living or were once living? What are the criteria for assigning something to the living world or the nonliving world? Biologists have established that living things share seven characteristics of life. These characteristics are organization and the presence of one or more cells, response to a stimulus (plural, stimuli), homeostasis, metabolism, growth and development, reproduction, and change through time. Organization and Cells Organization is the high degree of order within an organismâs internal and external parts and in its interactions with the living world. For example, compare an owl to a rock. The rock has a spe- cific shape, but that shape is usually irregular. Furthermore, differ- ent rocks, even rocks of the same type, are likely to have different shapes and sizes. In contrast, the owl is an amazingly organized individual, as shown in Figure 1-2. Owls of the same species have the same body parts arranged in nearly the same way and interact with the environment in the same way. Copyright Š by Holt, Rinehart and Winston. All rights reserved. ORGANISM (Barn Owl) ORGAN (Owlâs Ear) TISSUE (Nervous Tissue Within the Ear) CELL (Nerve Cell) Every living organism has a level of organization. The different levels of organization for a complex multicellular organism, such as an owl, are shown in the figure below. FIGURE 1-2 THE SCIENCE OF LIFE 7 All living organisms, whether made up of one cell or many cells, have some degree of organization. A cell is the smallest unit that can perform all lifeâs processes. Some organisms, such as bacteria, are made up of one cell and are called unicellular (YOON-uh-SEL-yoo-luhr) organisms. Other organisms, such as humans or trees, are made up of multiple cells and are called multicellular (MUHL-ti-SEL-yoo-luhr) organisms. Complex multicellular organisms have the level of orga- nization shown in Figure 1-2. In the highest level, the organism is made up of organ systems, or groups of specialized parts that carry out a certain function in the organism. For example, an owlâs ner- vous system is made up of a brain, sense organs, nerve cells, and other parts that sense and respond to the owlâs surroundings. Organ systems are made up of organs. Organs are structures that carry out specialized jobs within an organ system. An owlâs ear is an organ that allows the owl to hear. All organs are made up of tissues. Tissues are groups of cells that have similar abilities and that allow the organ to function. For example, nervous tissue in the ear allows the ear to detect sound. Tissues are made up of cells. A cell must be covered by a membrane, contain all genetic information necessary for replication, and be able to carry out all cell functions. Within each cell are organelles. Organelles are tiny structures that carry out functions necessary for the cell to stay alive. Organelles contain biological molecules, the chemical compounds that provide physical structure and that bring about movement, energy use, and other cellular functions. All biological molecules are made up of atoms. Atoms are the simplest particle of an ele- ment that retains all the properties of a certain element. Response to Stimuli Another characteristic of life is that an organism can respond to a stimulusâa physical or chemical change in the internal or external environment. For example, an owl dilates its pupils to keep the level of light entering the eye constant. Organisms must be able to respond and react to changes in their environment to stay alive. ORGANELLE (Mitochondrion) BIOLOGICAL MOLECULE (Phospholipid) ATOM (Oxygen) cell from the Latin, cella meaning âsmall room,â or âhutâ Word Roots and Origins www.scilinks.org Topic: Characteristics of Life Keyword: HM60257 mb06se_bios01.qxd 5/18/07 10:37 AM Page 7 8 CHAPTER 1 Homeostasis All living things, from single cells to entire organisms, have mecha- nisms that allow them to maintain stable internal conditions. Without these mechanisms, organisms can die. For example, a cellâs water content is closely controlled by the taking in or releas- ing of water. A cell that takes in too much water will rupture and die. A cell that doesnât get enough water will also shrivel and die. Homeostasis (HOH-mee-OH-STAY-sis) is the maintenance of a stable level of internal conditions even though environmental conditions are constantly changing. Organisms have regulatory systems that maintain internal conditions, such as temperature, water content, and uptake of nutrients by the cell. In fact, multi- cellular organisms usually have more than one way of maintain- ing important aspects of their internal environment. For example, an owlâs temperature is maintained at about 40°C (104°F). To keep a constant temperature, an owlâs cells burn fuel to produce body heat. In addition, an owlâs feathers can fluff up in cold weather. In this way, they