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Object detection in slef driving cars
Quiz by Vishal Divakaran
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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.
## 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.
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.
Understanding Quantum Theory of Electrons in Atoms The goal of this section is to understand the electron orbitals (location of electrons in atoms), their different energies, and other properties. The use of quantum theory provides the best understanding to these topics. This knowledge is a precursor to chemical bonding. As was described previously, electrons in atoms can exist only on discrete energy levels but not between them. It is said that the energy of an electron in an atom is quantized, that is, it can be equal only to certain specific values and can jump from one energy level to another but not transition smoothly or stay between these levels. The energy levels are labeled with an n value, where n = 1, 2, 3, …. Generally speaking, the energy of an electron in an atom is greater for greater values of n. This number, n, is referred to as the principal quantum number. The principal quantum number defines the location of the energy level. It is essentially the same concept as the n in the Bohr atom description. Another name for the principal quantum number is the shell number. The shells of an atom can be thought of concentric circles radiating out from the nucleus. The electrons that belong to a specific shell are most likely to be found within the corresponding circular area. The further we proceed from the nucleus, the higher the shell number, and so the higher the energy level (Figure 9.4.1). The positively charged protons in the nucleus stabilize the electronic orbitals by electrostatic attraction between the positive charges of the protons and the negative charges of the electrons. So the further away the electron is from the nucleus, the greater the energy it has. This quantum mechanical model for where electrons reside in an atom can be used to look at electronic transitions, the events when an electron moves from one energy level to another. If the transition is to a higher energy level, energy is absorbed, and the energy change has a positive value. To obtain the amount of energy necessary for the transition to a higher energy level, a photon is absorbed by the atom. A transition to a lower energy level involves a release of energy, and the energy change is negative. This process is accompanied by emission of a photon by the atom. The following equation summarizes these relationships and is based on the hydrogen atom: The values nf and ni are the final and initial energy states of the electron. The principal quantum number is one of three quantum numbers used to characterize an orbital. An atomic orbital, which is distinct from an orbit, is a general region in an atom within which an electron is most probable to reside. The quantum mechanical model specifies the probability of finding an electron in the three-dimensional space around the nucleus and is based on solutions of the Schrödinger equation. In addition, the principal quantum number defines the energy of an electron in a hydrogen or hydrogen-like atom or an ion (an atom or an ion with only one electron) and the general region in which discrete energy levels of electrons in a multi-electron atoms and ions are located. Another quantum number is l, the angular momentum quantum number. It is an integer that defines the shape of the orbital, and takes on the values, l = 0, 1, 2, …, n – 1. This means that an orbital with n = 1 can have only one value of l, l = 0, whereas n = 2 permits l = 0 and l = 1, and so on. The principal quantum number defines the general size and energy of the orbital. The l value specifies the shape of the orbital. Orbitals with the same value of l form a subshell. In addition, the greater the angular momentum quantum number, the greater is the angular momentum of an electron at this orbital. Orbitals with l = 0 are called s orbitals (or the s subshells). The value l = 1 corresponds to the p orbitals. For a given n, p orbitals constitute a p subshell (e.g., 3p if n = 3). The orbitals with l = 2 are called the d orbitals, followed by the f-, g-, and h-orbitals for l = 3, 4, 5, and there are higher values we will not consider. There are certain distances from the nucleus at which the probability density of finding an electron located at a particular orbital is zero. In other words, the value of