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Calculate Unit Rates
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Calculate unit rates with fractions7.4.B Calculate unit rates from mathematical and real-world problems6.4(B) apply qualitative and quantitative reasoning to solve prediction and comparison of real‐ world problems involving ratios and rates 6.4(G) generate equivalent forms of fractions, decimals, and percents using real‐world problems, including problems that involve money 6.4(H) convert units within a measurement system, including the use of proportions and unit rates 7.4(B) calculate unit rates from rates in mathematical and real‐world problemsIntroduction 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.Figure 18-11 represents the amount of energy stored as organic material in each trophic level in an ecosystem. The pyramid shape of the diagram indicates the low percentage of energy transfer from one level to the next. On average, 10 percent of the total energy consumed in one trophic level is incor- porated into the organisms in the next. Why is the percentage of energy transfer so low? One reason is that some of the organisms in a trophic level escape being eaten. They eventually die and become food for decomposers, but the energy contained in their bodies does not pass to a higher trophic level. Even when an organism is eaten, some of the molecules in its body will be in a form that the consumer cannot break down and use. For example, a cougar cannot extract energy from the antlers, hooves, and hair of a deer. Also, the energy used by prey for cellu- lar respiration cannot be used by predators to synthesize new bio- mass. Finally, no transformation or transfer of energy is 100 percent efficient. Every time energy is transformed, such as during the reactions of metabolism, some energy is lost as heat. Limitations of Trophic Levels The low rate of energy transfer between trophic levels explains why ecosystems rarely contain more than a few trophic levels. Because only about 10 percent of the energy available at one trophic level is transferred to the next trophic level, there is not enough energy in the top trophic level to support more levels. Organisms at the lowest trophic level are usually much more abundant than organisms at the highest level. In Africa, for exam- ple, you will see about 1,000 zebras, gazelles, and other herbivores for every lion or leopard you see, and there are far more grasses and shrubs than there are herbivores. Higher trophic levels con- tain less energy, so, they can support fewer individuals.A population is a group of organisms that belong to the same species and live in a particular place at the same time. All of the bass living in a pond during a certain period of time make up a pop- ulation because they are isolated in the pond and do not interact with bass living in other ponds. The boundaries of a population may be imposed by a feature of the environment, such as a lake shore, or they can be arbitrarily chosen to simplify a study of the population. The humans shown in Figure 19-1 are part of the pop- ulation of a city. The properties of populations differ from those of individuals. An individual may be born, it may reproduce, or it may die. A population study focuses on a population as a whole—how many individuals are born, how many die, and so on. Population Size A population’s size is the number of individuals that the population contains. Size is a fundamental and important population property but can be difficult to measure directly. If a population is small and composed of immobile organisms, such as plants, its size can be determined simply by counting individuals. Often, though, individ- uals are too abundant, too widespread, or too mobile to be counted easily, and scientists must estimate the number of individuals in the population. Suppose that a scientist wants to know how many oak trees live in a 10 km2 patch of forest. Instead of searching the entire patch of forest and counting all the oak trees, the scientist could count the trees in a smaller section of the forest, such as a 1 km2 area. The scientist could then use this value to estimate the population of the larger area. SECTION 1 OBJECTIVES ● Describe the main properties that scientists measure when they study populations. ● Compare the three general patterns of population dispersion. ● Identify the measurements used to describe changing populations. ● Compare the three general types of survivorship curves. VOCABULARY population population density dispersion birth rate death rate life expectancy age structure survivorship curve FIGURE 19-1 A population can be widely distributed, as Earth’s human population is, or confined to a small area, as species of fish in a lake are. Copyright © by Holt, Rinehart and Winston. All rights reserved. 