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Point, line, plane, ray, line segment, angle
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Tobruk, a small town on the Libyan coast, was central to much of the fighting that took place in the Western Desert during the Second World War. It had originally been developed by the Italians during their colonisation of eastern Libya during the early decades of the 20th century. With a sheltered deep water harbour it became a key naval outpost. It was fortified during the 1930s with both coastal defence batteries and a 50 kilometre-long perimeter of reinforced concrete platoon posts, and other supporting infrastructure such as gun positions, headquarters bunkers, underground supply dumps, and observation towers. When British and Commonwealth forces advanced out of Egypt and into Libya in January 1941, Tobruk was their second objective. The Italian defence perimeter was attacked by the 6th Australian Division on the morning of 22 January and the town fell the next morning. The operation resulted in approximately 27,000 Italian prisoners and the capture of over 200 artillery pieces, but cost 49 Australian lives. The 6th Division's advance pressed on beyond Tobruk and eventually they were withdrawn from Libya to be deployed to Greece.The 9th Australian Division was moved in to Libya in February 1941 to garrison the territory captured by the 6th. By this time, however, German troops had arrived in Libya to reinforce their Italian allies and they launched an offensive that the British Commonwealth forces were ill-disposed to hold back. A retreat towards Egypt commenced. The 9th Division was ordered to fall back upon Tobruk, hold it in order deny its port facilities to the Germans, and delay their advance so as to provide time for defences on the Egyptian frontier to be prepared. Tobruk and the 9th Division were subsequently encircled, beginning what became known as "the siege of Tobruk". Reinforced by the 18th Brigade of the 7th Australian Division and other British and Commonwealth troops, and resupplied by the sea, the 9th Division held Tobruk from April to September 1941. During this period it repelled two major German attacks. In September and October the 9th Division, its condition steadily declining, was relieved by the British 70th Division, which continued to defend Tobruk until the siege was finally lifted by Operation Crusader in December. The defence of Tobruk resulted in 749 Australian deaths, and another 604 became prisoners of war. Tobruk was the scene of further heavy fighting in June 1942 when the fortunes of war again saw a British Commonwealth force seeking to deny the port to the enemy. The Axis forces, however, were in no mood for another siege and launched a massive attack to capture it on 20 June. It remained in their hands until their final retreat from Libya in November 1942.John Hurst Edmondson (1914-1941), soldier, was born on 8 October 1914 at Wagga Wagga, New South Wales, only child of native-born parents Joseph William Edmondson, farmer, and his wife Maude Elizabeth, née Hurst. The family moved to a farm near Liverpool when Jack was a child. Educated at Hurlstone Agricultural High School, he worked with his father and became a champion rifle-shooter. He was a council-member of the Liverpool Agricultural Society and acted as a steward at its shows. Having served (from March 1939) in the 4th Battalion, Militia, he enlisted in the Australian Imperial Force on 20 May 1940 and was posted to the 2nd/17th Battalion. Later that month he was promoted acting corporal (substantive in November). Well built and about 5 ft 9 ins (175 cm) tall, Edmondson settled easily into army life and was known as a quiet but efficient soldier. His battalion embarked for the Middle East in October and trained in Palestine. In March 1941 the 2nd/17th moved with other components of the 9th Division to Libya and reached Marsa Brega before an Axis counter-attack forced them to retreat to Tobruk. The siege of the fortress began on 11 April. Two days later the Germans probed the perimeter, targeting a section of the line west of the El Adem Road near Post R33. This strong-point was garrisoned by the 2nd/17th's No.16 Platoon in which Edmondson was a section leader. The enemy intended to clear the post as a bridgehead for an armoured assault on Tobruk.Under cover of darkness thirty Germans infiltrated the barbed wire defences, bringing machine-guns, mortars and two light field-guns. Lieutenant Austin Mackell, commanding No.16 Platoon, led Edmondson's five-man section in an attempt to repel the intruders. Armed with rifles, fixed bayonets and grenades, the party of seven tried to outflank the Germans, but were spotted by the enemy who turned their machine-guns on them. Unknown to his mates, Edmondson was severely wounded in the neck and stomach. Covering fire from R33 ceased at the pre-arranged time of 11.45 p.m. and Mackell ordered his men to charge. Despite his wounds, Edmondson accounted for several enemy soldiers and saved Mackell's life. When the remaining Germans fled, the Australians returned to their lines. Although Edmondson was treated for his wounds, he died before dawn on 14 April 1941. The Germans' armoured attack that morning was thwarted, partly due to the earlier disruption of their plans. Edmondson was buried in Tobruk war cemetery. He had not married. His Victoria Cross, gazetted on 4 July, was the first awarded to a member of Australia's armed forces in World War II. In April 1960 Mrs Edmondson gave her son's medals to the Australian War Memorial, Canberra, where they are displayed alongside his portrait (1958) by Joshua Smith. At Liverpool a public clock commemorates Edmondson, as do the clubrooms used by the sub-branch of the Returned Services League of Australia.Perhaps my nerves will be more under control when I am by myself. There were no entries in the diary until Friday April 18 when she wrote: Fighting terrific in Greece and North Africa…. I dread the casualty list also the heaviest air raid over London to date. Account …. of heavy fighting and much use of bayonet at Tobruk. Also gives an account of a charge in which a Lieutenant and a Corporal took prominent parts on Easter Sunday night. Of course, no names. When I read it …. I was sure the Corporal was Jack…. It said no casualties but …. I know … that all is not well with Jack. ….. (and) Stuffy ….has not come home yet. On Wednesday April 23 she received a letter from Jack dated March 30 and for the first time he said the conditions were bad. The food short, water one bottle for 48 hours. It worried me terribly so I posted a parcel (of) milk tablets, chocolate milk, biscuits (and) cigarettes.Tuesday April 15 I was feeling afraid of something while I was working and packing the cake (and) had a couple of brandys to (keep going).April 26 Received the following telegram in the mail, the bus man brought it in. “It is with deep regret that I have to inform you that Corporal John Hurst Edmondson was killed in action on the 14th April and desire to convey the profound sympathy of the Ministry for the Army and the Military Board.”Her final entry