trap an insulating layer of air next to the birdâs body to maintain its body temperature. Metabolism Living organisms use energy to power all the life processes, such as repair, movement, and growth. This energy use depends on metabolism (muh-TAB-uh-LIZ-uhm). Metabolism is the sum of all the chemical reactions that take in and transform energy and materials from the environment. For example, plants, algae, and some bacteria use the sunâs energy to generate sugar molecules during a process called photosynthesis. Some organisms depend on obtaining food energy from other organisms. For instance, an owlâs metabolism allows the owl to extract and modify the chemi- cals trapped in its nightly prey and use them as energy to fuel activities and growth. Growth and Development All living things grow and increase in size. Some nonliving things, such as crystals or icicles, grow by accumulating more of the same material of which they are made. In contrast, the growth of living things results from the division and enlargement of cells. Cell division is the formation of two new cells from an existing cell, as shown in Figure 1-3. In unicellular organisms, the primary change that occurs following cell division is cell enlargement. In multi- cellular life, however, organisms mature through cell division, cell enlargement, and development. Development is the process by which an organism becomes a mature adult. Development involves cell division and cell differen- tiation, or specialization. As a result of development, an adult organism is composed of many cells specialized for different func- tions, such as carrying oxygen in the blood or hearing. In fact, the human body is composed of trillions of specialized cells, all of which originated from a single cell, the fertilized egg. This unicellular organism, Escherichia coli, inhabits the human intestines. E. coli reproduces by means of cell division, during which the original cell splits into two identical offspring cells. FIGURE 1-3 Observing Homeostasis Materials 500 mL beakers (3), wax pen, tap water, thermometer, ice, hot water, goldfish, small dip net, watch or clock with a second hand Procedure 1. Use a wax pen to label three 500 mL beakers as follows: 27°C (80°F), 20°C (68°F), 10°C (50°F). Put 250 mL of tap water in each beaker. Use hot water or ice to adjust the tem- perature of the water in each beaker to match the temperature on the label. 2. Put the goldfish in the beaker of 27°C water. Record the number of times the gills move in 1 minute. 3. Move the goldfish to the beaker of 20°C water. Repeat observations. Move the goldfish to the beaker of 10°C. Repeat observations. Analysis What happens to the rate at which gills move when the temp- erature changes? Why? How do gills help fish maintain homeostasis? Quick Lab mb06se_bios01.qxd 5/18/07 10:37 AM Page 8 THE SCIENCE OF LIFE 9 Reproduction All organisms produce new organisms like themselves in a process called reproduction. Reproduction, unlike other characteristics, is not essential to the survival of an individual organism. However, because no organism lives forever, reproduction is essential for the continuation of a species. Glass frogs, as shown in Figure 1-4, lay many eggs in their lifetime. However, only a few of the frogsâ off- spring reach adulthood and successfully reproduce. During reproduction, organisms transmit hereditary informa- tion to their offspring. Hereditary information is encoded in a large molecule called deoxyribonucleic acid, or DNA. A short segment of DNA that contains the instructions for a single trait of an organism is called a gene. DNA is like a large library. It contains all the booksâgenesâthat the cell will ever need for making all the struc- tures and chemicals necessary for life. Hereditary information is transferred to offspring during two kinds of reproduction. In sexual reproduction, hereditary information recombines from two organisms of the same species. The resulting offspring are similar but not identical to their parents. For example, a male frogâs sperm can fertilize a femaleâs egg and form a single fer- tilized egg cell. The fertilized egg then develops into a new frog. In asexual reproduction, hereditary information from different organisms is not combined; thus the original organism and the new organism are genetically the same. A bacterium, for example, reproduces asexually when it splits into two identical cells. Change Through Time Although individual organisms experience many changes during their lifetime, their basic genetic characteristics do not change. However, populations of living organisms evolve or change through time. The ability of populations of organisms to change over time is important for survival in a changing world. This factor is also impor- tant in explaining the diversity of life-forms we see on Earth today.