382 CHAPTER 19 If the small patch contains 25 oaks, an area 10 times larger would likely contain 10 times as many oak trees. A similar kind of sampling technique might be used to estimate the size of the pop- ulation shown in Figure 19-2. To use this kind of estimate, the sci- entist must assume that the distribution of individuals in the entire population is the same as that in the sampled group. Estimates of population size are based on many such assumptions, so all esti- mates have the potential for error. Population Density Population density measures how crowded a population is. This measurement is always expressed as the number of individuals per unit of area or volume. For example, the population density of humans in the United States is about 30 people per square kilome- ter. Table 19-1 shows the population sizes and densities of humans in several countries in 2003. These estimates are calculated for the total land area. Some areas of a country may be sparsely popu- lated, while other areas are very densely populated. Dispersion A third population property is dispersion (di-SPUHR-zhuhn). Dispersion is the spatial distribution of individuals within the popu- lation. In a clumped distribution, individuals are clustered together. In a uniform distribution, individuals are separated by a fairly con- sistent distance. In a random distribution, each individual’s location is independent of the locations of other individuals in the popula- tion. Figure 19-3 illustrates the three possible patterns of dispersion. Clumped distributions often occur when resources such as food or living space are clumped. Clumped distributions may also occur because of a species’ social behavior, such as when animals gather into herds or flocks. Uniform distributions may result from social behavior in which individuals within the same habitat stay as far away from each other as possible. For example, a bird may locate its nest so as to maximize the distance from the nests of other birds. These migrating wildebeests in East Africa are too numerous and mobile to be counted. Scientists must use sampling methods at several locations to monitor changes in the population size of the animals. FIGURE 19-2 TABLE 19-1 Population Size and Density of Some Countries Population size Population density Country (in millions) (in individuals/km2) China 1,289 135 India 1,069 325 United States 292 30 Russia 146 8 Japan 128 337 Mexico 105 54 Kenya 32 54 Australia 20 3 dispersion from the Latin dis-, meaning “out,” and spargere, meaning “to scatter” Word Roots and Origins Copyright © by Holt, Rinehart and Winston. All rights reserved. POPULATIONS 383 The social interactions of birds called gannets, which are shown in Figure 19-3b, result in a uniform distribution. Each gannet chooses a small nesting area on the coast and defends it from other gannets. In this way, each gannet tries to maximize its distance from all of its neighbors, which causes a uniform distribution of individuals. Few populations are truly randomly dispersed. Rather, they show degrees of clumping or uniformity. The dispersion pattern of a population sometimes depends on the scale at which the popu- lation is observed. The gannets shown in Figure 19-3b are uni- formly distributed on a scale of a few meters. However, if the entire island on which the gannets live is observed, the distribution appears clumped because the birds live only near the shore. POPULATION DYNAMICS All populations are dynamic—they change in size and composition over time. To understand these changes, scientists must know more than the population’s size, density, and dispersion. One important measure is the birth rate, the number of births occur- ring in a period of time. In the United States, for example, there are about 4 million births per year. A second important measure is the death rate, or mortality rate, which is the number of deaths in a## Fill in the Blank Fill in the blank with the correct words: 1. Work is the use of _____________ to move an object. _____________ 2. Only the part of the applied force that acts in the same _____________ as motion does work. _____________ 3. Work transfers _____________ from one form to another. _____________ 4. The unit of work is the _____________ (J). _____________ 5. The formula for work is _____________ . _____________ [Word Bank]: force, direction, energy, joule, $W = F \times d$ ## Multiple Choice Questions Choose the correct answer from the choices for each question: 1. Which statement best describes when work is done on an object? * A. A force acts on the object while it moves in the same direction as the force. * B. A force acts on the object while it does not move. * C. The object moves but no force is applied. * D. A force acts opposite to the motion and the object does not move. 2. If a force of 10 N moves an object 3 m in the same direction as the force, how much work is done? * A. 3 J * B. 10 J * C. 30 J * D. 13 J 3. Which unit correctly measures work? * A. Newton (N) * B. Meter (m) * C. Joule (J) * D. Watt (W) 4. If a force acts perpendicular to the motion of an object, how much work does that component do? * A. Maximum work * B. Zero work * C. Negative work * D. Equal to force times distance 5. Using the example from the passage, the work done lifting a 30 N dog 2 m is: * A. 15 J * B. 60 J * C. 32 J * D. 90 J ## Open-Ended Questions Answer the following questions in complete sentences: 1. Explain in your own words why only the component of force in the direction of motion does work. 2. A student pushes a box with a 25 N horizontal force and the box moves 4 m horizontally. Calculate the work done and show your steps using the formula from the passage. 