Align Panel: This panel allows you to align one or more objects the the artboard or other objects. Alignment: Formatting the appearance of text with the margins of the text box. Anchor Point Tool: Allows you to add or remove handles to create a curve on an anchor point. Anchor Points: A point on a path indicates a change of direction. Appearance Panel: This panel shows you the fill, strokes, graphic styles, and effects that have been applied to an object, group or layer and are able to modify theses from this panel directly. Area Type Tool: This occurs when using the type tool and you click and drag a text box, the text will automatically wrap inside the box. Blend Tool: This tool allows you to combine shapes/colors between two or more objects to create a new object between the original, blending the colors and shapes by inserting the middle steps to get from one object to the next. Blob Brush Tool: This tool is used to create free-form objects that can have a more hand-drawn feel. Brushes: Allows you to set the appearance/style of a path, can be applied to existing paths or used to create new paths. Clipping Masks: This command allows you to mask objects to a shape so that only areas that lie within the shape are visible, the mask and objects that are masked are called a clipping set. Closed Path: A path that has the same beginning and ending point. It forms a complete shape that can be filled with color or text. Create Outlines: This command converts text to closed paths and can be found in the Type Menu. Curvature Pen Tool: Allows you to simply create paths with curved anchor points. Curves: Can be applied to an anchor point using handles to create an arched line. Direct Select Tool: Allows you to select individual points of any path. Effects: These can be added to objects to create quick dynamic characteristics. Eraser Tool: This tool allows you to remove anchor points and cut through paths. Expand Objects: This allows you to divide a single object into multiple objects that make up its appearance. Eyedropper Tool: This tool allows you to sample the color or text from an existing part of the artwork. Global Swatches: This is a color swatches that will be automatically updated throughout your artwork when you change them, indicated with a white triangle in the corner of the swatch. Graphic Styles: A set of reusable appearance attributes that allow you to quickly change the look of an object. Grouping: This command allows you to link objects together so that they can be moved, scaled, rotator, or copy. Groups can be nested inside other groups. Hierarchy: To create visual order in design, controlling what the viewer looks at in order using size, color, contrast, etc… Image Trace: This command allows you to convert a raster image into a vector artwork. Isolation Mode: This mode allows you to adjust single objects/groups inside a group without ungrouping the group. Join Tool: This tool joins paths and anchor points together quickly. Kerning: This is the adjustment of the space between two individual letters. Knife Tool: This tool allows you to split an object into 2 pieces along a freehand path you draw. Leading: This is the adjustment of the space between lines of text. Live Corners: This widget appears when using the Direct Select tool and a corner is selected, when used this will create a rounded corner. Live Paint: This command allows you to quickly apply colors to objects in a complex design. Open Path: A path that does not end, not connected back to the original anchor point. Overflow Text: This occurs when the text box is too small to house all the text and is indicated by a small red plus sign in the bottom right corner of the text box. Paintbrush Tool: This tool is used to create free-form paths that can have a more hand-drawn feel. Paragraph Spacing: The space that occurs between lines of text. Pathfinder Panel: This panel allows you to create complex shapes by selecting 2 or more objects and using the buttons in the panel to cut, combine, or divide the objects. Paths: These are created when 2 or more points are connected, these are created using the pen tool. Pen Tool: It allows you to create and edit anchor points and paths. Pencil Tool: This tool is used to create free-form shapes or lines, the accuracy of the lines can be adjusted. Perspective Tool: This tool allows you to place elements on a perspective grid to adjust objects on a different perspective, automatically snapping to the perspective grid. Placeholder Text: Text that is placed in a text box that "holds a place" in a design to allow for creating a layout or adjust the text design. Point Type Tool: This occurs when when using the type tool and you click once, the text will continue without wrapping. Readability: The characteristics of fonts and styles that make test easy to identify and read. Scale, Shear, Distort Objects: This set of commands allows you to adjust the size and perspective of objects. Scissors Tool: This tool allows you to split a path into 2 pieces. Selection Tool: Allows you to select paths, objects or groups by click or dragging over them. Shape Builder Tool: This interactive tool allows you to create complex shapes by merging and erasing simpler objects. Shapes Tools: A group of tools to create basic shapes without using the pen tool (rectangle, ellipse, polygon, star, etc…). Smooth Tool: This tool will smooth a complex path and reduce the number of anchor points. Swatches: This is a saved color that can be applied in a design via the swatches panel and can be grouped, these can include gradients and patterns. Text Wrapping: This is when the text in a text box automatically wraps to the next line when it reaches the edge of the box. Threading Text: This is the ability to create 2 or more text boxes that are linked, when text is added/adjusted in one box, it will affect the other(s). Touch Type Tool: This tool allows you to adjust individual letter in a previously created text box. Tracking: This is the adjustment of the overall spacing between letters. Transform Objects: This allows you to change the size of objects. Type on a Path Tool: This tool allows you to add text along any previously created path. Type Tool: This tool allows you to create text in a design. View Modes: Ability to view projects and adjust the display on the screen. Modes include Outlines, Presentation, & Full Screen.
