
Located in West China, Xi'an is one of the most notable ancient Chinese cities. More than 3,100 years have passed since it was built. Xi'an is the starting point of the Silk Road. It used to be a commercial hub connecting the west and the Orient during its prosperous period. Xi'an has the largest bell tower, which was built in the Ming Dynasty. The largest collection of ancient stone tablets and the Terracotta Warriors with a history of 2,000 years. Now it is a rapidly developing modern city. Xi'an is an undying city in Chinese culture.
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The Benin Kingdom is located in the southern forest of West Africa (Modern Nigeria) and formed by the Edo people, flourished from the 13th to 19th century CE. The capital also is called Benin, it was the hub of a trade network exclusively controlled by the king or Oba and which included relations with Portuguese traders who sought gold and slaves. Benin went into decline during the 18th century. The Benin territory is a mixture of rainforest, dry forest, and mangrove swamp. The heartland was a circle around the capital, also called Benin, extending some 60 kilometers in all directions and was ruled directly by the king. THE OBA OF BENIN This is the traditional ruler and the custodian of the culture of the Edo people and all Edoid people. THE FOLLOWING ARE THE PAST OBAS OF BENIN KINGDOM AND YEAR OF REIGNING. 1. Pre – Imperial Benin (1200- 1440) Eweka 1 (1200 -1235), Uwakhuahen (1235 – 1243), Ehenmihen (1243 – 1255). 2. Imperial Benin ((1440 – 1897) Ewuare the Great (1440 – 1473), Olua (1473 -1480), Ozolua(1483 – 1504). 3. Post- Imperial Benin (1914- Present) Eweka II(1914 – 1933) Akenzua II (1933 – 1978) Erediauwa (1976 – 2016)
Continental Drift Theory. From the discussion of the rock cycle, it has been pointed out that through Earth's external and internal processes. Earth's surface is constantly changing. However, this idea of a changing environment did not conform with the belief of earlier scientists. Rather, they thought that the geographic positions of ocean basins and continents have been static since the beginning of time. It was around the 1500s when Leonardo da Vinci, upon his discovery of fossil seashells found at the high mountains of Italy, first thought of the idea that the areas where mountains are located may have been oceans in the past. Through time, other fossils of marine organisms found far above the current sea level further supported the idea that mountains were uplifted and weathering wore them down. At around the 1800s, most scientists have accepted the idea that Earth's crust is undergoing large vertical movements or uplifting. There was also evidence of possible horizontal movements, but the scientists then were not convinced about it. Alfred Wegener showed evidence of horizontal or lateral movement of the continents in his continental drift theory. According to him, the continents have drifted around the world and have once formed a giant landmass or supercontinent called Pangaea. To support his theory, Alfred Wegener presented a set of geographical, biological, and climatic evidence.Wegener's geographical evidence included the jigsaw puzzle fit of the current continents. He pointed out that the coastlines of South America and Africa seem to fit together. He also pointed the presence of mountain ranges having similar rock types and age but separated by vast oceans, like that of the folded rocks of the Caledonian mountains. The same folded rocks run through West Africa, North America, Newfoundland, Ireland, Wales, Scotland, Greenland, and Norway, all of which are now separated by the Atlantic Ocean. A geographical evidence on the similar rock types in West Africa, North America, Greenland, and Europe is found. The biological evidence came in the discovery of similar plant and animal fossils in different continents separated by oceans. The animal fossils of Mesosaurus and Lystrosaurus indicate that they were not capable of crossing the oceans to reach the other continents. If they were, the fossils should have been more widely distributed Africa, Australia, India, and South America were too large to be carried by wind. This indicates that the areas where the fossils were found were closely linked. It has also been found out that the plant only grew in areas with subpolar climate, which would indicate that the landmasses were located near the South Pole.Lastly, for his climatic evidence, Wegener discovered that a glacial period occurred during the late Paleozoic era in Southern Africa, South America, Australia, and India. The initial explanation for this event was global cooling, but it was rejected because large tropical swamps with so much vegetation were found at the same time in the Northern Hemisphere. This further supported the idea that the supercontinent was indeed near the South Pole, and the continents in Northern Hemisphere were once near the equator. The glacial period also left glacial striations, or the scratches glaciers make as they move across on the underlying bedrock, on the aforementioned continents. For such an event to happen, the continents would have to be connected. SCIENCE PIONEER. Alfred Wegener (1880-1930). Alfred Wegener was a German polar researcher, geophysicist, and meteorologist. He was known for his work on the continental drift theory. In his effort to defend his work, he went to the Greenland ice sheet where he died.Even with all the compelling evidence, the continental drift