the wavefunction ψ is zero at this distance for this orbital. Such a value of radius r is called a radial node. The number of radial nodes in an orbital is n – l – 1. Consider the examples in Figure 9.4.2. The orbitals depicted are of the s type, thus l = 0 for all of them. It can be seen from the graphs of the probability densities that there are 1 – 0 – 1 = 0 places where the density is zero (nodes) for 1s (n = 1), 2 – 0 – 1 = 1 node for 2s, and 3 – 0 – 1 = 2 nodes for the 3s orbitals. The s subshell electron density distribution is spherical and the p subshell has a dumbbell shape. The d and f orbitals are more complex. These shapes represent the three-dimensional regions within which the electron is likely to be found. Principal quantum number (n) & Orbital angular momentum (l): The Orbital Subshell: https://youtu.be/ms7WR149fAY If an electron has an angular momentum (l ≠ 0), then this vector can point in different directions. In addition, the z component of the angular momentum can have more than one value. This means that if a magnetic field is applied in the z direction, orbitals with different values of the z component of the angular momentum will have different energies resulting from interacting with the field. The magnetic quantum number, called ml, specifies the z component of the angular momentum for a particular orbital. For example, for an s orbital, l = 0, and the only value of ml is zero. For p orbitals, l = 1, and ml can be equal to –1, 0, or +1. Generally speaking, ml can be equal to –l, –(l – 1), …, –1, 0, +1, …, (l – 1), l. The total number of possible orbitals with the same value of l (a subshell) is 2l + 1. Thus, there is one s-orbital for ml = 0, there are three p-orbitals for ml = 1, five d-orbitals for ml = 2, seven f-orbitals for ml = 3, and so forth. The principal quantum number defines the general value of the electronic energy. The angular momentum quantum number determines the shape of the orbital. And the magnetic quantum number specifies orientation of the orbital in space, as can be seen in Figure 9.4.3. Figure 9.4.4 illustrates the energy levels for various orbitals. The number before the orbital name (such as 2s, 3p, and so forth) stands for the principal quantum number, n. The letter in the orbital name defines the subshell with a specific angular momentum quantum number l = 0 for s orbitals, 1 for p orbitals, 2 for d orbitals. Finally, there are more than one possible orbitals for l ≥ 1, each corresponding to a specific value of ml. In the case of a hydrogen atom or a one-electron ion (such as He+, Li2+, and so on), energies of all the orbitals with the same n are the same. This is called a degeneracy, and the energy levels for the same principal quantum number, n, are called degenerate energy levels. However, in atoms with more than one electron, this degeneracy is eliminated by the electron–electron interactions, and orbitals that belong to different subshells have different energies. Orbitals within the same subshell (for example ns, np, nd, nf, such as 2p, 3s) are still degenerate and have the same energy. While the three quantum numbers discussed in the previous paragraphs work well for describing electron orbitals, some experiments showed that they were not sufficient to explain all observed results. It was demonstrated in the 1920s that when hydrogen-line spectra are examined at extremely high resolution, some lines are actually not single peaks but, rather, pairs of closely spaced lines. This is the so-called fine structure of the spectrum, and it implies that there are additional small differences in energies of electrons even when they are located in the same orbital. These observations led Samuel Goudsmit and George Uhlenbeck to propose that electrons have a fourth quantum number. They called this the spin quantum number, or ms. The other three quantum numbers, n, l, and ml, are properties of specific atomic orbitals that also define in what part of the space an electron is most likely to be located. Orbitals are a result of solving the Schrödinger equation for electrons in atoms. The electron spin is a different kind of property. It is a completely quantum phenomenon with no analogues in the classical realm. In addition, it cannot be derived from solving the Schrödinger equation and is not related to the normal spatial coordinates (such as the Cartesian x, y, and z). Electron spin describes an intrinsic electron “rotation” or “spinning.” Each electron acts as a tiny magnet or a tiny rotating object with an angular momentum, even though this rotation cannot be observed