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constant rat eof change
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Constant rate of change using tablesConstant Rate of Change - 7HConstant rate of changeConstant Rate of Change/ Constant of Proportionality Marine and Coastal Processes.What are the hazards that usually occur along marine and coastal areas? Coastal processes, such as waves, tides, sea level changes, crustal movement, and storm surges will result to coastal erosion, submersion, and saltwater intrusion. Coastal Erosion. Coastal erosion is the wearing down of the coastlines by the movement of wind and water. It is not a constant process; instead, the rate of erosion depends on other events such as cyclones. When cyclones occur along coastal areas, the winds and waves carry the sediment away from the shoreline. Shorelines play an important role to society. They are used in transportation, fishing, and tourism. Therefore, preventing coastal erosion is of utmost priority. There are three main classifications of stabilizing the shoreline: hard stabilization, soft stabilization, and retreat. 1. Hard stabilization is done by building structures that will slow down the erosion on areas that are prone to erosion. Examples of hard stabilization structures are jetties, sea walls, and breakwaters. Though they may slow down the erosion in one area, it may hasten the erosion in other areas. 2. Soft stabilization includes the process of beach nourishment, wherein sand from an offshore location is brought to an area with a receding shoreline. It does not make use of structures like the ones used in hard stabilization. 3. Retreat is the option taken by residents near areas where coastal erosion is already severe. At this point, the authorities no longer attempt to save the shoreline but rather limit the amount of human interference in the area. Submersion. Coastal erosion happens because of the interaction of the winds and waves on the shoreline. Submersion, on the other hand, happens because of the changes in the sea level, specifically, when it rises dangerously above the normal level. This is all due to the increase in the global temperature, which, in turn, melts the glacial deposits and increases the overall sea level. Another factor that may cause submersion is the vertical movement of the plates. Landmasses can be uplifted, which can also cause changes in the sea level. It can also be caused by tsunamis and storm surges. Submersion will most likely occur in reclaimed lands. These are the areas that were originally part of oceans, riverbeds, or lakebeds. They are low-lying flatlands, so even a small rise in sea level can cause great damage on the land. To prevent this from happening not only in reclaimed lands but also in coastal areas, a hard stabilization technique is used. Sea walls are built along the coastline to protect the land from being easily flooded. Aside from sea walls, dikes can also help prevent flooding. The government can also upgrade the infrastructures built in coastal areas, regenerate mangroves, or relocate the people. There are also other proposed strategies to mitigate coastal submersion, such as imposing of setback policies and construction regulations and creating adaptive plans for coastal management. Saltwater Intrusion. In coastal areas where there is an interaction between saltwater and fresh water, saltwater intrusion is one of the hazards that are evident in that area. Saltwater intrusion is the movement of saltwater into the freshwater aquifer. The natural flow is that the fresh water, which is less dense, moves towards the denser saltwater. But if the fresh water is being withdrawn faster than it is being replenished, then there will be a change in pressure and saltwater intrusion will occur.There are a few ways of preventing saltwater intrusion. One is to stop using the well where fresh water has been depleted and let the groundwater replenish naturally via the water cycle. The other method is to build two wells: a pumping well-built farther inland and an injection well-built closer to the coast. Using the injection well, fresh water is pumped into the aquifer to prevent the saltwater from intruding. The different marine and coastal hazards often occur in the Philippines, being an archipelago with the longest coastline. Manila Bay is one of the coastal areas of the Philippines that is facing various threats from both natural and anthropological causes. Saltwater intrusion occurs due to uncontrolled withdrawal of groundwater to be used by residential, commercial, and industrial areas built around the bay. It is also frequently flooded due to poor drainage systems and improper disposal of waste. Since Manila Bay is shared by four coastal provinces, four noncoastal provinces, and the National Capital Region, each local government unit and national agencies need to collaborate in planning, developing, and managing its marine and coastal resources. And it is not only Manila Bay but other parts as well, for as long as they are in coastal areas, hazards will mostly likely occur if not immediately addressed.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.Soils Southeast Asia, on balance, has a higher proportion of relatively fertile soils than most tropical regions, and soil erosion is less severe than elsewhere. Much of the region, however, is covered by tropical soils that generally are quite poor in nutrients. Often the profusion of plant life is more related to heat and moisture than to soil quality, even though these climatic conditions intensify both chemical weathering and the rate of bacterial action that usually improve soil fertility. Once the vegetation cover is removed, the supply of humus quickly disappears. In addition, the often heavy rainfall leaches the soils of their soluble nutrients, hastens erosion, and damages the soil texture. The leaching process in part results in laterites of reddish clay that contain hydroxides of iron and alumina. Laterite soils are common in parts of Myanmar, Thailand, and Vietnam and also occur in the islands of the Sunda Shelf, notably Borneo. The most fertile soils occur in regions of volcanic activity, where the ejecta is chemically alkaline or neutral. Such soils are found