3. Describe one real-life example (not in the passage) where work transfers energy from one form to another. Identify the force, the distance, and the energy transfer.① Simplify to the simplest form the expression: 2x (2 x + 1) + 3𝑥 (𝑥 + 2), then find the numerical value of the expression when 𝑥 = 1 ② Find by inspection: (2𝑥 + 1)(𝑥 + 4) ③ Find the expansion of: (𝑥 − 5)2 ④ If (𝑥 − 5)(𝑥 + 5) = 𝑥2 − 𝑐 , then what is the value of c ? Assessment ④ (B) Time: 15 min ① Simplify to the simplest form the expression: 2𝑥 (2𝑥 + 1) + 3𝑥 (𝑥 + 2), then find the numerical value of the expression when 𝑥 = −1 ② Find by inspection: (𝑥 + 3)(𝑥 + 4) ③ Find the expansion of: (𝑥 + 1) 2 ④ If (𝑥 − 5)(𝑥 + 5) = 𝑥2+ 𝑏𝑥 + 𝑐 , then what is the value of 𝑏 ? Assessment ⑤(C) Time: 15 min ① If: (𝑥 + 𝑦 ) = 3, (𝑥 – 𝑦 ) = 9 , then what is the value of: (𝑥2 − 𝑦2 )? ② If: (3𝑥 − 4) 2 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 , then what is the value of b ? ③ A square its side’s length is (𝑥 + 3) length unit, calculate its area in terms of 𝑥 . ④ (𝑥 + 3)(𝑥 + 2) = 𝑥2 + 𝑏𝑥 + 6 , then what is the value of b ? ⑤ Find the solution set of the following inequality in Z: 5 − 3𝑥 ≥ 14Electrostatics The section of CBSE Class 12 Physics electrostatic potential and capacitance notes mainly deals with the in-depth analysis of electromagnetic phenomena when they are not performing any movements. Additionally, it is divided into ten further sub-topics to study the companion processes of reaching the state. These are - 1. Electric charge In this section of Physics ch 2 Class 12 notes, you get to learn about the basic features of electric charge and its expression in Physics. Along with its basics, the sections help to understand the full potential of charge. Different aspects of Charge included in Class 12 Physics Chapter 2 notes are - Definition Type: Positive and Negative Charge Unit and dimensional formula Point Charge Properties of Charge Comparison of Charge and Mass Methods of Charging Electroscope 2. Coulomb's Law Force is created when charges of opposite signs attract each other, and they repulse if the signs are the same. Coulomb's law tries to define this phenomenon through a mathematical formula, explicitly mentioned in Physics Class 12 notes Chapter 2. Moreover, there is key information about the variation of the constant k and its effect on a medium. Coulomb's law's vector form and the principle of superimposition are also explained in ch 2 Physics Class 12 notes. (Image will be uploaded soon) 3. Electric Field As stated in Class 12 Physics Chapter 2 notes, every positively or negatively charged particle has their respective electric fields. It feels a force at the time of interaction which might be attraction or repulsion. As it arises from electric charge, it is crucial to know about its different parts like - Electric field intensity Relation between electric force and electric field Super imposition of electric field Point charge Continuous charge distributions Properties of Electric Field Lines Motion of Charged Particles in an Electric field Learning more about the electric field from electric potential and capacitance notes Class 12 helps a student to get a grasp of upcoming chapters. 4. Electric Potential Energy When energy helps a charge to move from an electric field, it is known as the Electric Potential Energy. This section of electrostatic chapter Class 12 notes requires a student to study the Electron volt (eV), and the potential energy that an n number of charges can hold. 5. Electric Potential This section of Class 12 Physics Chapter 2 notes focuses on in-depth learning of Electric Potential or Voltage. Basically, it defines the potential movement of energy. 6. Relation between Electric Field and Potential Apart from knowing more about the relationship between the two values, Physics Class 12 Chapter 2 notes also discuss equipotential surfaces. 