in terms of the spatial coordinates. The magnitude of the overall electron spin can only have one value, and an electron can only “spin” in one of two quantized states. One is termed the α state, with the z component of the spin being in the positive direction of the z axis. This corresponds to the spin quantum number ms=12. The other is called the β state, with the z component of the spin being negative and ms=−12. Any electron, regardless of the atomic orbital it is located in, can only have one of those two values of the spin quantum number. The energies of electrons having ms=−12 and ms=12 are different if an external magnetic field is applied. Figure 9.4.5 illustrates this phenomenon. An electron acts like a tiny magnet. Its moment is directed up (in the positive direction of the z axis) for the 12 spin quantum number and down (in the negative z direction) for the spin quantum number of −12. A magnet has a lower energy if its magnetic moment is aligned with the external magnetic field (the left electron) and a higher energy for the magnetic moment being opposite to the applied field. This is why an electron with ms=12 has a slightly lower energy in an external field in the positive z direction, and an electron with ms=−12 has a slightly higher energy in the same field. This is true even for an electron occupying the same orbital in an atom. A spectral line corresponding to a transition for electrons from the same orbital but with different spin quantum numbers has two possible values of energy; thus, the line in the spectrum will show a fine structure splitting. The Pauli Exclusion Principle An electron in an atom is completely described by four quantum numbers: n, l, ml, and ms. The first three quantum numbers define the orbital and the fourth quantum number describes the intrinsic electron property called spin. An Austrian physicist Wolfgang Pauli formulated a general principle that gives the last piece of information that we need to understand the general behavior of electrons in atoms. The Pauli exclusion principle can be formulated as follows: No two electrons in the same atom can have exactly the same set of all the four quantum numbers. What this means is that electrons can share the same orbital (the same set of the quantum numbers n, l, and ml), but only if their spin quantum numbers ms have different values. Since the spin quantum number can only have two values (±12), no more than two electrons can occupy the same orbital (and if two electrons are located in the same orbital, they must have opposite spins). Therefore, any atomic orbital can be populated by only zero, one, or two electrons. The properties and meaning of the quantum numbers of electrons in atoms are briefly
Here'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.
Align Panel: This panel allows you to align one or more objects the the artboard or other objects. Alignment: Formatting the appearance of text with the margins of the text box. Anchor Point Tool: Allows you to add or remove handles to create a curve on an anchor point. Anchor Points: A point on a path indicates a change of direction. Appearance Panel: This panel shows you the fill, strokes, graphic styles, and effects that have been applied to an object, group or layer and are able to modify theses from this panel directly. Area Type Tool: This occurs when using the type tool and you click and drag a text box, the text will automatically wrap inside the box. Blend Tool: This tool allows you to combine shapes/colors between two or more objects to create a new object between the original, blending the colors and shapes by inserting the middle steps to get from one object to the next. Blob Brush Tool: This tool is used to create free-form objects that can have a more hand-drawn feel. Brushes: Allows you to set the appearance/style of a path, can be applied to existing paths or used to create new paths. Clipping Masks: This command allows you to mask objects to a shape so that only areas that lie within the shape are visible, the mask and objects that are masked are called a clipping set. Closed Path: A path that has the same beginning and ending point. It forms a complete shape that can be filled with color or text. Create Outlines: This command converts text to closed paths and can be found in the Type Menu. Curvature Pen Tool: Allows you to simply create paths with curved anchor points. Curves: Can be applied to an anchor point using handles to create an arched line. Direct Select Tool: Allows you to select individual points of any path. Effects: These can be added to