in parts of Sumatra and much of Java in Indonesia. The alluvial soils of the river valleys also are highly fertile and are intensively cultivated. Climate All of Southeast Asia falls within the warm, humid tropics, and its climate generally can be characterized as monsoonal (i.e., marked by wet and dry periods). Changing seasons are more associated with rainfall than with temperature variations. There is, however, a high degree of climatic complexity within the region. Temperatures Regional temperatures at or near sea level remain fairly constant throughout the year, although monthly averages tend to vary more with increasing latitude. Thus, with the exception of northern Vietnam, annual average temperatures are close to 80 °F (27 °C). Increasing elevation acts to decrease average temperatures, and such locations as the Cameron Highlands in peninsular Malaysia and Baguio in the Philippines have become popular tourist destinations in part because of their relatively cooler climates. Proximity to the sea also tends to moderate temperatures. Precipitation Much of Southeast Asia receives more than 60 inches (1,500 millimeters) of rainfall annually, and many areas commonly receive double and even triple that amount. The rainfall pattern is distinctly affected by two prevailing air currents: the northeast (or dry) monsoon and the southwest (or wet) monsoon. The northeast monsoon occurs roughly from November to March and brings relatively dry, cool air and little precipitation to the mainland. As the southwestward-flowing air passes over the warmer sea, it gradually warms and gathers moisture. Precipitation is especially heavy where the airstream is forced to rise over mountains or encounters a landmass. The east coast of peninsular Malaysia, the Philippines, and parts of eastern Indonesia receive the heaviest rains during this period. The southwest monsoon prevails from May to September, when the air current reverses and the dominant flow is to the northeast. The mainland receives the bulk of its rainfall during this period. Over much of the southern Malay Peninsula and insular Southeast Asia there is little or no prolonged dry season. This is especially marked in much of the equatorial region and along the east coast of the Philippines. While the dry and wet monsoons are important in explaining rainfall patterns, so too are such factors as relief, land and sea breezes, convectional overturning and cyclonic disturbances. These factors often are combined with monsoonal effects to produce highly variable rainfall patterns over relatively short distances. While many of the cyclonic disturbances produce only moderate rainfall, others mature into tropical storms—called cyclones in the Indian Ocean and typhoons in the Pacific—that bring heavy rains and destruction to the areas over which they pass. The Philippines are particularly affected by these storms. Plant life Tropical forests in Southeast Asia Tropical forests in Southeast Asia The seasonal nature and pattern of Southeast Asia’s rainfall, as well as the region’s physiography, have strongly affected the development of natural vegetation. The hot, humid climate and enormous variety of habitats have given rise to an abundance and diversity of vegetative forms unlike that in any other area of the world. Much of the natural vegetation has been modified by human action, although large areas of relatively untouched land still can be found. The vegetation can be grouped into two broad categories: the tropical-evergreen forests of the equatorial lowlands and the open type of tropical-deciduous, or “monsoon,” forests in areas of seasonal drought. The evergreen forests are characterized by multiple stories of vegetation, consisting of a variety of trees and plants. Although a large diversity of tree species is found in these forests, members of the Dipterocarpaceae family account for roughly half of the varieties. Deciduous forests are found in eastern Indonesia and those parts of the mainland where annual rainfall does not exceed 80 inches. Just as in the equatorial forest, a wide variety of species is normally the rule. Certain species, such as teak, have become highly valued commercially. Teak is found in parts of Indonesia, Myanmar, Thailand, and Laos. In addition to these two basic types of vegetation, other regional patterns reflect topography. Especially noteworthy are coastal and highland plant communities. Mangrove belts, of which there are more than 30 varieties, occur where silt is deposited in coastal areas. Upland forests dominated by maples, oaks, and magnolias are found especially on mainland mountain slopes. Human activity has been rapidly altering the stands of virgin forest in Southeast Asia. Most deforestation results from removal for fuelwood and clearing for agriculture and grazing. Although only a relatively small portion of the total land area has been permanently cleared for cultivation—e.g., in Java (Indonesia) and western Luzon (the Philippines)—in some areas shifting cultivation has brought about the replacement of virgin forest with secondary growth. In addition, nearly all countries have commercial logging industries; notable are those in Indonesia, Malaysia, Thailand, and Myanmar. A growing problem has been illegal logging. Thus, timber harvesting has come to contribute significantly to deforestation. Programs in social forestry and reforestation have yet to halt the rapid denuding of the landscape. Animal life Southeast Asia is situated where two major divisions of the world’s fauna meet. The region itself constitutes the eastern half of what is called the Oriental, or Indian, zoogeographic region (part of the much larger realm of Megagaea). Bordering along the south and east is the Australian zoogeographic region, and the eastern portion of insular Southeast Asia—Celebes (Sulawesi), the Moluccas, and the Lesser Sunda Islands—constitutes a transition zone between these two faunal regions. a classroom in Brazil More From Britannica education: Southeast Asia Southeast Asia is notable, therefore, for a considerable diversity of wildlife throughout the region. These differences are especially striking between the species of the eastern and western fringes as well as between those of the archipelagic south