theory hardly convinced the scientific community at that time because Wegener was unable to identify a credible mechanism that drives the continental drift. He was unable to clearly explain how the continents moved and how the larger continents broke through the ocean floor. Eventually, critics of the continental drift began to accept the theory when new evidence supporting the theory was discovered. The new evidence led to a more encompassing theory the theory of plate tectonics. This theory provided a more convincing explanation as to how the continents moved. The evidence that paved the way for the theory of plate tectonics was the idea of wandering poles. Scientists began studying volcanic rocks to determine the location of the magnetic poles. When volcanic rocks crystallize, the minerals with magnetic properties align themselves parallel to Earth's magnetic field at the time the minerals were formed. This finding allowed scientists to determine the polarity of Earth's magnetic field and the magnetic inclination that showed the location of the poles. Upon studying the paleomagnetism of the rocks, geophysicists found out that rocks from various locations point to different magnetic north poles, suggesting that the poles have wandered. Since movement of magnetic poles is very unlikely, scientists have accepted the idea that the continents are indeed moving. And if the continents are moving, scientists thought that maybe the ocean basins are moving too. They also discovered that some rocks showed magnetic reversals, which led them to believe that the magnetic north pole now was not always the magnetic north pole. Seafloor Spreading. After World War II, exploration on the ocean floor became the focus of many geologic studies. It was only then that the ocean ridge system was discovered. A geologist in Princeton University named Harry Hess, along with other scientists, studied this ocean ridge system and hypothesized that the oceanic crust was moving away from the ridge. His hypothesis, known as seafloor spreading, showed that the ocean floor is split along the ridge where the magma rises to form the new ocean floor.Because of this, rocks located near the ridge are younger than those that are located magnetic polarity of Earth is also preserved in those rocks. Withe ridge scientists were able to see the magnetic reversals in the ocean floor, and they were able to make use of information to determine that the ocean floor is moving at a rate of about 10 cm per year. Plate Tectonics. Confirmation of the seafloor spreading hypothesis proved that continents are not moving above the ocean floor. Rather, it is the fragments of the lithosphere. The lithosphere is the rigid layer that is composed of the uppermost mantle and the crust that carry the continents and the ocean basins along. These fragments of the lithosphere are called plates. Underneath the lithosphere is a weaker region in the mantle known as asthenosphere that behaves like a fluid. Thus, the lithosphere floats above the asthenosphere, making it detached and free to move. This became the basis of the theory of plate tectonics. Now that it has been made clear that it is the plates which are moving, the question as to how they move remained. Sir Arthur Holmes proposed the driving force for this plate movement in 1919. He suggested that the movement in the mantle carries the plates along. It was previously discussed that Earth's interior is very hot due to the heat produced by radioactive decay. Convection takes place in the mantle, keeping the asthenosphere hot and weak. The convection currents produced in the asthenosphere are the ones carrying the lithospheric plates and making them move. However, convection currents are not enough. Mechanisms such as ridge push and slab pull aid the convection currents to slowly move the lithospheric plates. Ridge push occurs at mid ocean ridges which are higher in elevation than the surrounding trenches and abyssal plains. The new ocean floor from the ridge is hot and relatively thin. As it moves away from the ridge, it cools down and gets denser, heavier, and thicker. Below this cooling ocean floor is the asthenosphere, which is less dense. This area becomes a massive shear zone and the new ocean floor will effectively slide down the slope of the asthenosphere. When the plate collides with another plate with lesser density, the denser plate sinks and a subduction zone is formed. When the subducting plate sinks, it pulls on the rest of the plate behind it. These mechanisms explain the movement of the plates.Earth has seven major lithospheric plates that account for 94% of Earth's surface. These are the North American Plate, South American Plate, Pacific Plate, African Plate, Eurasian Plate, Indo-Australian Plate, and Antarctic Plate. These plates are constantly moving relative to the other plates. Thus, the interaction of plates occurs mostly along the boundaries. These movements are plotted using information from earthquakes and volcanic activities. There are three main types of plate boundaries: convergent, divergent, and transform boundaries Convergent boundaries are boundaries where two plates move towards each other A convergent boundary is also known as destructive margin since this is where the collision between two plates occhins. There are three types of convergence-oceanic oceanic, oceanic-continental, and continental-continental. Trenches are features of the ocean floor that are present in both oceanic-oceanic boundary and