A Brief History of Washington’s Crossing of the Delaware River, Christmas Night 1776... In the fall of 1776, General George Washington and his army had suffered a series of defeats at the hands of the British Army. The Continental Army had lost every battle with the British in the New York campaign: Long Island, Manhattan, Brooklyn Heights, Harlem and White Plains and had surrendered Fort Washington and Fort Lee. At Fort Lee, the army barely escaped and was forced to leave behind its store of provisions, ammunition, and many of its weapons. A sense of defeat had settled around Washington as he was forced to retreat across New Jersey in November and finally to Pennsylvania on December 8, 1776. The British, at least, considered the war over. By December 11th, the only reason the British had not taken Philadelphia, the seat of the Continental Congress, was that Washington had ordered every boat in the Delaware River on the New Jersey side to be brought to the Pennsylvania side, thus denying the British army transportation. Washington knew that the British would be capable of resuming an offensive by crossing the Delaware once it iced over. As the harsh winter set in, the morale of the American troops was at an all-time low. The soldiers were forced to deal with a lack of both food and warm clothing, while Washington watched his army shrink because of desertions and expiring enlistments. Now, more than ever, a victory was desperately needed. Washington devised a courageous plan to take the offensive and cross the Delaware River on Christmas night and attack the Hessian garrison at Trenton, New Jersey, nine miles south of his encampment near McConkey's Ferry. The original plan called for three divisions to cross the Delaware under the cover of darkness. Lt. Col. John Cadwalader's division was to cross at Bristol and engage the southern most contingent of British forces — Hessian troops under the command of Colonel von Donop. General James Ewing's division was to cross at Trenton Ferry and take a position south of Assunpink Creek below Trenton and hold the bridge over that stream. Washington's division was to cross at McConkey's Ferry and then divide into two corps under General Nathanael Greene and General John Sullivan. Their point of attack was Trenton and the Hessian troops quartered there under the command of Colonel Johann Gottlieb Rall. The boats to be used for the crossing were gathered earlier in the month in compliance with General Washington's orders, primarily as a defensive measure. Various types of boats had been collected, most notably the large Durham boats used to carry pig iron down the Delaware to the Philadelphia markets. There were a number of problems in moving a large number of men, cannons, and supplies in an age when overland transportation was by foot and animal power. The roads were rutted and winding. There were no bridges over major rivers because the technology did not exist to span great distances. A river like the Delaware was crossed by ferry, sometimes out of service because of ice floes or floods, and certainly not designed to carry masses of men and equipment across quickly. A river could be a formidable natural barrier to an army on the move. Washington had several logistical concerns for the crossing. In addition to the troops were the cannon; each of which required at least two horses to pull it. The heavier twelve pounders, and probably the eight pounders, had four horses. There would have been between four and six ammunitions wagons. Officers of the rank of colonel or higher may have had horses. In sum, Washington had to move 2,400 men, eighteen cannons, at least four ammunition wagons and fifty to seventy-five horses across the Delaware River the night of December 25, 1776. Fully expecting to be supported by Cadwalader's and Ewing's divisions south of Trenton, Washington assembled his own troops near McKonkey's Ferry in preparation for the crossing. By 6:00 pm, 2,400 men had begun crossing the ice-chocked river. There was an abrupt change in the weather, forcing the men to fight their way through sleet and a blinding snowstorm. The river was flooded with sheets of ice moving at eleven or twelve miles per hour. These obstacles proved to be too much for the two supporting divisions led by Generals Cadwalader and Ewing, who did not cross at their assigned points along the river. It was Washington's pure force of will and determination that led to his troops' successful crossing of the river. Increasing Washington's odds were the sailors of Marblehead, Massachusetts. This group of hardened seamen, led by Col. John Glover, were used to the Nor'easters of New England. Sheer determination and muscles conditioned to the demands of rowing under the weather conditions now facing the Continental army enabled the Marbleheaders to row back and forth across the Delaware countless times. During the time of the Revolution, American soldiers marched single file along the margins of the roads. They were only assembled into a battle line (three deep) when they reached the battlefield. The battle plan had Washington's army marching in two divisions... General Greene's and General Sullivan's. They made a night march in two columns on separate roads, a very tricky operation that was prone to failure since the columns needed to arrive at the battlefield at the same time to carry out the surprise attack planned by Washington. The American army carried out the march flawlessly. Against all odds, Washington and his men successfully completed the crossing and marched to Trenton on the morning of December 26th and, in the resulting battle, achieved a resounding victory over the Hessians. By moving ahead with his bold and daring plan, General Washington reignited the cause of freedom and gave new life to the American Revolution.