objects to create quick dynamic characteristics. Eraser Tool: This tool allows you to remove anchor points and cut through paths. Expand Objects: This allows you to divide a single object into multiple objects that make up its appearance. Eyedropper Tool: This tool allows you to sample the color or text from an existing part of the artwork. Global Swatches: This is a color swatches that will be automatically updated throughout your artwork when you change them, indicated with a white triangle in the corner of the swatch. Graphic Styles: A set of reusable appearance attributes that allow you to quickly change the look of an object. Grouping: This command allows you to link objects together so that they can be moved, scaled, rotator, or copy. Groups can be nested inside other groups. Hierarchy: To create visual order in design, controlling what the viewer looks at in order using size, color, contrast, etc… Image Trace: This command allows you to convert a raster image into a vector artwork. Isolation Mode: This mode allows you to adjust single objects/groups inside a group without ungrouping the group. Join Tool: This tool joins paths and anchor points together quickly. Kerning: This is the adjustment of the space between two individual letters. Knife Tool: This tool allows you to split an object into 2 pieces along a freehand path you draw. Leading: This is the adjustment of the space between lines of text. Live Corners: This widget appears when using the Direct Select tool and a corner is selected, when used this will create a rounded corner. Live Paint: This command allows you to quickly apply colors to objects in a complex design. Open Path: A path that does not end, not connected back to the original anchor point. Overflow Text: This occurs when the text box is too small to house all the text and is indicated by a small red plus sign in the bottom right corner of the text box. Paintbrush Tool: This tool is used to create free-form paths that can have a more hand-drawn feel. Paragraph Spacing: The space that occurs between lines of text. Pathfinder Panel: This panel allows you to create complex shapes by selecting 2 or more objects and using the buttons in the panel to cut, combine, or divide the objects. Paths: These are created when 2 or more points are connected, these are created using the pen tool. Pen Tool: It allows you to create and edit anchor points and paths. Pencil Tool: This tool is used to create free-form shapes or lines, the accuracy of the lines can be adjusted. Perspective Tool: This tool allows you to place elements on a perspective grid to adjust objects on a different perspective, automatically snapping to the perspective grid. Placeholder Text: Text that is placed in a text box that "holds a place" in a design to allow for creating a layout or adjust the text design. Point Type Tool: This occurs when when using the type tool and you click once, the text will continue without wrapping. Readability: The characteristics of fonts and styles that make test easy to identify and read. Scale, Shear, Distort Objects: This set of commands allows you to adjust the size and perspective of objects. Scissors Tool: This tool allows you to split a path into 2 pieces. Selection Tool: Allows you to select paths, objects or groups by click or dragging over them. Shape Builder Tool: This interactive tool allows you to create complex shapes by merging and erasing simpler objects. Shapes Tools: A group of tools to create basic shapes without using the pen tool (rectangle, ellipse, polygon, star, etc…). Smooth Tool: This tool will smooth a complex path and reduce the number of anchor points. Swatches: This is a saved color that can be applied in a design via the swatches panel and can be grouped, these can include gradients and patterns. Text Wrapping: This is when the text in a text box automatically wraps to the next line when it reaches the edge of the box. Threading Text: This is the ability to create 2 or more text boxes that are linked, when text is added/adjusted in one box, it will affect the other(s). Touch Type Tool: This tool allows you to adjust individual letter in a previously created text box. Tracking: This is the adjustment of the overall spacing between letters. Transform Objects: This allows you to change the size of objects. Type on a Path Tool: This tool allows you to add text along any previously created path. Type Tool: This tool allows you to create text in a design. View Modes: Ability to view projects and adjust the display on the screen. Modes include Outlines, Presentation, & Full Screen.