oceanic-continental boundary. Subduction occurs at the trenches, therefore, these are characterized as the deepest parts of Earth. A divergent boundary is the opposite of convergent boundary: two plates move away from each other. Divergent boundaries create new crust; thus, they are also known as constructive margins. The ocean ridge system is a divergent boundary where new ocean floor is produced as magma rises, pushing the older rocks aside.Transform boundary is also known as conservative plate margin since two plates just move past one another, neither creating nor destroying land. Earthquake epicenters are usually detected at transform boundaries because the rocks tend to break and not fold or sink, like in convergent boundaries. Evolution of the Ocean Basins. Both the movement of the plates and seafloor are responsible for the evolution of ocean basins. Along the divergent boundary where ocean ridge systems are found, magma is released and new ocean floor is created. Along convergent boundaries, the ocean floor is being destroyed. The evolution of the ocean basins started during the time when Pangaea was still present and was surrounded by the vast ocean or superocean known as Panthalassa, also called Paleo-Pacific or "old Pacific." Upon the initial break up of Pangaea into Laurasia and Gondwanaland, the Tethys Sea began to form. Then, the Eurasian and North about, forming the North Atlantic. The South Atlantic only started to form when the African Plate and South American Plate separated. The continued movement of the plates created the Himalayas at one side and separated the Pacific Ocean and Atlantic Ocean at the other side, which consequently formed the current ocean basins. Both the movement of the plates and seafloor are responsible for the evolution of ocean basins. Along the divergent boundary where ocean ridge systems are found, magma is released and new ocean floor is created. Along convergent boundaries, the ocean floor is being destroyed. The evolution of the ocean basins started during the time when Pangaea was still present and was surrounded by the vast ocean or superocean known as Panthalassa, also called Paleo-Pacific or "old Pacific." Upon the initial break up of Pangaea into Laurasia and Gondwanaland, the Tethys Sea began to form. Then, the Eurasian and North about, forming the North Atlantic. The South Atlantic only started to form when the African Plate and South American Plate separated. The continued movement of the plates created the Himalayas at one side and separated the Pacific Ocean and Atlantic Ocean at the other side, which consequently formed the current ocean basins.Continents do not immediately end at the point where the ocean meets the land. They may extend slightly into the oceans. The portion of the continent that is submerged is called continental margin. There are two types of continental margin: passive margin and active margin. A passive continental margin consists of a continental shelf, continental slope, and continental rise. It is not associated with plate boundaries; thus, there are very little tectonic activities. An active continental margin only has a continental shelf and a continental slope. It is associated with plate boundaries; thus, a main feature of this boundary is a trench. The different features of a continental margin are the following: 1. The continental shelf is the gently-sloping submerged portion of the continent. 2. The continental slope is the steep slope after the continental shelf. It is still part of the continent. 3. The continental rise is the gently-sloping area after the continental slope and before the ocean floor. 4. The trenches are the deepest parts of the ocean. These are narrow depressions caused by the subduction of the ocean floor along the convergent boundaries. 5. The mid-oceanic ridge is the mountain range system in the ocean. It is responsible for the production of new ocean floor. This is the region where new magma constantly emerges from. SCIENCE CAREER. A scientific illustrator uses art to inform and communicate complex details and concepts of science. He/She makes use of scientifically informed observations and research along with his/her technical art and aesthetic skills to make accurate representations. In Natural History, the scientific illustrators recreate how the extinct species look like by working with scientists and fossil records. Moreover, with the advances in technology, illustrators are now into 3D modelling, animation, and video making. Earth's History. All the processes that have been discussed require long periods of time to create a noticeable change on Earth's surface. You can just imagine how long it would take to create an oceanas vast as the Pacific Ocean if the ocean floor moves only at about 10 cm/year. It is then important to know the history of Earth to learn the complexities of its past and be able to use it to understand the present. Just like learning the history of a country that requires one to read a lot of books, learning the history of Earth involves studying a lot of rocks. Rocks, especially sedimentary rocks, contain a lot of information about Earth's past. It holds the key to most of the geologic processes that happened on Earth and the key to uncovering how life on Earth evolved. But these discoveries are worthless if there is no time perspective. Thus, one of the most important contributions of geologists to mankind is the geologic time scale, which holds a history that is exceedingly long.