Land warfare is a complex domain that involves the application of military power on the ground to achieve political and strategic objectives. Modern military doctrine, such as that used by the U.S. Army and the Indian Army, categorizes these elements into Combat Power and the Principles of War. 1. The 8 Elements of Combat Power Combat power is the total means of destructive, constructive, and information capabilities that a military unit can apply. It is typically broken down into eight key elements: ElementDescriptionLeadershipThe "multiplier" of all other elements. It provides purpose, direction, and motivation to soldiers.InformationEnables commanders to make informed decisions and creates opportunities to achieve results.Mission CommandThe system used to integrate the other elements. It focuses on decentralized execution based on the commander's intent.Movement & ManeuverThe movement of forces to gain a positional advantage over the enemy to deliver lethal or non-lethal effects.IntelligenceThe understanding of the enemy, terrain, weather, and civil considerations.FiresThe use of weapon systems (artillery, mortars, air support) to create specific lethal or non-lethal effects.SustainmentThe logistics required to maintain operations, including ammunition, fuel, food, and medical support.ProtectionThe preservation of the force so that the commander can apply maximum combat power.2. The Principles of War These are the enduring "rules of thumb" that guide how land forces are employed strategically and tactically: Objective: Direct every operation toward a clearly defined and attainable goal. Offensive: Seize, retain, and exploit the initiative. You cannot win by defending alone. Mass: Concentrate the effects of combat power at the most advantageous place and time. Economy of Force: Allocate the minimum essential combat power to secondary efforts so you can "mass" elsewhere. Maneuver: Place the enemy in a position of disadvantage through flexible movement. Unity of Command: Ensure all forces operate under a single responsible commander toward a common objective. Security: Prevent the enemy from gaining an unexpected advantage. Surprise: Strike the enemy at a time, place, or in a manner for which they are unprepared. Simplicity: Prepare clear, uncomplicated plans to minimize confusion in the "fog of war." 3. The Modern Legal Framework Land warfare is also governed by the Law of Land Warfare (International Humanitarian Law), which rests on four pillars: Military Necessity: Actions must be necessary to achieve a legitimate military goal. Distinction: Forces must distinguish between combatants and non-combatants (civilians). Proportionality: The anticipated harm to civilians must not be excessive in relation to the concrete military advantage gained. Unnecessary Suffering: Weapons and methods must not cause gratuitous or superfluous injury. Note: Contemporary land warfare is increasingly "Multi-Domain," meaning land forces must now integrate with cyber, space, and electronic warfare to be effective. , While land warfare uses many tools, the two primary "philosophies" of how to win a war are Attrition and Maneuver. Most modern conflicts are a spectrum of both, but understanding the pure form of each helps explain military strategy. 1. Attrition Warfare: The "Sledgehammer" Attrition warfare is a strategy where one side attempts to win by wearing down the enemy to the point of collapse through continuous losses in personnel, equipment, and supplies. Core Logic: "I have more than you." It assumes that if you can destroy the enemy’s resources faster than they can replace them, you will eventually win. Focus: Firepower and mass. Success is measured by "body counts," equipment destroyed, and the steady seizing of terrain. Command Style: Usually centralized and methodical. It requires strict synchronization of massive resources (artillery, logistics, manpower). Historical Example: The Battle of Verdun (WWI). German Chief of Staff Erich von Falkenhayn famously stated his goal was to "bleed France white" by forcing them to defend a position they could not afford to lose, regardless of the cost in lives. 2. Maneuver Warfare: The "Scalpel" Maneuver warfare seeks to shatter the enemy’s moral and physical cohesion—their ability to act as a unified force—rather than simply destroying every soldier. Core Logic: "I am faster and more unpredictable than you." It aims to create a state of chaos where the enemy's leadership can no longer make effective decisions. Focus: Speed, surprise, and dislocation (forcing the enemy to be in the wrong place at the wrong time). The OODA Loop: Developed by Col. John Boyd, this is the heart of maneuver theory. It stands for Observe, Orient, Decide, Act. The goal is to cycle through these steps faster than the enemy, essentially "getting inside" their decision-making process until they collapse from confusion. Historical Example: The 1940 Invasion of France (Blitzkrieg). Instead of fighting a line-by-line battle of attrition, German forces used speed and concentrated armor to bypass strongpoints, cut communication lines, and cause a total systemic collapse of the French military in weeks. 