Describe and discuss: . the significance of teaching for social justice.

Describe and discuss: . the significance of teaching for social justice.

By accepting that diverse societies have dissimilar cultures, they appreciate … Read More...
Paper 2. Cultural Analysis Essay Overview: The second paper for our class will be a cultural analysis. You will choose an artifact common in American culture and use the object as a way to understand an aspect of American culture. Analysis: The word analysis derives from a Greek word meaning “to dissolve, loosen, or undo.” In a sense, analysis means to divide the whole into its parts, to examine those parts carefully, to look at the relationships among the parts, and then to use this understanding of the parts to better understand the whole. Synonyms for writing to analyze would be writing to interpret, clarify and explain. Purpose: When you write to analyze, you apply your own critical thinking to a puzzling object to offer your own ideas. Your goal is raise interesting questions and make interesting observations about the object or event being analyzed. These may be questions that your audience hasn’t even thought to ask. And you will need to provide tentative answers to those questions, supported by your close examination. Analysis—not Evaluation: Please keep in mind that your purpose is to explore and explain why an artifact is important in American culture and what it says about American culture. However, you are not writing about whether or not this aspect of the culture is good or bad. You should not talk about whether or not the people who engage with this artifact are good or bad for doing so. Format: The first draft will be three typed pages, and you will bring to class 2 copies. It should have your name on every page. Ideally, it will be in MLA (Modern Language Association) format, though this is not important at that stage. See the format of the paper in the example below. Details: For your analysis, you must choose a cultural artifact and explain what this artifact says about an aspect of American culture. Please look through the class PowerPoint on Blackboard entitles “Analysis of a Cultural Artifact” to see a thorough description of the process of analysis. Assessment: In this essay, I am looking for • Superior essays will chose a good artifact, a take readers through a thorough investigation to the artifact to interpret, clarify and explain what the artifact demonstrates about American culture. • well-formed, clear sentences • unified and coherent paragraphs • use of standard grammar, diction, and mechanics of American English. Sample Paper Format (on back) Last Name 1 Your Full Name Dr. Riley-Brown ENG 110: Composition Essay 2 Date Title of Paper Centered This is where the first line of your paper will go. Double space beneath your title and indent the first line of each paragraph five (5) spaces. The essay should have margins that are one each on each side. You should use Times New Roman font in 12 point font size.

Paper 2. Cultural Analysis Essay Overview: The second paper for our class will be a cultural analysis. You will choose an artifact common in American culture and use the object as a way to understand an aspect of American culture. Analysis: The word analysis derives from a Greek word meaning “to dissolve, loosen, or undo.” In a sense, analysis means to divide the whole into its parts, to examine those parts carefully, to look at the relationships among the parts, and then to use this understanding of the parts to better understand the whole. Synonyms for writing to analyze would be writing to interpret, clarify and explain. Purpose: When you write to analyze, you apply your own critical thinking to a puzzling object to offer your own ideas. Your goal is raise interesting questions and make interesting observations about the object or event being analyzed. These may be questions that your audience hasn’t even thought to ask. And you will need to provide tentative answers to those questions, supported by your close examination. Analysis—not Evaluation: Please keep in mind that your purpose is to explore and explain why an artifact is important in American culture and what it says about American culture. However, you are not writing about whether or not this aspect of the culture is good or bad. You should not talk about whether or not the people who engage with this artifact are good or bad for doing so. Format: The first draft will be three typed pages, and you will bring to class 2 copies. It should have your name on every page. Ideally, it will be in MLA (Modern Language Association) format, though this is not important at that stage. See the format of the paper in the example below. Details: For your analysis, you must choose a cultural artifact and explain what this artifact says about an aspect of American culture. Please look through the class PowerPoint on Blackboard entitles “Analysis of a Cultural Artifact” to see a thorough description of the process of analysis. Assessment: In this essay, I am looking for • Superior essays will chose a good artifact, a take readers through a thorough investigation to the artifact to interpret, clarify and explain what the artifact demonstrates about American culture. • well-formed, clear sentences • unified and coherent paragraphs • use of standard grammar, diction, and mechanics of American English. Sample Paper Format (on back) Last Name 1 Your Full Name Dr. Riley-Brown ENG 110: Composition Essay 2 Date Title of Paper Centered This is where the first line of your paper will go. Double space beneath your title and indent the first line of each paragraph five (5) spaces. The essay should have margins that are one each on each side. You should use Times New Roman font in 12 point font size.