Chapter 8: The Worlds of North and South Geography Geography refers to the seasons, climate, soil, and physical features of a region (mountains, rivers, etc.) The differences in geography b/t the N and S is one of the major reasons slavery b/c entrenched in the S while it died out in the N. Geography of the North The N has diverse geography and experiences four distinct seasons including long, harsh winters. The Great Plains region has some of the best farmland in the country. New England has rocky, hilly wilderness, not well suited for farming. It has hundreds of bays and harbors along its coastline. States farther S had rich soil and coastal access through rivers. The N also experienced mass deforestation b/c of the need for lumber and to make room for farms. Geography of the South Climate: the S had mild winters, and a long, hot, humid growing season. It has fertile lowlands, marshes and swamps. It's ideal for growing tobacco, sugar, rice, indigo, and cotton (cash crops). B/c of the geography of the S, their whole way of life was based on agriculture and geography is one of the major reasons why slavery took off in the S. Economies Economy basically refers to the way people make and spend money. The Northern economy was far more diversified than the Southern. Economy of the North The North experienced the Industrial Revolution—the shift from handmade goods to machine-made goods. This resulted in new jobs, increased production, and improved efficiency in agriculture. IOW, you can make things faster, easier, and cheaper. More ppl get more stuff. Factories were almost always located next to rivers. The Reaper The Indust. Rev. changed northern agriculture with Cyrus McCormick’s reaper. It could cut 28xs more grain than a single man. The Sewing Machine Elias Howe's sewing machine; At 250 stitches a minute, Howe's lockstitch mechanism out-stitched the output of five hand seamstresses with a reputation for speed, completing in one hour what took the sewers 14.5 hours. The Textile Mill Francis Cabot Lowell's textile mill: essentially the first factory in the US, Lowell set the model for all future factories. Interchangeable Parts Eli Whitney's interchangeable parts; considered the "dawning of a new age" of machinery. This concept was applied to pretty much all manufacturing. Economy of the South The South's economy was based on AGRICULTURE. Most southerners were agrarians. Most had small farms, some owned plantations. Slavery beginning to decline in late 1700s; prices went down (tobacco, indigo) and cotton was difficult. King Cotton Cotton was South’s most important crop. Earned more money than all other exports combined. The S would go on to supply 75% of the world's cotton demand. Cotton Gin Eli Whitney invented the cotton gin in 1794 and forever changed the US. The gin made cotton incredibly profitable. We start to see the effects of the cotton gin around 1820. Slavery and Cotton Southerners put all their money into slaves and land, and almost none into building factories. With the spread of cotton, demand for slaves increased. 1790 to 1850, number of slaves rose 600%. Transportation Again, the N was far more inventive in their approach to transportation than the S. Transportation in the North National Road National Road stretched from the East (the Potomac), over the Appalachians, to the West (Illinois), over 620 miles. Steamboat In 1807, Robert Fulton invented the steamboat. It traveled 150 miles UP the Hudson River at a speed of 5 mph. Erie Canal Built b/t 1817 and 1825, the canal spanned 363 miles and connected Lake Erie to the Hudson River. This connected farms in the W to cities in the E and the Atlantic Ocean. Clipper Ship Clippers were narrow w massive sails that were built for speed. They cut the time it took to cross the Atlantic in half. Locomotive The fastest and cheapest way to move goods was by steam-powered trains. The first RR was the B&O which was built in 1827. Transportation in the South Most people and goods in the South traveled by rivers in steamboats. The South had trains, but less than half the amount of railroad track than the North had. Society (The People) The people who made up the N and S could not have been more different. The S was primarily agrarian while the N was b/c urbanized. The S was holding on to the past, while the N was embracing change. Society in the South Society was organized into 3 distinct classes of people: rich plantation owners at the top; then white farmers and workers; slaves on the bottom. This rigid social class system was the result of a slave-based agricultural system. Power Structure Only 1 in 4 whites owned a slave. Plantation owners, who owned more than 20 slaves, dominated politics and the economy. Society in the North 7 of 10 Northerners still lived on farms by the 1840s (6 of 10 by 1860), but urbanization was growing fast in the N. The N relied on wage labor as opposed to slave labor, so most blacks in the N were free. N blacks were not treated equally and the N was about as racist as the S. Immigration Compared to the S, the N population was exploding, in large part bc of immigration. Between 1845 and 1860, 4 million immigrants came to the North. Most were German and Irish. Irish--a potato famine; German--a failed revolution. Ethnic neighborhoods developed as a result.