3. Key Differences at a Glance FeatureAttrition WarfareManeuver WarfareObjectivePhysical destruction of the enemy army.Functional/Psychological collapse of the enemy.TargetThe enemy's strength (mass).The enemy's weakness (vulnerability).Primary ToolMassed Firepower.Movement and Tempo.Command"Command Push" (Top-down, rigid)."Recon Pull" (Decentralized, flexible).Success MetricExchange ratios (Kill counts).Disruption and loss of enemy control.4. The Modern Synthesis: "Schwerpunkt" In practice, no army is purely "maneuver" or "attrition." To maneuver successfully, you often need a period of attrition to punch a hole in the enemy's line. A critical concept here is the Schwerpunkt (Center of Gravity/Focus of Effort). A commander identifies the single most important place to strike and concentrates all available "elements of power" there. While the rest of the front might look like attrition, the Schwerpunkt is where the maneuver happens to achieve a breakthrough. Modern Reality: In high-intensity conflicts today (like the war in Ukraine), we see a "return to attrition" because modern sensors (drones, satellites) make it very difficult to achieve the surprise needed for pure maneuver warfare. When you can see everything, it's hard to be "unexpected."
Can you create an evaluation using this information PHONETICS VS. PHONOLOGY Whereas phonetics is the study of sounds that occur in language, phonology is the study of how these sounds are organized and how they function in language. It uses the classifications of sounds derived from phonetics to describe and analyze how sounds occur in speech. STRUCTURALIST PHONEMICS STRUCTURALIST PHONEMICS As linguists began to study sounds in fine detail, they recognized increasingly complex aspects of phonetic organization. For example, the sound /p/ appears in different varieties in English. STRUCTURALIST PHONEMICS One of the varieties of /p/ is indicated by [ph]. This sound is produced with an accompanying puff of air called aspiration, as in the words “pill,” and “peace.” Another sound, indicated by [p•], is produced when there is little or no aspiration; this sound occurs in a word like “spill.” A third major variety for the /p/ sound is the unreleased [p– ], which may occur at the end of a word like “stop.” To deal with these variations for the /p/ sound, the structuralists suggested the existence of an abstract unit which they termed a phoneme. STRUCTURALIST PHONEMICS A phoneme was defined by the structuralists as an abstract phonological unit that represents a class of real sounds, termed the allophones of a phoneme. The phoneme /p/ in English, then, is represented by the allophones [ph], [p•], and [p– ]. STRUCTURALISTS: MINIMAL PAIRS How do we know what these abstract units of sound called phonemes are? In order to find the phonemes of a language, the structuralists developed the concept of the minimal pair, defined as any two words that: a) Contain the same number of segments b) Differ in meaning c) Exhibit only one phonetic difference. STRUCTURALISTS: MINIMAL PAIRS In practical terms, phonemes distinguish meanings; and a phoneme can also be defined as the smallest meaning-distinguishing unit of sound. For instance, the words “pin” /pɪn/ and “bin” /bɪn/ mean different things, and the only one difference in these words occurs in the initial sounds. STRUCTURALISTS: MINIMAL PAIRS By using the concept of a minimal pair, we can determine that the three variations of the /p/ sound do not represent three phonemes. Certainly, it is possible to pronounce the word cap with either an aspirated [ph ] or unreleased [p– ]; however, the two forms [kæph ] and [kæp– ] are not a minimal pair, even though they involve different sounds, because they are identical in meaning. STRUCTURALISTS: FREE VARIATION The two forms [kæph ] and [kæp– ] are, therefore, said to exhibit free variation: that is, the pronunciation may vary without signifying a change in meaning. In other words, we may conclude that the unreleased [p– ] and the aspirated [ph ] are not representations of different phonemes in English; they are, in fact, allophones of one phoneme, /p/. STRUCTURALISTS: COMPLEMENTARY DISTRIBUTION When phonemes have more than one allophone in a language, the allophones are said to be in complementary distribution. Complementary distribution means that the allophones of a phoneme occur in different phonetic environments (that is, with different sounds surrounding them). TRANSFORMATIONAL- GENERATIVE PHONOLOGY TRANSFORMATIONAL-GENERATIVE PHONOLOGY Transformational-generative phonology is a relatively recent development in linguistic theory. Chomsky launched Transformational-Generative Grammar in 1957, but the earliest studies within this framework were largely concerned with syntax. A decade later, the first comprehensive transformational-generative treatment of English phonology appeared: Chomsky and Halle’s The Sound Pattern of English (1968). TRANSFORMATIONAL-GENERATIVE PHONOLOGY Transformational-generative phonologists strongly oppose the structuralists’ phonemic level. They replace this level by a series of rules that directly relate underlying representations to observed phonetic representations. The central mechanisms in transformational-generative phonology, then, are underlying representations and phonological rules. PHONOLOGICAL RULES A rule is an operational statement in which some linguistic entity is modified, resulting