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Travelling Part 2 - short test 8th grades
Quiz by Joanna Lange
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Longman exam activator. Travelling (part 2)1. How many pillars of Islam are there in total? A) Three B) Five (Correct) C) Six D) Seven Hint: Think about the famous Hadith of Gabriel where he asks about the basic practices of Islam. 2. What is the first pillar of Islam? A) Salah (Prayer) B) Zakat (Charity) C) Shahadah (Declaration of faith) (Correct) D) Sawm (Fasting) Hint: It is the declaration that there is no god but Allah and Muhammad is His messenger. 3. How many times a day must a Muslim perform Salah (prayer)? A) Three times B) Five times (Correct) C) Four times D) Six times Hint: Count Fajr, Dhuhr, Asr, Maghrib, and Isha. 4. What does the word 'Zakat' mean in terms of practice? A) Fasting all day B) Giving charity to the poor (Correct) C) Traveling to Makkah D) Reading the Quran Hint: It involves sharing a small part of your saved wealth to purify the rest of it. 5. During which Islamic month do Muslims fast (Sawm)? A) Muharram B) Ramadan (Correct) C) Shawwal D) Dhul-Hijjah Hint: It is the month in which the Quran was first revealed. 6. Where must a Muslim go to perform Hajj? A) Madinah B) Jerusalem C) Makkah (Correct) D) Cairo Hint: This city contains the Kaaba, the direction Muslims face during prayer. 7. Which pillar of Islam directly trains a Muslim in patience and feeling the hunger of the poor? A) Shahadah B) Sawm (Fasting) (Correct) C) Salah D) Hajj Hint: It is done during the daylight hours of Ramadan. Part 2: Pillars of Iman (أركان الإيمان) 8. How many pillars of Iman (faith) are there? A) Five B) Six (Correct) C) Four D) Eight Hint: It is one more than the number of pillars of Islam. 9. What is the first and most important pillar of Iman? A) Belief in Angels B) Belief in Allah (Correct) C) Belief in the Books D) Belief in the Last Day Hint: This is the belief in Monotheism (Tawhid). 10. Muslims believe that angels are created from what? A) Fire B) Clay C) Light (Correct) D) Water Hint: Think of a bright source that illuminates the dark, which matches their pure and luminous nature. 11. Which angel was responsible for bringing the revelation (Quran) to the Prophet Muhammad? A) Angel Mikaeel (Michael) B) Angel Jibreel (Gabriel) (Correct) C) Angel Israfeel D) Angel Malak al-Mawt Hint: He is the leader of all angels and visited the Prophet in the cave of Hira. 12. Belief in the Holy Books is a pillar of Iman. Which book was given to Prophet Isa (Jesus)? A) The Torah B) The Zabur C) The Injeel (Correct) D) The Quran Hint: The English translation often links this word closely to the 'Gospel'. 13. Who is the final Prophet and Messenger sent by Allah to mankind? A) Prophet Ibrahim (Abraham) B) Prophet Musa (Moses) C) Prophet Muhammad (Correct) D) Prophet Nuh (Noah) Hint: He was born in Makkah and received the Quran. 14. What does 'Belief in the Last Day' mean? A) Belief in the last day of Ramadan B) Belief in the Day of Judgment (Correct) C) Belief in the weekend D) Belief that the sun will never set Hint: It is the day when people will be rewarded with Paradise or punished based on their deeds. 15. What is the sixth pillar of Iman? A) Belief in Qadar (Divine Decree/Fate) (Correct) B) Belief in Hellfire C) Belief in the Companions D) Belief in Charity Hint: It relates to destiny and accepting whatever Allah has written for us. 16. What is the Arabic word for the 'Divine Decree' or destiny in the pillars of Iman? A) Zakat B) Qadar (Correct) C) Injeel D) Tawhid Hint: It sounds like 'Al-Qadr', as in the night of decree (Laylat al-Qadr). 17. Belief in Prophets includes believing in messengers mentioned in other scriptures. Who did Allah speak to directly? A) Prophet Musa (Moses) (Correct) B) Prophet Nuh (Noah) C) Prophet Yusuf (Joseph) D) Prophet Yunus (Jonah) Hint: He is the prophet associated with Mount Sinai and parting the sea. Part 3: Ihsan (الإحسان) 18. What is the meaning of 'Ihsan' according to the famous Hadith? A) To give all your money away B) To worship Allah as if you see Him (Correct) C) To memorize the whole Quran D) To fast twice a week Hint: It is the highest level of religion, focusing on absolute perfection and sincerity in worship. 19. If you cannot see Allah during worship, what must you always remember according to Ihsan? A) That other people are watching you B) That Allah sees you (Correct) C) That you should finish quickly D) That the angels will pray for you Hint: Allah is All-Seeing (Al-Baseer) and All-Knowing (Al-Aleem). 20. Which of the following represents the correct order of levels in religion from lowest to highest? A) Ihsan, then Iman, then Islam B) Islam, then Iman, then Ihsan (Correct) C) Iman, then Islam, then Ihsan D) Islam, then Ihsan, then Iman Hint: Every Muhsin (person of Ihsan) is a Mu'min (person of Iman) and a Muslim, but not vice versa.PASSIVE TRANSPORT Cell membranes help organisms maintain homeostasis by controlling what substances may enter or leave cells. Some substances can cross the cell membrane without any input of energy by the cell in a process known as passive transport. DIFFUSION The simplest type of passive transport is diffusion. Diffusion is the movement of molecules from an area of higher concentration to an area of lower concentration. This difference in the concentration of molecules across a distance is called a concentration gradient. Consider what happens when you add a sugar cube to a beaker of water. As shown in Figure 5-1, the sugar cube sinks to the bottom of the beaker. This sinking makes the concentration of sugar mole- cules greater at the bottom of the beaker than at the top. As the cube dissolves, the sugar molecules begin to diffuse slowly through the water, moving towards the lower concentration at the top. Diffusion is driven entirely by the molecules’ kinetic energy. Molecules