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.
WHERE ARE THE PARTS LOCATED IN ROBOTS AND MECHS
Nucleus pulposus • Located in the mid-to-posterior part of the disc. • Pulplike gel or Hydrophilic like gel consistency large branching proteoglycans. • Collagen forms an infrastructure that support the proteoglycan network. • Hydrated proteoglycans are thin type II collagen fibers, elastin fibers, and other proteins • Proteoglycan is water-binding glycosaminoglycans linked to core proteins • In youth the nucleus pulposus within the lumbar discs consists of 70% to 90% water. • The hydrated nucleus allows the disc to function as a hydraulic shock absorption system, dissipating and transferring loads across vertebrae
Understanding Quantum Theory of Electrons in Atoms The goal of this section is to understand the electron orbitals (location of electrons in atoms), their different energies, and other properties. The use of quantum theory provides the best understanding to these topics. This knowledge is a precursor to chemical bonding. As was described previously, electrons in atoms can exist only on discrete energy levels but not between them. It is said that the energy of an electron in an atom is quantized, that is, it can be equal only to certain specific values and can jump from one energy level to another but not transition smoothly or stay between these levels. The energy levels are labeled with an n value, where n = 1, 2, 3, …. Generally speaking, the energy of an electron in an atom is greater for greater values of n. This number, n, is referred to as the principal quantum number. The principal quantum number defines the location of the energy level. It is essentially the same concept as the n in the Bohr atom description. Another name for the principal quantum number is the shell number. The shells of an atom can be thought of concentric circles radiating out from the nucleus. The electrons that belong to a specific shell are most likely to be found within the corresponding circular area. The further we proceed from the nucleus, the higher the shell number, and so the higher the energy level (Figure 9.4.1). The positively charged protons in the nucleus stabilize the electronic orbitals by electrostatic attraction between the positive charges of the protons and the negative charges of the electrons. So the further away the electron is from the nucleus, the greater the energy it has. This quantum mechanical model for where electrons reside in an atom can be used to look at electronic transitions, the events when an electron moves from one energy level to another. If the transition is to a higher energy level, energy is absorbed, and the energy change has a positive value. To obtain the amount of energy necessary for the transition to a higher energy level, a photon is absorbed by the atom. A transition to a lower energy level involves a release of energy, and the energy change is negative. This process is accompanied by emission of a photon by the atom. The following equation summarizes these relationships and is based on the hydrogen atom: The values nf and ni are the final and initial energy states of the electron. The principal quantum number is one of three quantum numbers used to characterize an orbital. An atomic orbital, which is distinct from an orbit, is a general region in an atom within which an electron is most probable to reside. The quantum mechanical model specifies the probability of finding an electron in the three-dimensional space around the nucleus and is based on solutions of the Schrödinger equation. In addition, the principal quantum number defines the energy of an electron in a hydrogen or hydrogen-like atom or an ion (an atom or an ion with only one electron) and the general region in which discrete energy levels of electrons in a multi-electron atoms and ions are located. Another quantum number is l, the angular momentum quantum number. It is an integer that defines the shape of the orbital, and takes on the values, l = 0, 1, 2, …, n – 1. This means that an orbital with n = 1 can have only one value of l, l = 0, whereas n = 2 permits l = 0 and l = 1, and so on. The principal quantum number defines the general size and energy of the orbital. The l value specifies the shape of the orbital. Orbitals with the same value of l form a subshell. In addition, the greater the angular momentum quantum number, the greater is the angular momentum of an electron at this orbital. Orbitals with l = 0 are called s orbitals (or the s subshells). The value l = 1 corresponds to the