in a new linguistic entity. Rules may add elements, remove elements, or change elements. By using phonological rules, linguists attempt to demonstrate that there is order in linguistic phenomena and that linguistic patterns are systematic. PHONOLOGICAL DERIVATION A phonological derivation is an operation that begins with an underlying representation and, through the application of a set of specific rules, yields the actual sound the speaker produces. The representation of a phonological rule has the following general appearance. /A/ → [B] / C “A” changes to “B” under condition “C” PHONOLOGICAL RULE – EXAMPLE In most Southern dialects, the word ten is pronounced like the word tin. This is not an isolated fact, for den is pronounced like din and Ben is pronounced like bin, and so on. This very general fact can be represented by the phonological rule: /ɛ/ → [I] / ___ [n] den /dɛn/ → /dIn/ Ben /bɛn/ → /bIn/ ten /tɛn/ → /tIn/ /ɛ/ → [I] / ___ [n] - high - low - tense + front + high - tense + front + sonorant + anterior + coronal - continuant NOTATIONAL DEVICES IN PHONOLOGICAL RULES The statement of phonological rules can be complex, and linguists have developed several notational devices for writing them. Often, the following symbols will be necessary for stating the conditions under which rules apply: # indicates a word boundary + indicates an intraword boundary $ indicates a syllable boundary UNDERLYING REPRESENTATIONS AND RELATED ISSUES The transformational-generative description of phonology relates underlying representations to phonetic representations by rules. This can be represented in a simple example: In English, there are certain pairs of words like sign / signature, and malign / malignant that exhibit a regular alternation in their phonetic representations: [g] is present in the second member of the pairs but absent in the first member. UNDERLYING REPRESENTATIONS AND RELATED ISSUES To explain the relatedness of words such as sign / signature, we could claim that the underlying representation of the segment in all such pairs is /g/ and that a rule operates to delete /g/ before syllable-final nasals. Thus, the rule “/g/ is deleted before syllable-final nasal” would appear formally as: + voice - anterior →∅ ____ [+ nasal] $ - coronal UNDERLYING REPRESENTATIONS AND RELATED ISSUES On the left-hand side of the arrow, we place the features needed to uniquely specify /g/ among the consonants; that is, no other consonant has the features [+ voice], [- anterior], and [- coronal]. The symbols → mean that the sound /g/ changes to nothing or more properly “/g/ is deleted.” The horizontal line following the slash mark refers to the position of /g/ - namely, before a segment that is [+nasal]. Finally, this [+nasal] segment occurs before a syllable boundary, as indicated by $. A less formal way of writing this rule would be: /g/ → / _ [+nasal] $ Notice that this rule also helps describe such alternations as phlegm/phlegmatic and paradigm/paradigmatic. Application Activity: Think of other words in which this rule can be applied. Write the sound segments to prove /g/ is deleted. Another example is the process through which the prefix meaning “not” is added to words. This prefix alternates among the forms /Im/, /In/, and /Iŋ/, depending on the point of articulation of the initial segment of the following word. -If the segment begins in the extreme front part of the mouth (labials), the form is /Im/, as in improper. -If the segment begins in the extreme back part of the mouth (velars), the form is /Iŋ/, as in incomplete. -If the segment begins in the mid-region of the mouth (all other sounds), the form is /In/, as in indecent. *Exceptions:Words beginning with /r/ or /l/. Analyze the Word “in + complete,” for example. /n/ → [ŋ] / __ [k] - continuant - continuant - continuant + sonorant → + sonorant - sonorant + anterior - anterior - strident + coronal - coronal - coronal + tense THE VELAR SOFTENING RULE Still another example of alternation in English is found in pairs of words like “electric / electricity,” in which the segments /k/ and /s/ alternate. /k/ changes to [s] only before non- low, front vowels. THE VELAR SOFTENING RULE /k/ → [s] / __ - continuant + continuant - strident → - sonorant V - anterior + anterior - low - coronal + coronal - back
Make mcq quiz with 4 option in which one is correct -'10 Basis of Material Science • .....;;;";;;"~~;;,,;;,,,,;.;.,,;;,,,;,,;.;,.,------------ 6. Temporary materials: Some materials are meant to be placed in the oral cavity for a short period of time for different reasons. • Temporary crowns: While a permanent crown is prepared in the dental laboratory, the patient must wait for few days before it can be fabricated and cemented into place. Does patient experience any problems during this time period? If the tooth is vital (the pulp is alive), the patient is likely to experience pain and sensitivity while eating and drinking, also it looks unesthetic. What can be done to solve this problem? A temporary crown is placed before the patient leaves the clinic. It is constructed and luted in the same appointment in which the crown preparation is done. Temporary crowns are not very strong or esthetic but they serve adequately till the permanent crown is ready to be cemented. • Temporary