Commas Directions: Correct the sentences by adding commas where needed. 1. After the sound of the bell we realized it was a false alarm. 2. Mr. Yoshino the head of the department resigned yesterday. 3. The gentleman with the black umbrella who is an ambassador to the United States said hello to us as we were entering the hotel. 4. Even though we won the game the players unfortunately did not play their best. 5. Heather walked quickly up to the door and knocked hoping that someone would answer. Authorâs Purpose 6. An author writes a story about a boy who saves his town from a flood by using his quick thinking. The author includes exciting descriptions of the boy's bravery. What is the authorâs most likely purpose for writing this story? A. To inform readers about the dangers of floods B. To entertain readers with a heroic tale C. To explain how to prevent floods D. To persuade readers to prepare for emergencies 7. Which of the following is an example of an author writing to persuade? A. A science textbook chapter explaining the water cycle B. A commercial encouraging people to adopt shelter pets C. A short story about a girl who finds a magical necklace D. A recipe for making chocolate chip cookies 8. Read the following sentence: "Studies show that students who read for 20 minutes a day score higher on tests. Reading is one of the best habits you can develop for success in school and life." What is the authorâs purpose in this passage? A. To entertain readers with a fun story B. To persuade readers to read more often C. To inform readers about how books are written D. To explain how to find books to read 9. An author writes a how-to guide titled 10 Easy Steps to Plant a Garden. What is the authorâs primary purpose? A. To persuade readers to grow their own vegetables B. To inform readers how to plant a garden C. To entertain readers with funny garden tips 10. Read the excerpt: "Long ago, in a village surrounded by mountains, the people discovered a secret about their water well. Every full moon, the well water turned to gold for just one night. But no one knew why. This mystery brought travelers from far and wide, hoping to uncover the truth." What is the authorâs purpose in this excerpt? A. To persuade readers to visit the village B. To inform readers about a historical event C. To entertain readers with a mysterious tale D. To explain the science behind the water Main Idea When I stepped out into the bright sunlight from the darkness of the movie house, I had only two things on my mind: Paul Newman and a ride home. I was wishing I looked like Paul Newman--- he looks tough and I don't--- but I guess my own looks aren't so bad. I have light-brown, almost-red hair and greenish-gray eyes. I wish they were more gray because I hate most guys that have green eyes, but I have to be content with what I have. My hair is longer than a lot of boys wear theirs, squared off in back and long at the front and sides, but I am a greaser and most of my neighborhood rarely bothers to get a haircut. Besides, I look better with long hair. 11. What is the main idea? The narrator likes movies. The narrator wishes he was Paul Newman. The narrator is content with his appearance. The narrator looks better with long hair. 12. The narrator believes. . . looks are important. he should get a haircut. green eyes are bad. that he has red hair. Once there were four girls who shared a pair of pants. The girls were all different sizes and shapes, and yet the pants fit each of them. You may think this is a suburban myth. But I know it's true, because I am one of them, one of the sisters of the Traveling Pants. We discovered their magic last summer, purely by accident. The four of us were splitting up for the first time in our lives. Carmen had gotten them from a secondhand place without even bothering to try them on. She was going to throw them away, but by chance, Tibby spotted them. First Tibby tried them; then me, Lena; then Bridget; then Carmen. By the time Carmen pulled them on, we knew something extraordinary was happening. If the same pants fit and I mean really fit the four of us, they aren't ordinary. They don't belong completely to the world of things you can see and touch. My sister, Effie, claims I don't believe in magic, and maybe I didn't then. But after the first summer of the Traveling Pants, I do. 13. What is the main idea? Four friends were connected through a special pair of pants. A pair of pants called the Traveling Pants. Carmen finding a pair of pants from a second-hand shop. The girls believing in magic. 14. The narrator included that the pants fit all of them to emphasize how the girls become friends. the girls are different sizes. why the pants are special. where the pants came from. If you are interested in stories with happy endings, you would be better off reading some other book. In this book, not only is there no happy ending, there is no happy beginning and very few happy things in the middle. This is because not very many happy things happened in the lives of the three Baudelaire youngsters. Violet, Klaus, and Sunny Baudelaire were intelligent children, and they were charming, and resourceful, and had pleasant facial features, but they were extremely unlucky, and most everything that happened to them was rife with misfortune, misery, and despair. I'm sorry to tell you this, but that is how the story goes. 15. What is the main idea? description about the story to come. A warning about the story and its sad content. A declaration about the Baudelaire family. A beginning for the end of the story. 16. The narrator believes the reader does not like sad stories. likes stories with happy endings. canât enjoy the story. will find the story unhappy. 