are in constant motion because they have kinetic energy. Molecules move randomly, traveling in a straight line until they hit an object, such as another molecule. When they hit some- thing, they bounce off and move in a new direction, traveling in another straight line. If no object blocks their movement, they con- tinue on their path. Thus, molecules tend to move from areas where they are more concentrated to areas where they are less concentrated, or “down” their concentration gradient. In the absence of other influences, diffusion will eventually cause the molecules to be in equilibrium—the concentration of molecules will be the same throughout the space the molecules occupy. Returning to the example in Figure 5-1, if the beaker of water is left undisturbed, at some point the concentration of sugar molecules will be the same throughout the beaker. The sugar concentration will then be at equilibrium. SECTION 1 OBJECTIVES ● Explain how an equilibrium is established as a result of diffusion. ● Distinguish between diffusion and osmosis. ● Explain how substances cross the cell membrane through facilitated diffusion. ● Explain how ion channels assist the diffusion of ions across the cell membrane. VOCABULARY passive transport diffusion concentration gradient equilibrium osmosis hypotonic hypertonic isotonic contractile vacuole turgor pressure plasmolysis cytolysis facilitated diffusion carrier protein ion channel Sugar Water 1 2 3 FIGURE 5-1 Sugar molecules, initially in a high concentration at the bottom of a beaker, , will move about randomly through diffusion, , and eventually reach equilibrium, . At equilibrium the sugar concentration will be the same throughout the beaker. Diffusion occurs naturally because of the kinetic energy the molecules possess. 3 2 1 Copyright © by Holt, Rinehart and Winston. All rights reserved. 98 CHAPTER 5 It is important to understand that even at equilibrium the ran- dom movement of molecules continues. But because there is an equal concentration of molecules everywhere, molecules are just as likely to move in one direction as in any other. The random movements of many molecules in many directions balance one another, and equilibrium is maintained. Diffusion Across Membranes Cell membranes allow some molecules to pass through, but not others. If a molecule can pass through a cell membrane, it will diffuse from an area of higher concentration on one side of the membrane to an area of lower concentration on the other side. Diffusion across a membrane is also called simple diffusion, and only allows certain molecules to pass through the membrane. The simple diffusion of a molecule across a cell membrane depends on the size and type of molecule and on the chemical nature of the membrane. A membrane can be made, in part, of a phospho- lipid bilayer, and certain proteins can form pores in the membrane. Molecules that can dissolve in lipids may pass directly through the membrane by diffusion. For example, because of their nonpolar nature, both carbon dioxide and oxygen dissolve in lipids. Molecules that are very small but not soluble in lipids may diffuse across the membrane by moving through the pores in the membrane.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.6 class_Module 8_Holidays and travelling_Part 1Longman exam activator. Travelling (part 1)How does a nomad child go to class every day, though? A new idea is the travelling school, a big ger which goes where the herders go in the warmer months. These are the months when nomad parents really need their children's help. Mongolia is developing. There are more jobs now, and new towns. The travelling school allows Mongolian nomads to keep their traditional lifestyle, but learn about the modern world, too. The nomads hope their children can be part of both. Most people nowadays live in cities and towns crowded with people in towering skyscrapers or blocks of flats and heavy traffic on the streets. There are schools, shopping centres, cinemas, theatres and restaurants. The life of a Mongolian nomad couldn't be more different. B A family's nearest neighbours are kilometres away and the only building is the ger, their traditional tent. But it is thanks to their animals that these nomads manage to survive. Their horses carry them and their equipment over long distances. Their cows, yaks, goats and sheep give them food, wool and leather. The herder goes where the animals go, moving to find new grass. It's a hard life. The nomads' skin is dark and tough from the sun and wind, and their eyes are narrow to protect them from the harsh winter. If it gets too cold, there isn't enough to eat and they lose many animals. Nomads are proud to be herders and proud to live more freely than city people. They want their children to have an easier life, however, and try to send them to school. Transport is usually just a way to get from one destination to another when you are traveling. There are different ways to travel- by car, by plane, by train, by us, by cruise and by ship. Each way has its advantage and disadvantages. The most popular type of transport are cars. Cars are convenient, getting you exactly when you want to be without additional buses, trains or walking needed to get to your final destination. Buses are available in all cities, towns and most villages. They are ideal for short journeys from one part of a town to another. Cars and buses are not comfortable ways to travel for long distances. It is difficult to sleep, your body is in one position and there is not enough pace too move. Planes are the fastest way of travelling but they are expensive. Besides you can’t enjoy the beauty of landscapes for a long time as it’s usually possible while taking off and landing. Trains are slower but they are more comfortable. They allow you to relax and enjoy the scenery. Traveling by ship is also very comfortable and pleasant, especially during good weather.