p orbitals. For a given n, p orbitals constitute a p subshell (e.g., 3p if n = 3). The orbitals with l = 2 are called the d orbitals, followed by the f-, g-, and h-orbitals for l = 3, 4, 5, and there are higher values we will not consider. There are certain distances from the nucleus at which the probability density of finding an electron located at a particular orbital is zero. In other words, the value of the wavefunction ψ is zero at this distance for this orbital. Such a value of radius r is called a radial node. The number of radial nodes in an orbital is n – l – 1. Consider the examples in Figure 9.4.2. The orbitals depicted are of the s type, thus l = 0 for all of them. It can be seen from the graphs of the probability densities that there are 1 – 0 – 1 = 0 places where the density is zero (nodes) for 1s (n = 1), 2 – 0 – 1 = 1 node for 2s, and 3 – 0 – 1 = 2 nodes for the 3s orbitals. The s subshell electron density distribution is spherical and the p subshell has a dumbbell shape. The d and f orbitals are more complex. These shapes represent the three-dimensional regions within which the electron is likely to be found. Principal quantum number (n) & Orbital angular momentum (l): The Orbital Subshell: https://youtu.be/ms7WR149fAY If an electron has an angular momentum (l ≠ 0), then this vector can point in different directions. In addition, the z component of the angular momentum can have more than one value. This means that if a magnetic field is applied in the z direction, orbitals with different values of the z component of the angular momentum will have different energies resulting from interacting with the field. The magnetic quantum number, called ml, specifies the z component of the angular momentum for a particular orbital. For example, for an s orbital, l = 0, and the only value of ml is zero. For p orbitals, l = 1, and ml can be equal to –1, 0, or +1. Generally speaking, ml can be equal to –l, –(l – 1), …, –1, 0, +1, …, (l – 1), l. The total number of possible orbitals with the same value of l (a subshell) is 2l + 1. Thus, there is one s-orbital for ml = 0, there are three p-orbitals for ml = 1, five d-orbitals for ml = 2, seven f-orbitals for ml = 3, and so forth. The principal quantum number defines the general value of the electronic energy. The angular momentum quantum number determines the shape of the orbital. And the magnetic quantum number specifies orientation of the orbital in space, as can be seen in Figure 9.4.3. Figure 9.4.4 illustrates the energy levels for various orbitals. The number before the orbital name (such as 2s, 3p, and so forth) stands for the principal quantum number, n. The letter in the orbital name defines the subshell with a specific angular momentum quantum number l = 0 for s orbitals, 1 for p orbitals, 2 for d orbitals. Finally, there are more than one possible orbitals for l ≥ 1, each corresponding to a specific value of ml. In the case of a hydrogen atom or a one-electron ion (such as He+, Li2+, and so on), energies of all the orbitals with the same n are the same. This is called a degeneracy, and the energy levels for the same principal quantum number, n, are called degenerate energy levels. However, in atoms with more than one electron, this degeneracy is eliminated by the electron–electron interactions, and orbitals that belong to different subshells have different energies. Orbitals within the same subshell (for example ns, np, nd, nf, such as 2p, 3s) are still degenerate and have the same energy. While the three quantum numbers discussed in the previous paragraphs work well for describing electron orbitals, some experiments showed that they were not sufficient to explain all observed results. It was demonstrated in the 1920s that when hydrogen-line spectra are examined at extremely high resolution, some lines are actually not single peaks but, rather, pairs of closely spaced lines. This is the so-called fine structure of the spectrum, and it implies that there are additional small differences in energies of electrons even when they are located in the same orbital. These observations led Samuel Goudsmit and George Uhlenbeck to propose that electrons have a fourth quantum number. They called this the spin quantum number, or ms. The other three quantum numbers, n, l, and ml, are properties of specific atomic orbitals that also define in what part of the space an electron is most likely to be located. Orbitals are a result of solving the Schrödinger equation for electrons in atoms. The electron spin is a different kind of property. It is a