7. Electric Dipole Essentially, 'Dipoles' are two opposite points of charge represented with q and –q, with their distance between each other being 2a. Electric Dipoles are crucial in your study of Physics Class 12 Chapter 2 notes to learn more about electric fields and their potential. Additionally, Class 12 Physics Chapter 2 notes focus on the influence of electric dipoles on a uniform electric field mainly through Force and Torque, Work, and Potential Energy. In the last part of Electrostatics, further focus is on using the formulas to their fullest potential. It includes subsections of Electric Field, Electric Potential Energy, Electric Potential, and Electric Dipole. In the notes for electrostatic potential and capacitance, you will find proper solutions accompanied by clear and crisp diagrams for better understanding. 8. Gauss's Law Apart from just discussing the Gauss's Law, in Physics Class 12 ch 2 notes there is a thorough explanation of its properties and applications. The Gauss' Law states that net electric flux passing through a hypothetical closed surface is equal to the net electric charge present within the same closed surface. Being a broad part of the whole chapter, you may need to spend a little more time on it. Moving forward, it starts discussing the properties of conductors in relation to Gauss's Law. The Class 12 Physics notes Chapter 2 perfectly defines the journey to Gauss' Law from Coulomb's Law. Here is the Gauss's Law present in the Class 12 Physics ch 2 notes, (image will be uploaded soon) 9. Capacitors There is a dedicated section about Capacitors in the Class 12 Physics Chapter 2 notes elucidating its functions and importance as storage of potential electric energy. After explaining the structure of a capacitor, it points out the different types, parallel plate, spherical and cylindrical. The section of Chapter 2 notes of Physics Class 12 is further divided into subheads like: Properties of an ideal battery Grouping of capacitors Simple circuits (Series and Parallel) Dielectric Van de Graaff generator Combination of drops Charge distribution method Wheatstone Bridge-based circuit Extended Wheatstone Bridge Infinite network of capacitors Redistribution of charge between two capacitors Vedantu prepares the Class 12 Physics Chapter 2 notes with help from subject matter experts. In the PDF, you get a comprehensive idea of the topic along with potential answers to the most asked questions. Furthermore, the detailed explanation on each section and subsections are written in a simple language allows a student to ace their exams with wholesome knowledge. These Physics Chapter 2 Class 12 notes are going to be one of the best supplementary study materials besides a student’s textbooks. Visit the Vedantu website or download the app to get your hands on all important notes! Important Questions A charge of 4 × 10–8C is uniformly distributed on the surface of a spherical conductor, having a radius of 15 cm. Determine the electric field just outside this sphere at a point that is 15 cm from the centre of this sphere. Determine the capacitance given that the distance between the two plates has been reduced by half and the parallel plate capacitor holds a capacitance of 20 pF (where 1pF = 10-12 F) having air between the two plates. What will be the total capacitance of a combination where three capacitors, each having a capacitance of 20 pF, are connected in series. A square having a side of 10 cm has a 500 µC charge at its centre. Determine the work done to move a charge of 10 µC between two points that are diagonally opposite each other on the square. At an equatorial point, what will be the electrostatic potential because of an electric dipole? Calculate the work done to move a test charge, q, through a length of 1 cm along the equatorial axis of an electric dipole? Polarisation A capacitor has its plates enclosed in a medium that can be filled by insulating substances. A net dipole moment is then induced by an electric field in the dielectric. This event causes the field in an opposite direction. Equipotential Surface An equipotential surface is a type of surface where the potential always has a constant value. If considered as a point charge, the concentric spheres that are centred at a particular area of this charge are basically equipotential surfaces. Advantages of Vedantu's Revision Notes: A Comprehensive Resource for Effective Learning There are several reasons why one may refer to Vedantu's revision notes for studying a subject like Electrostatic Potential and Capacitance. Here are some key points: Comprehensive Coverage: Vedantu's revision notes provide a comprehensive coverage of the entire topic, ensuring that all important concepts and subtopics are included. Concise and Organized: The notes are designed to be concise, focusing on the key points and core ideas. They are organized in a structured manner, making it easy for students to navigate and revise the content. Simplified Explanation: The revision notes offer simplified explanations of complex concepts, making them more accessible and easier to understand. This helps students grasp the material more effectively. 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