Assignment 8 Due: 11:59pm on Friday, April 4, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 10.3 Part A If a particle’s speed increases by a factor of 5, by what factor does its kinetic energy change? ANSWER: Correct Conceptual Question 10.11 A spring is compressed 1.5 . Part A How far must you compress a spring with twice the spring constant to store the same amount of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct = 25 K2 K1 cm x = 1.1 cm Problem 10.2 The lowest point in Death Valley is below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 . Part A What is the change in potential energy of an energetic 80 hiker who makes it from the floor of Death Valley to the top of Mt.Whitney? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.3 Part A At what speed does a 1800 compact car have the same kinetic energy as a 1.80×104 truck going 21.0 ? Express your answer with the appropriate units. ANSWER: Correct Problem 10.5 A boy reaches out of a window and tosses a ball straight up with a speed of 13 . The ball is 21 above the ground as he releases it. 85m m kg U = 3.5×106 J kg kg km/hr vc = 66.4 km hr m/s m Part A Use energy to find the ball’s maximum height above the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Use energy to find the ball’s speed as it passes the window on its way down. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C Use energy to find the speed of impact on the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Hmax = 30 m v = 13 ms v = 24 ms Problem 10.8 A 59.0 skateboarder wants to just make it to the upper edge of a “quarter pipe,” a track that is one-quarter of a circle with a radius of 2.30 . Part A What speed does he need at the bottom? Express your answer with the appropriate units. ANSWER: Correct Problem 10.12 A 1500 car traveling at 12 suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll. Part A kg m 6.71 ms kg m/s What will be the car’s speed as it coasts into the gas station on the other side of the valley? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Ups and Downs Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy transformations that take place involve the object’s kinetic energy and its gravitational potential energy . The law of conservation of energy for such cases implies that the sum of the object’s kinetic energy and potential energy does not change with time. This idea can be expressed by the equation , where “i” denotes the “initial” moment and “f” denotes the “final” moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. First, let us consider an object launched vertically upward with an initial speed . Neglect air resistance. Part A As the projectile goes upward, what energy changes take place? ANSWER: v = 6.8 ms K = (1/2)mv2 U = mgh Ki + Ui = Kf + Uf v Correct Part B At the top point of the flight, what can be said about the projectile’s kinetic and potential energy? ANSWER: Correct Strictly speaking, it is not the ball that possesses potential energy; rather, it is the system “Earth-ball.” Although we will often talk about “the gravitational potential energy of an elevated object,” it is useful to keep in mind that the energy, in fact, is associated with the interactions between the earth and the elevated object. Part C The potential energy of the object at the moment of launch __________. ANSWER: Both kinetic and potential energy decrease. Both kinetic and potential energy increase. Kinetic energy decreases; potential energy increases. Kinetic energy increases; potential energy decreases. Both kinetic and potential energy are at their maximum values. Both kinetic and potential energy are at their minimum values. Kinetic energy is at a maximum; potential energy is at a minimum. Kinetic energy is at a minimum; potential energy is at a maximum. Correct Usually, the zero level is chosen so as to make the relevant calculations simpler. In this case, it makes good sense to assume that at the ground level–but this is not, by any means, the only choice! Part D Using conservation of energy, find the maximum height to which the object will rise. Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: Correct You may remember this result from kinematics. It is comforting to know that our new approach yields the same answer. Part E At what height above the ground does the projectile have a speed of ? Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: is negative is positive is zero depends on the choice of the “zero level” of potential energy U = 0 hmax v g hmax = v2 2g h 0.5v v g h = 3 v2 8g Correct Part F What is the speed of the object at the height of ? Express your answer in terms of and . Use three significant figures in the numeric coefficient. Hint 1. How to approach the problem You are being asked for the speed at half of the maximum height. You know that at the initial height ( ), the speed is . All of the energy is kinetic energy, and so, the total energy is . At the maximum height, all of the energy is potential energy. Since the gravitational potential energy is proportional to , half of the initial kinetic energy must have been converted to potential energy when the projectile is at . Thus, the kinetic energy must be half of its original value (i.e., when ). You need to determine the speed, as a multiple of , that corresponds to such a kinetic energy. ANSWER: Correct Let us now consider objects launched at an angle. For such situations, using conservation of energy leads to a