completely quantum phenomenon with no analogues in the classical realm. In addition, it cannot be derived from solving the Schrödinger equation and is not related to the normal spatial coordinates (such as the Cartesian x, y, and z). Electron spin describes an intrinsic electron “rotation” or “spinning.” Each electron acts as a tiny magnet or a tiny rotating object with an angular momentum, even though this rotation cannot be observed in terms of the spatial coordinates. The magnitude of the overall electron spin can only have one value, and an electron can only “spin” in one of two quantized states. One is termed the α state, with the z component of the spin being in the positive direction of the z axis. This corresponds to the spin quantum number ms=12. The other is called the β state, with the z component of the spin being negative and ms=−12. Any electron, regardless of the atomic orbital it is located in, can only have one of those two values of the spin quantum number. The energies of electrons having ms=−12 and ms=12 are different if an external magnetic field is applied. Figure 9.4.5 illustrates this phenomenon. An electron acts like a tiny magnet. Its moment is directed up (in the positive direction of the z axis) for the 12 spin quantum number and down (in the negative z direction) for the spin quantum number of −12. A magnet has a lower energy if its magnetic moment is aligned with the external magnetic field (the left electron) and a higher energy for the magnetic moment being opposite to the applied field. This is why an electron with ms=12 has a slightly lower energy in an external field in the positive z direction, and an electron with ms=−12 has a slightly higher energy in the same field. This is true even for an electron occupying the same orbital in an atom. A spectral line corresponding to a transition for electrons from the same orbital but with different spin quantum numbers has two possible values of energy; thus, the line in the spectrum will show a fine structure splitting. The Pauli Exclusion Principle An electron in an atom is completely described by four quantum numbers: n, l, ml, and ms. The first three quantum numbers define the orbital and the fourth quantum number describes the intrinsic electron property called spin. An Austrian physicist Wolfgang Pauli formulated a general principle that gives the last piece of information that we need to understand the general behavior of electrons in atoms. The Pauli exclusion principle can be formulated as follows: No two electrons in the same atom can have exactly the same set of all the four quantum numbers. What this means is that electrons can share the same orbital (the same set of the quantum numbers n, l, and ml), but only if their spin quantum numbers ms have different values. Since the spin quantum number can only have two values (±12), no more than two electrons can occupy the same orbital (and if two electrons are located in the same orbital, they must have opposite spins). Therefore, any atomic orbital can be populated by only zero, one, or two electrons. The properties and meaning of the quantum numbers of electrons in atoms are briefly
Historical Markets These were markets where social and economic activities took place in the past. These markets were usually located close to the palaces of the traditional rulers. Some markets operated daily others were held once in a week or on a specified day. SOME IMPORTANT HISTORICAL MARKETS IN NIGERIA The following are some of the important historical markets in Nigeria: 1. Kurmi Market: It is the oldest and largest market where all the trans- Saharan traders met for trading purposes in the pre-colonial period. It is located in Kano State. 2. Oyingbo Market: It is one of the oldest markets in Lagos. It was established by the Awori before the coming of the British in 1851. The market is located at Ebute – Metta, Lagos. It was mainly established as a depot for farm produce. 3. Oja – Oba Market: It is one of the oldest markets in Ibadan. The market is located around Mapo Hill. It is one of the biggest foodstuff markets in the Ibadan. 4. Oba Market: It is one of the oldest markets in Benin City. It was established as far back as the 15th century. The market is known for the sale of food items, clothes and fabrics, traditional beads and jewelries. 5.Nkwo Igbo Market: It is the oldest and biggest market in Eastern Nigeria during the pre- colonial period. The market held every Nkwo day. Locally produced foodstuffs such as fruits, palm fruits, cocoyam and okra were sold in the market.