restorations: Sometimes it is difficult to decide immediately the best line of treatment for a particular tooth. The exact condition of the pulp may not be obvious to the dentist from the patient's symptoms. A dentist removes all or part of the decay and then places a temporary restoration to have time to observe the behaviour of the pulp or to give the pilip time to heal before deciding the further treatment required. Classification based on Location of Fabrication 4,9 Materials can be classified based on the location of fabrication into: • Direct restorative materials. • Indirect restorative materials Direct restorative materials: They include those materials which are used to restore cavity preparations directly in the oral cavity (Box 1.5). Box 1.5: Examples of direct restorative materials Amalgam, composites, glass ionomer and other materials, which set by chemical reactions in the mouth. Indirect restorative materials: It includes those restorations which must be fabricated outside the mouth, indirectly on a cast/ model/ die, because their processing condition would harm oral tissues. Materials used in the construction of such prosthesis are called indirect restorative materials (Box 1.6). Box 1.6: Examples of indirect restorative materials Gold inlays, crowns of metal, ceramic and polymers, which are processed at elevated temperatures. Some indirect composite restorations can be processed under specific wavelength of light, e.g. Ceramage. Classification based on Longevity of Use 1. Permanent restorations: These restorations are not planned to be replaced for a particular time period. Though they are referred to as permanent, actually they are not, e.g. fillings, crowns, bridges and dentures do not last forever (Fig. 1.5). 2. Temporary restorations: These restorations are planned to be replaced in a short period of time, such as few days to weeks. For ~ Permanent C/) c c -.2 0 c- :;::; Cll co Interim ~ Q; 0 .8ll::1iJ C/) o~ Cll a:: c:=:J Temporary Time period Fig. 1.5: Diagram depicting the time period of use of a restoration. (Arrow in permanent restoration depicts that such restorations are not planned to be replaced for a long period of time.) Introducton to Dental Materials Dental materials Box 1.7: Characteristics of metals 1. High thermal and electrical conductivity 2. Ductility (pure metals are very soft and they can be bent without breaking) 3. Opacity (they do not transmit light) 4. Luster (they have a surface that strongly reflects light and appears bright and shiny) 5. They tend to dissolve to some extent in water or other aqueous solutions, producing cations. 6. All metals are white (actually gray) except for gold, which is yellow, and copper, which is reddish. 7. All metals are solid at room temperature except mercury, which is liquid at room temperature and is used with silver alloys as amalgam. 8. All metals have high melting temperatures because of high strength of the metallic bond that holds the atoms together. 3. Polymers 4. Composites Composites are mixtures of two or more of the first three classes in which the different components remain distinct from one another in the final structure. A common example is composite resin. Fig. 1.7a: Three-dimensional structure of iron (metal) Metals Metals are the oldest of the three classes of materials that have been used as dental materials. Metals are characterized by metallic bonds (Box 1.7) which will be discussed in the next chapter. Metals solidify with their atoms in a regular or crystalline arrangement (see Chapter 2), often in the form of a cube (Fig. 1.7a). example, temporary fillings done in a tooth during root canal treatment, which have to be replaced within 2-4 days during subsequent visits. They are used to protect the tooth and provide function till the final restoration is done. 3. Interim restoration: At times, dental treatment requires "long-term" definite temporary restorations or "interim" restorations. For examle, a 7-year-old child, met with trauma and fractured one of his central incisors. A large composite build- up may serve his immediate requirement until the root formation is completed and a permanent crown is placed. 5 Classification based on the Chemical Nature of the Material These are the atoms that make up a material and the way they are bonded together determine the properties of that materiaLS Weak bonds make for weak materials and vice versa (Table 1.4). Materials can be classified into different categories based on their primary atomic bonds (Fig. 1.6): 1. Metals 2. Ceramics Fig. 1.6: Classification of dental materials based on chemical nature 12 Basis of Material Science Box 1.9: Benefits of ceramics in dentistry 1. Many ceramic oxides are used as pigmenting agents. These oxides produce good range of colors. Due to this characteristic, we are able to match almost any tooth color with good esthetic results. 2. They are inert, i.e. not chemically reactive. This quality provides ceramics with good bio- compatibility. 