and the mainland north. The differences stem largely from the isolation, over varying lengths of geologic time, of species following their migration from the Asian continent. In addition, the tropical rain forests in many parts of the region, with their great diversity of vegetation, have made possible the development of complex communities of animals that fill specialized ecological niches. Especially numerous are arboreal and flying creatures. orangutans orangutansOrangutans (Pongo pygmaeus) in Sumatra, Indonesia. The distinction between the two faunal regions is best depicted by their mammal populations. In general, Australia is inhabited largely by marsupials (pouched mammals) and monotremes (egg-laying mammals), while Southeast Asia contains placental mammals and such hybrid species as the bandicoot of eastern Indonesia. Small mammals such as monkeys and shrews are the most numerous, while in many areas the larger mammals have been pushed into more remote areas and national preserves. Bears, gibbons, elephants, deer, civets, and pigs are found in both mainland and insular Southeast Asia, as are diminishing numbers of tigers. The Malayan tapir, a relative of the rhinoceros, is native to the Malay Peninsula and Sumatra, while the tarsier is found in the Philippines and parts of Indonesia. A number of rare endemic species are found in Indonesia and East (insular) Malaysia, including the Sumatran and Javan rhinoceros, the orangutan, the anoa (a dwarf buffalo), the babirusa (a wild swine), and the palm civet. As the pace of development accelerates and populations continue to expand in Southeast Asia, concern has increased regarding the impact of human activity on the region’s environment. A significant portion of Southeast Asia, however, has not changed greatly and remains an unaltered home to wildlife. The nations of the region, with only few exceptions, have become aware of the need to maintain forest cover not only to prevent soil erosion but to preserve the diversity of flora and fauna. Indonesia, for example, has created an extensive system of national parks and preserves for this purpose. Even so, such species as the Javan rhinoceros face extinction, with only a handful of the animals remaining in western JavaHere's how scientists figured out the age of the universe It took some cosmic detective work. by Passant Rabie Oct. 20, 2021 You never ask a cosmic being its age. But if that cosmic being encompasses all of space, time, and matter, you could get a little curious. Scientists have long been curious about the age of the universe and how much time has elapsed since the Big Bang. Today, scientists estimated the age of the universe to be approximately 13.8 billion years old. But how did scientists estimate how old the universe is, and are they sure of that number? It all comes down to ancient stars and the ever-expanding cosmos. How do astronomers calculate the age of the universe? To estimate the age of the universe, scientists rely on two main methods. Calculating the expansion rate of the universe Determining the ages of the oldest stars The Hubble Constant: Since its conception, the universe has been expanding at an accelerating rate. The universe’s expansion rate is known as the Hubble Constant, which is estimated at 46,200 mph per million light-years. The Hubble Constant was first calculated in the 1920s by American astronomer Edwin Hubble after discovering that several galaxies were moving away from Earth. Scientists looked to distant galaxies to measure how fast the universe was expanding. Hubble also noted that the further a galaxy was, the faster it was moving away. Based on Hubble’s observations, the astronomer came up with Hubble’s law which showed a correlation between how far an object is and the speed at which it’s receding. Using Hubble law, scientists were able to estimate the expansion rate of the universe. Scientists were then able to use the Hubble Constant to estimate the age of the universe by working backward, all the way back to the Big Bang. This extrapolation depends on the current density and composition of the universe, which shows the history of its expansion. In 2012 NASA’s Wilkinson Microwave Anisotropy Probe used that data to estimate the universe's age to be 13.772 billion years old, give or take 59 million years. A year later, The European Space Agency’s Planck spacecraft estimated the universe's age to be 13.82 billion years. Ancestral stars: Another way to determine the age of the universe is to look to the oldest stars. The universe can’t be younger than its oldest stars. Therefore, to narrow down the age of the universe, scientists measure the ages of the very first stars that formed in the cosmos. The lifecycle of a star depends on its mass, with high mass stars burning fuel at a faster rate and therefore dying out faster while low mass stars can live up to 20 billion years. Globular clusters are a dense stellar collection of around a million stars which all formed roughly around the same time. These clusters can then serve as timekeepers for the universe. By determining the masses of their stars, scientists can estimate when the globular cluster formed. The oldest globular clusters contain stars that are 0.7 times less massive than the Sun, which suggests that they are between 11 to 18 billion years old. What came before the Big Bang? Scientists can trace the universe back to its explosive birth, the Big Bang. But what happened before this theoretical birth of the cosmos? The universe may have been a singularity, all compact within a form that is smaller than a subatomic particle. It’s difficult to imagine what caused this matter to exist, but one theory even suggests that our universe was born from another universe while another imagines a series of universes being born out of one another like a formation of bubbles. Meanwhile, another theory suggests that the universe goes through an endless cycle of death and rebirth, born from its own demise. How old is the universe in seconds? If the universe is indeed cyclical, then time becomes irrelevant. But just in case you’re still attached to the modern way in which we measure the progression of life, then the age of the universe comes up to about 436,117,076,900,000,000 seconds.