What is Electric Force? Electric force is just one of many types of forces in the world of physics. Forces are how and why things move, and can be explained by Newton's Laws of Motion. On the smallest scale, electric force is the resulting interaction between two charged particles. These charges can be either positive or negative. Larger objects can be charged by having an abundance of either of these particles, and therefore can create an electric force on a larger scale. Electric force is the reason why hair will sometimes stand up on its own and is also why we have electricity, allowing us to live in the modern world with lights and technology. Even out in nature electric force is present, as electric force causes lightning to strike. Electric force is fundamental to our everyday way of living. Reviewing Newton's Laws of Motion Newton's Laws of motion are the basic principles or ground rules that are applied all across physics. They describe how objects move and can be used to describe the interaction of charges. They are the following: An object in motion will stay in motion unless an external force is applied The force exerted on an object is equal to the mass times the acceleration of the object. ( ) Every force has an equal and opposite force Newton's laws explain how and why charged particles move. Since there is a force involved (e.g. electric force), particles will move around, which is explained by the first law. The second law describes how acceleration of charges can be calculated once the electric force is known. The third law explains how attractive and repulsive forces between charged objects are equal and opposite. Electric Force Examples and Types of Charge As previously mentioned, there are only two types of charges; positive and negative. Two like charges will repel (or move away from) each other, and two opposite charges will attract (or move towards) each other. In other words, two positive or two negative charges will repel, while a positive and a negative charge will attract. Opposite charges will attract while like charges will repel. Attraction versus Repelling Forces Notice how the forces acting upon each other are equal and opposite, as Newton's third law states. Both charges are exerting forces onto each other. Charges in Atoms An atom is made up of three types of particles; protons, neutrons, and electrons. Protons have a positive charge, neutrons have no charge, and electrons have a negative charge. There are no positive or negative charges smaller than protons and electrons. Objects on a larger scale result in an overall positive or negative charged due to an uneven distribution of protons to electrons. An atom consisting of more protons than electrons would be considered positive, and an atom with more electrons than protons would be considered negative. Protons are held close to the nucleus and are tightly bound to an atom, so it's difficult for protons to escape an atom. Electrons, on the other hand, are much further away from the nucleus of an atom. This makes it much easier for them to be removed from an atom. Electrons can leave or join atoms, making them positive or negative depending on the amount of protons. Similarly, for the bigger picture, overall materials and objects with more electrons than protons would be considered negative, and vice versa. Electric Force Examples Hair standing up: When hair is brushed, the hairbrush can strip electrons from hair strands, resulting in the hair being positively charged. This addition of electrons to the hairbrush in turn makes the hairbrush negatively charged. Since the hair is now positively charged, and like forces repel, hair strands will move away from each other, resulting in the hair standing up. Current electricity: All of our everyday technology is powered through current electricity, which is the consistent flow of electrons through conductive materials. This flow is caused by the electric force, as the electrons flow from a negative source to a positive source. Lightning: During a storm, it is common for an abundance of electrons to build up on the bottom of a cloud, making that part of the cloud negatively charged. Positive charges in the ground start to gather on the surface or even on tall objects such as trees as they are attracted towards the negatively charged undersides of clouds. Lightning strikes as a result of these charges becoming extremely built up. Lightning is caused by electric force Lightning Electric Force Equation: Coulomb's Law The magnitude of the electric force, or the amount of force in which objects repel or attract, depends on the distance between the two charged objects and the amount of charge each object carries. The electric force is stronger the closer together the two charges are, and weaker as the two charges move apart. Electric force is also stronger with more charge, and weaker with less charge. This effect on electric force is predictable, and is known as Coulomb's Law. It can be calculated using a mathematical equation, and the resulting magnitude of electric force is measured in Newtons. Coulomb's Law Electric force can be calculated using the following equation known as Coulomb's Law: In this equation, F is the electric force measured in newtons, K is a constant known as the electrostatic constant, and are charges one and two measured in coulombs, and is the radial distance in meters between the two charges. Since the distance is squared and on the denominator, the electric force drops off exponentially as charges move away from each other. This means that the Electric force is inversely proportional to distance. As charges move away from each other, the electric force between them gets smaller and smaller, until the force is negligible. The amount of charges are in the numerator of this equation, making the magnitude of the force larger with more charge. This means that the force is directly proportional to the amount of charge. When the charges are smaller, the amount of force will be smaller. When there is a lot of charge, the force will be much greater. When calculating the electric force using Coulomb's law, the resulting answer only gives the magnitude