3. Ceramic materials are translucent, like natural teeth. This translucency gives the ceramic crown a more natural appearance than any other dental material. Fig. 1.7b: Internal arrangement of tetrahedral structure of ceramic (silica) four large oxygen atoms surround smaller silicon atom Ceramics A ceramic is a compound formed by the union of a metallic and a non-metallic element (Box 1.8). Most of these materials are oxides, formed by the union of oxygen with metals such as silicon, aluminum, calcium and magnesium (Fig.1.7b). Ceramics may be simple or complex. Examples of simple ceramics are alumina and silica. Examples of complex ceramics are feldspar (potassium aluminum silicate) and kaolin (hydrated aluminum silicate). Ceramics may be crystalline or non- crystalline (i.e. amorphous). Porcelain is a specific type of ceramic used extensively in dentistry (Box 1.9). Box 1.8: Characteristics of ceramics 1. High melting points. 2. Brittleness, which means they cannot be bent or deformed (no sliding) to any extent without actually cracking and breaking. 3. They are poor conductor of heat and electricity. 4. They are chemically inert. 5. They have excellent esthetic result in terms of matching natural teeth. Fig. 1.8: Stucture of synthetic polymer Polymers They are the latest addition (early to mid- 1900s) to dental materials. Most of the polymers are nowadays synthesized by humans. Polymers are giant, long-chain organic molecules (Fig. 1.8). Polymers are characterized by covalent bonds within each molecule, giving them tremendous strength in a single direction. Try to break a nylon rope by pulling it! They are poor conductors of heat and electri- city. Most polymers have a structure containing thousands of carbon atoms linked together like beads on a string. Others, such as silicone polymers are formed with silicon-oxygen bonds. Introducton to Dental Materials Table 1.4: Characteristics of different materials 13 Characteristics Bond Properties Crystal structure Metals Metallic bonding High strength and hardness, high electrical and thermal conductivity BCC, FCC, or HCP unit cells Ceramics Ionic or covalent bonding, or both High hardness and stiffness, electrically insulating, refractory, and chemically inert Crystalline or amorphous Polymers Covalent bonding Low sensitivity, high electrical resistivity, and low thermal conductivity, strength and stiffness vary widely Amorphous and crystalline Composites Composites are combinations of any of the basic ceramic, metallic and polymeric materials (Box 1.10). Each material that makes up composites is called a phase. Their properties tend to be somewhere between those of their basic constituents and are used to enhance their performance, longevity and handling chracterstics. Box 1.10: Types of composites in dentistry 1. Ceramic - metallic composite: Tungsten carbide bur. 2. Metal - polymer composite: Die materials in dental laboratory. 3. Ceramic - polymer composite: Enamel, dentin, bone and restorative composites. A composite is a kind of "combination" of materials, which compliment each other. The properties lacking in one material are compensated by those of the other material. For example, restorative composite has two phases, namely resin and fillers. Teeth and bones are examples of natural composites. Enamel is a composite of hydroxyapatite (which is a ceramic material) and protein (which is a polymer). EVALUATION OF DENTAL MATERIALS Most manufacturers of dental materials maintain a quality assurance programme (As per international standard like ADA specifications) and materials are thoroughly tested before being released into the market for dental practitioner (Fig. 1.9). Laboratory Evaluations Most ADA/ ANSI specifications involve laboratory tests. The tests performed as per these specifications are useful but they all are performed in vitro, (carried out in the laboratory away from the clinical conditions) which have a lot of limitations in clinical practice.lO Clinical Notes 1. For example, most of the direct restorative materials are tested for their compressive strength but ultimately the material is subjected to a combination of compressive, tensile and shear stresses, which may decide the final success or failure of the material under masticatory load. 2. Similarly upper dentures mostly fracture along the midline because of bending. Hence a bending or transverse strength ~B-a-s-is-o-f-M-a-t-e-ria-I-S~c-ie-n-c-e-------------- ---------. test is far more meaningful for denture base materials than a compression test. Clinical Trials The majority of new materials are subjected to extensive clinical trials normally in co-operation with a dental college or hospital departments prior to their release. CONCLUSION As the number of available materials is going up, it is important that the dentist remains more aware about new products so that their judgement about the selection of material remains successful. Materials which have not been thoroughly evaluated should be avoided, specially with clinical dentistry falling under Consumer Protection Act (CPA). I Research and development I iI Manufacturer/analysis Ideal requirements for clinical use: Thermal, optical, mechanical, chemical, biological Available materials and their properties are evaluated Launch of new I product Choice and selection of material by the dentist Critical assessment based on clinical performance I I H feedback to I
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 •(-9.8m/s2) •d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) •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.