17. Read the following sentence: Of course you can exaggerate your story, but what you say must be based on truth. Which word means the same as exaggerate? repeat reveal overstate increase 18. What is the meaning of the word inaugurated, used in the following sentence: Less than two months after Abraham Lincoln was inaugurated President in 1861, he encountered one of the most difficult tasks ever experienced by a United States leader: civil war. elected by a vote brought into office identified by name viewed as an authority 19. What does the phrase âpractice your presentation so much that you could do it in your sleepâ suggest in the following sentence: The best advice is to practice your presentation so much that you could do it in your sleep. get plenty of sleep the night before giving a presentation give their presentations in front of a small audience first take advice from their teachers on how to write a presentation memorize their presentations before they give them 20. Read the following sentence: The Phoenix Mars Lander is a NASA spacecraft that landed on the Red Planet in May 2009 to study the history of water and potential for life on the planet. What is another word for potential? existence situation possibility qualification
Create quiz based on this information Who is the author of Letter 1, and who is the intended recipient? The author of Letter 1 is Robert Walton. The intended recipient is his sister, Mrs. Saville. What is the author's purpose in writing this letter? The author's purpose in writing this letter is to update his sister on his progress and feelings regarding his upcoming Arctic expedition. Where is the author currently located, and what is the significance of the setting? The author is currently in St. Petersburg, Russia. The significance of the setting is that it is the starting point of his journey towards the Arctic, and it sets the tone for the novel's exploration of extreme environments. Describe the author's feelings about the natural world and the northern journey. The author expresses excitement and confidence about his journey. He is inspired by the cold northern breeze, which fills him with delight and a sense of adventure. What is the author's fascination with the pole, and how does he describe it? The author is fascinated by the idea of the North Pole as a land of beauty and eternal light. He envisions it as a region of wonder and hopes to make groundbreaking discoveries there. What are some of the author's hopes and expectations for his journey? The author hopes to make significant discoveries, including a passage near the pole to shorten travel times and the secret of the magnet's power. He also wants to explore uncharted lands. How does the author's enthusiasm change as he writes the letter? At the beginning of the letter, the author is enthusiastic and confident. However, as he reflects on the challenges and uncertainties of his journey, his enthusiasm becomes mixed with doubt and a sense of the unknown. What role has reading played in the author's life, and how does it relate to his journey? Reading has played a significant role in the author's life, sparking his early interest in exploration. He initially wanted to embark on a seafaring life, but reading led him to poetry and later to his current expedition. How has the author prepared for his upcoming expedition? The author has prepared by enduring hardships, accompanying whale-fishers, studying mathematics, medicine, and physical science, and even working as an under-mate on a Greenland whaler to gain practical experience. What does the author express about his feelings, courage, and hopes for the future? The author expresses a strong desire to achieve a great purpose and a willingness to face the challenges and uncertainties of his expedition with courage. He hopes to return triumphant but acknowledges that success may take a long time, if ever.
Filmic Techniques Based on the work of Brad Smilanich Mis-en-Scene: originally a French theatrical term arrangements of all the visual elements of the stage area in film â âthe contents of the frame and the way those contents are organizedâ include: lighting, costume, dĂŠcor, props, camera movement or distance . . . all photographic decisions etc. Proxemics: Spatial relationship among characters within the mis-en-scene Rule of Thirds: a compositional rule of thumb in painting, design, photography etc. suggests image divided into 9 equal parts with two vertical and two horizontal lines important elements of the mis-en-scene should be placed along these lines and their intersections some suggest aligning with intersections makes for more interesting pictures than just centreing the subject Proxemics Camera Distance: Quite literally, how far the camera is from the subject being filmed The Hand Camera Camera Distance: Quite literally, how far the camera is from the subject being filmed Extreme Close Up: Singles out one small portion of the body or object Used to intensify emotion, or show reaction Camera Distance: Close up Shot: Shows head of character or small significant object Used to show emotions Camera Distance: Medium Shot: shows figures from the waist up allows character to be seen within background Camera Distance: Long Shot: shows figures from feet up similar to the âstageâ in live theatre orients audience to figures within a location or surrounding Camera Distance: Extreme Long Shot: Sometimes called an âestablishing shotâ Panoramic view of an exterior location orients audience to a location Camera Distance: Camera Angle: Cameraâs angle of view relative to the subject being photographed High Angle Shot: looks down on the subject often used to make the subject look small and insignificant (in combination with camera distance) puts