of the force and not the direction. In order to know the direction, you must know the types of charges. Once again, like forces repel, and unlike forces attract. It helps to draw a visual representation, or a free-body diagram, of the charges and forces acting upon them in order to understand the resulting force direction. Electric Field versus Electric Force An electric field is a direct result of an electric force. Its pure definition is electric force per unit charge, and can be thought of as a mapping of the force vectors. An electric field is present anytime there is an electric force. Therefore, when there are two or more charged particles, there is a surrounding electric field. The direction of the electric field is the direction a positive charge would flow if it were placed within the field. The electric field moves out from a positive charge and goes into a negative charge. Particles with unlike charges move towards each other, and their corresponding electric field lines move out from the positive charge and into the negative charge. The strength of the force at any given point can be seen through the spacing of the electric field lines. The electric force is strongest where the electric field lines are closest together, and weaker as these lines move apart. Like Coulomb's law expresses, electric field lines show how the electric force is strongest with a minimum distance between the two charges. Unlike charges will result in a repelling force, and the resulting electric field is a visual representation of this effect. Electric fields of two positive charges have the electric field moving out away from both of them. As with two negative charges, the field lines move in towards each negative. Lesson Summary An electric force is created when there are two or more charged particles or objects. These charges can be either positive or negative. Like charges will attract (move towards each other) while unlike charges will repel (move away from each other). As Newton's third law suggests, the forces acting upon each other are both equal and opposite. Electrons and protons within an atom are the two smallest types of charges there are. Electrons carry a negative charge while protons carry a positive charge. Electrons can be easily removed or added to atoms, making the overall charge positive or negative. Objects with more electrons than protons are negatively charged. Electric force is strengthened with increased charge and a shorter distance between the charges. This effect is known as Coulomb's law and can be calculated with the Coulomb's law equation. The magnitude of the force is measured in Newtons, and the direction can be determined by knowing whether the charges are attracting or repelling each other. An electric field is present wherever there is an electric force. The direction of this electric field is the direction a positive charge would flow if it where to be dropped in the field, which is from the positive to the negative.
Positive/Negative/Neutral Objects - How are they different? Positive: has fewer electrons than protons Negative: Has more electrons than protons Neutral: has equal numbers of protons and electrons Laws of Electric Charges - What are they? How are they applied? Like charges repel, opposites charges attract, charged AND neutral objects attract Induced Charge Separation - Explain this process. A shift of the position of electrons when a charged object is brought near it. If the charged object is positive, the electrons will move toward it. If the charged object is negative, the electrons will move away from it. Charging by Friction (What is happening with the charges? - Know electrostatic series examples) Process in which objects made from different materials rub against each other, producing a net static charge on each object. When charged by friction, one material will have a stronger attraction to electrons and will pull the electrons off the other material Charging by Conduction (Be able to explain what the electrons are doing) Charging by contact with a charged object. An object that becomes charged by contact always gets the same type of charge that is on the object that charges it. Grounding (Be able to explain how it happens) A method of removing static charges from an object. Electrons from the ground move up to the charged object. If the object is negative, electrons leave the object. If the object is positive, electrons enter the object. The ground always remains neutral Conductors/Insulators/Semiconductors (Know examples for each and characteristics) Conductor: A material that allows electrons to flow through it easily. GOOD CONDUCTORS: Silver, copper, gold, aluminium, magnesium, iron, usually metals Insulator: A material that prevents electrons from flowing through it. Plastic, wood and glass are examples. To prevent electric shocks, conductive wires are wrapped in insulators. Semiconductors: Have special properties that make them fair conductors, they are the foundation of modern electronics, including radios, computers and telephones. Charging by Induction (How do we induce a PERMANENT charge?) You can permanently charge an object using induction by attaching a conducting wire to the neutral object that goes to the ground Electric Discharge - What causes it? Know everyday examples. How is lightning formed? When two objects that have a charge imbalance are brought close together or come in contact with each other, electrons are transferred rapidly. Electrons move from the object with a more negative charge to the object with the less negative charge. Lightning occurs through an imbalance of charge between clouds and the ground. Negative charge at the bottom of a cloud repels the electrons at the earth's surface which move away, causing the ground to become positively charged Current Electricity: Refers to the electrons that flow in a controlled way through a conductor Forms of Current Electricity - Alternating Current (AC) vs Direct Current (DC) - How do they differ? AC: Electrons move back and forth, alternating their direction. produced in generating stations and is then distributed over long distances ex. Something plugged into a wall outlet