the camera (audience) in âpowerâ position Camera Angle: Low Angle Shot: looks up at the subject often used to make the subject look large and powerful puts the camera (audience) in a âsubmissiveâ position Camera Angle: Flat Angle Shot: camera on same plane as the subject feels most ânormalâ to an audience Camera Angle: Canted Shot: frame is unbalanced in relation to the subject may indicate a symbolic unbalance in the character Camera Angle: Camera Movement literally the camera moving with or around or to follow the subjects in the mis-en-scene or frame Camera Movement: Tilting Movement camera moves up or down on a horizontal axis similar to head nodding movement may be used to show subjects relation to surroundings Camera Movement: Panning Movement camera moves side to side on a vertical axis similar to head shaking movement may be used to establish setting Camera Movement: Dolly Movement camera mounted on a vehicle that moves along with the subject (camera moves, not pivots) follows the subject to signify something important Camera Movement: Crane Shot camera mounted on a crane or boom permits camera to move in & out, up & down, backward & forward often used for high aerial establishing shots Misc. Shots: Hand Held: camera carried to seem jerky, giving ârealistic feelâ Push In: camera moves up to a characterâs face to indicate an epiphany (realization) Spiral: camera circles subject for effect End for ELA 20-2 and 10-1 Shot Transitions/Editing: artificial editing done to string together multiple shots to create a narrative scene or sequence a cut is the change from one shot to another usually separated in to âsoftâ and âhardâ cuts Jump Cut: an instantaneous change from one shot to another this can be very natural or may disorient the audience, depending on how it is used Transitions/Editing Swish Pan: A pan where the speed of the camera is so fast that images are blurry used often to connect events in different settings that are connected by time Transitions/Editing Dissolve: transition where one shot gradually dissapears while another shot gradually appears often used to suggest change of setting or long time passage i.e. flashbacks Transitions/Editing Fade In/Out: transition where the shot gradually overexposes to white or underexposes to black often used to suggest a lengthy passage of time or change in location Transitions/Editing Wipe: transition where one shot is gradually eliminated as another shot moves onto the screen can be vertically or horizontally often suggests movement of the camera to another location Transitions/Editing Iris In/Out: transition where one shot gradually appears as an expanding circle in the middle of an old image suggests . . .??? Transitions/Editing Shot-Reverse Shot: one character is shown looking (often off-screen) at another character, and then the other character is shown looking "back" at the first character. Since the characters are shown facing in opposite directions, the viewer unconciously assumes that they are looking at each other. Transitions/Editing Two-Shot: Face-up shot of two people. Often used in interviews, or when two presenters are hosting a show. A "One-Shot" could be a mid-shot of either of these subjects. A "Three-Shot", unsurprisingly, contains three people. Transitions/Editing Shot Transitions/Editing: Sound: used to reflect or enhance what is shown visually on the screen can include dialogue, music, sound effects, voiceover etc. Diegetic Sound: sound that has a source in the world of the story dialogue spoken by characters, sound made by objects, or music coming from a source grounded in the story of the film Non-diegetic Sound: sound that has a source outside the world of the story usually part of the score or the soundtrack Parallel Sound: sound that complements the image shown i.e. romantic music during a love scene Counterpoint Sound: sound that contradicts the âfeelingâ of the image a happy song played while images of graphic violence are portrayed Voiceover: voice of a non-visible narrator laid over the scene often provides some comment about the narrative of the film Sound Bridge: used to âsoftenâ the transition between one scene and another takes sound from the next shot and overlays it on the current shot 2-3 seconds earlier than we see the image Examples of Diegetic/Non-Diegetic: In the first clip, the non-diegetic music changes to diegetic music when the main character moves inside of the convenience store. In the second clip, the âduhn duhn duuuuhâ which often is non-diegetic becomes diegetic because it is the band in the passing bus playing that music! End for ELA 20-1 Lighting: Can be used by a director to: Control the mood of a scene guide a viewerâs eye to a specific place in mis-en-scene Emphasize and de-emphasize elements in frame Add texture and color Make people look beautiful, ugly, sinister, or angelic Standard 3-Point Lighting: uses three lights called the key light, fill light and back light forms the basis of most lighting. once you understand three point lighting you are well on the way to understanding all lighting. Key Light: main light usually the strongest and has the most influence on the look of the scene. it is placed to one side of the camera/subject so that side is well lit and other side has shadow. Fill Light: secondary light is placed on the opposite side of the key light used to fill the shadows created by key softer and less bright than key Back Light: placed behind the subject ; lights it from the rear. provides definition and subtle highlights around the subject's outlines. Separates subject from background provides a three-dimensional look. Standard 3-Point Lighting: http://www.zvork.fr/vls/ Try using this simulator to play with lighting with those 3 points.
Introduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: ⢠Free-falling objects do not encounter air resistance. ⢠All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs ⢠Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 ⢠a ⢠d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 ⢠(-8.00 m/s2) ⢠d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) ⢠d (16.0 m/s2) ⢠d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi ⢠t + ½ ⢠a ⢠t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) ⢠(4.1 s) + ½ ⢠(6.00 m/s2) ⢠(4.10 s)2 d = (0 m) + ½ ⢠(6.00 m/s2) ⢠(16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: ⢠An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. ⢠If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. ⢠If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. ⢠If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi ⢠t + ½ ⢠a ⢠t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) ⢠(t) + ½ ⢠(-9.8 m/s2) ⢠(t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) ⢠(t)2 -8.52 m = (-4.9 m/s2) ⢠(t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 ⢠a ⢠d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 â˘(-9.8m/s2) â˘d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) â˘d (-19.6 m/s2) ⢠d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) ⢠d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.
What do an ancient Greek philosopher and a 19th century Quaker have in common with Nobel Prize-winning scientists? Although they are separated over 2,400 years of history, each of them contributed to answering the eternal question: what is stuff made of? It was around 440 BCE that Democritus first proposed that everything in the world was made up of tiny particles surrounded by empty space. And he even speculated that they vary in size and shape depending on the substance they compose. He called these particles "atomos," Greek for indivisible. His ideas were opposed by the more popular philosophers of his day. Aristotle, for instance, disagreed completely, stating instead that matter was made of four elements: earth, wind, water and fire, and most later scientists followed suit. Atoms would remain all but forgotten until 1808, when a Quaker teacher named John Dalton sought to challenge Aristotelian theory. Whereas Democritus's atomism had been purely theoretical, Dalton showed that common substances always broke down into the same elements in the same proportions. He concluded that the various compounds were combinations of atoms of different elements, each of a particular size and mass that could neither be created nor destroyed. Though he received many honors for his work, as a Quaker, Dalton lived modestly until the end of his days. Atomic theory was now accepted by the scientific community, but the next major advancement would not come until nearly a century later with the physicist J.J. Thompson's 1897 discovery of the electron. In what we might call the chocolate chip cookie model of the atom, he showed atoms as uniformly packed spheres of positive matter filled with negatively charged electrons. Thompson won a Nobel Prize in 1906 for his electron discovery, but his model of the atom didn't stick around long. This was because he happened to have some pretty smart students, including a certain Ernest Rutherford, who would become known as the father of the nuclear age. While studying the effects of X-rays on gases, Rutherford decided to investigate atoms more closely by shooting small, positively charged alpha particles at a sheet of gold foil. Under Thompson's model, the atom's thinly dispersed positive charge would not be enough to deflect the particles in any one place. The effect would have been like a bunch of tennis balls punching through a thin paper screen. But while most of the particles did pass through, some bounced right back, suggesting that the foil was more like a thick net with a very large mesh. Rutherford concluded that atoms consisted largely of empty space with just a few electrons, while most of the mass was concentrated in the center, which he termed the nucleus. The alpha particles passed through the gaps but bounced back from the dense, positively charged nucleus. But the atomic theory wasn't complete just yet. In 1913, another of Thompson's students by the name of Niels Bohr expanded on Rutherford's nuclear model. Drawing on earlier work by Max Planck and Albert Einstein he stipulated that electrons orbit the nucleus at fixed energies and distances, able to jump from one level to another, but not to exist in the space between. Bohr's planetary model took center stage, but soon, it too encountered some complications. Experiments had shown that rather than simply being discrete particles, electrons simultaneously behaved like waves, not being confined to a particular point in space. And in formulating his famous uncertainty principle, Werner Heisenberg showed it was impossible to determine both the exact position and speed of electrons as they moved around an atom. The idea that electrons cannot be pinpointed but exist within a range of possible locations gave rise to the current quantum model of the atom, a fascinating theory with a whole new set of complexities whose implications have yet to be fully grasped. Even though our understanding of atoms keeps changing, the basic fact of atoms remains, so let's celebrate the triumph of atomic theory with some fireworks. As electrons circling an atom shift between energy levels, they absorb or release energy in the form of specific wavelengths of light, resulting in all the marvelous colors we see. And we can imagine Democritus watching from somewhere, satisfied that over two millennia later, he turned out to have been right all along.