quicker solution than can be produced by kinematics. Part G A ball is launched as a projectile with initial speed at an angle above the horizontal. Using conservation of energy, find the maximum height of the ball’s flight. Express your answer in terms of , , and . Hint 1. Find the final kinetic energy Find the final kinetic energy of the ball. Here, the best choice of “final” moment is the point at which the ball reaches its maximum height, since this is the point we are interested in. u (1/2)hmax v g h = 0 v (1/2)mv2 h (1/2)hmax (1/4)mv2 h = (1/2)hmax v u = 0.707v v hmax v g Kf Express your answer in terms of , , and . Hint 1. Find the speed at the maximum height The speed of the ball at the maximum height is __________. ANSWER: ANSWER: ANSWER: Correct Part H A ball is launched with initial speed from ground level up a frictionless slope. The slope makes an angle with the horizontal. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of , , and . You may or may not use all of these quantities. v m 0 v v cos v sin v tan Kf = 0.5m(vcos( ))2 hmax = (vsin( ))2 2g v hmax v g ANSWER: Correct Interestingly, the answer does not depend on . The difference between this situation and the projectile case is that the ball moving up a slope has no kinetic energy at the top of its trajectory whereas the projectile launched at an angle does. Part I A ball is launched with initial speed from the ground level up a frictionless hill. The hill becomes steeper as the ball slides up; however, the ball remains in contact with the hill at all times. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of and . ANSWER: Correct The profile of the hill does not matter; the equation would have the same terms regardless of the steepness of the hill. Problem 10.14 A 12- -long spring is attached to the ceiling. When a 2.2 mass is hung from it, the spring stretches to a length of 17 . Part A What is the spring constant ? Express your answer to two significant figures and include the appropriate units. hmax = v2 2g v hmax v g hmax = v2 2g Ki + Ui = Kf + Uf cm kg cm k ANSWER: Correct Part B How long is the spring when a 3.0 mass is suspended from it? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.17 A 6.2 mass hanging from a spring scale is slowly lowered onto a vertical spring, as shown in . You may want to review ( pages 255 – 257) . For help with math skills, you may want to review: Solving Algebraic Equations = 430 k Nm kg y = 19 cm kg Part A What does the spring scale read just before the mass touches the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass before it touches the scale. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? ANSWER: Correct Part B The scale reads 22 when the lower spring has been compressed by 2.7 . What is the value of the spring constant for the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? Use these to determine the force on the mass by the spring, taking note of the directions from your picture. How is the spring constant related to the force by the spring and the compression of the spring? Check your units. ANSWER: F = 61 N N cm k = 1400 k Nm Correct Part C At what compression length will the scale read zero? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces on the mass. When the scale reads zero, what is the force on the mass due to the scale? What is the gravitational force on the mass? What is the force on the mass by the spring? How is the compression length related to the force by the spring and the spring constant? Check your units. ANSWER: Correct Problem 10.18 Part A How far must you stretch a spring with = 800 to store 180 of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: y = 4.2 cm k N/m J Correct Problem 10.22 A 15 runaway grocery cart runs into a spring with spring constant 230 and compresses it by 57 . Part A What was the speed of the cart just before it hit the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Spring Gun A spring-loaded toy gun is used to shoot a ball straight up in the air. The ball reaches a maximum height , measured from the equilibrium position of the spring. s = 0.67 m kg N/m cm v = 2.2 ms H Part A The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to . Hint 1. Potential energy of the spring The potential energy of a spring is proportional to the square of the distance the spring is compressed. The spring was compressed half the distance, so the mass, when launched, has one quarter of the energy as in the first trial. Hint 2. Potential energy of the ball At the highest point in the ball’s trajectory, all of the spring’s potential energy has been converted into gravitational potential energy of the ball. ANSWER: Correct A Bullet Is Fired into a Wooden Block A bullet of mass is fired horizontally with speed at a wooden block of mass resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed . H height = H 4 mb vi mw vf Part A Which of the following best describes this collision? Hint 1. Types of collisions An inelastic collision is a collision in which kinetic energy is not conserved. In a partially inelastic collision, kinetic energy is lost, but the objects colliding do not stick together. From this information, you can infer what completely inelastic and elastic collisions are. ANSWER: Correct Part B Which of the following quantities, if any, are conserved during this collision? Hint 1. When is kinetic energy conserved? Kinetic energy is conserved only in perfectly elastic collisions. ANSWER: perfectly elastic partially inelastic perfectly inelastic Correct Part C What is the speed of the block/bullet system after the collision? Express your answer in terms of , , and . Hint 1. Find the momentum after the collision What is the total momentum of the block/bullet system after the collision? Express your answer in terms of and other given quantities. ANSWER: Hint 2. Use conservation of momentum The momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for , this time expressed as the total momentum of the system before the collision. Express your answer in terms of and other given quantities. ANSWER: kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy vi mw mb ptotal vf ptotal = (mw + mb)vf ptotal vi ptotal = mbvi ANSWER: Correct Problem 10.31 Ball 1, with a mass of 150 and traveling at 15.0 , collides head on with ball 2, which has a mass of 340 and is initially at rest. Part A What are the final velocities of each ball if the collision is perfectly elastic? Express your answer with the appropriate units. ANSWER: Correct Part B Express your answer with the appropriate units. ANSWER: Correct Part C vf = mb vi mb+mw g m/s g (vfx) = -5.82 1 ms (vfx) = 9.18 2 ms What are the final velocities of each ball if the collision is perfectly inelastic? Express your answer with the appropriate units. ANSWER: Correct Part D Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.43 A package of mass is released from rest at a warehouse loading dock and slides down the = 2.2 – high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass , from the bottom of the chute. You may want to review ( pages 265 – 269) . For help with math skills, you may want to review: Solving Algebraic Equations (vfx) = 4.59 1 ms (vfx) = 4.59 2 ms m h m 2m Part A Suppose the packages stick together. What is their common speed after the collision? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are two parts to this problem: the block sliding down the frictionless incline and the collision. What conservation laws are valid in each part? In terms of , what are the kinetic and potential energies of the block at the top of the incline? What is the potential energy of the same block at the bottom just before the collision? What are the kinetic energy and velocity of block just before the collision? What is conserved during the collision? What is the total momentum of the two blocks before the collision? What is the momentum of the two blocks stuck together after the collision? What is the velocity of the two blocks after the collision? ANSWER: Correct Part B Suppose the collision between the packages is perfectly elastic. To what height does the package of mass rebound? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are three parts to this problem: the block sliding down the incline, the collision, and mass going back up the incline. What conservation laws are valid in each part? m m v = 2.2 ms m m What is an elastic collision? For an elastic collision, how are the initial and final velocities related when one of the masses is initially at rest? Using the velocity of just before the collision from Part A, what is the velocity of just after the collision in this case? What are the kinetic and potential energies of mass just after the collision? What is the kinetic energy of mass at its maximum rebound height? Using conservation of energy, what is the potential energy of mass at its maximum height? What is the maximum height? ANSWER: Correct Problem 10.35 A cannon tilted up at a 35.0 angle fires a cannon ball at 79.0 from atop a 21.0 -high fortress wall. Part A What is the ball’s impact speed on the ground below? Express your answer with the appropriate units. ANSWER: Correct Problem 10.45 A 1000 safe is 2.5 above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 48 . m m m m m h = 24 cm $ m/s m vf = 81.6 ms kg m cm Part A What is the spring constant of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.49 A 100 block on a frictionless table is firmly attached to one end of a spring with = 21 . The other end of the spring is anchored to the wall. A 30 ball is thrown horizontally toward the block with a speed of 6.0 . Part A If the collision is perfectly elastic, what is the ball’s speed immediately after the collision? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the maximum compression of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: = 2.5×105 k Nm g k N/m g m/s v = 3.2 ms Correct Part C Repeat part A for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D Repeat part B for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.4%. You received 120.28 out of a possible total of 121 points. x = 0.19 m v = 1.4 ms x = 0.11 m

Assignment 8 Due: 11:59pm on Friday, April 4, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 10.3 Part A If a particle’s speed increases by a factor of 5, by what factor does its kinetic energy change? ANSWER: Correct Conceptual Question 10.11 A spring is compressed 1.5 . Part A How far must you compress a spring with twice the spring constant to store the same amount of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct = 25 K2 K1 cm x = 1.1 cm Problem 10.2 The lowest point in Death Valley is below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 . Part A What is the change in potential energy of an energetic 80 hiker who makes it from the floor of Death Valley to the top of Mt.Whitney? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.3 Part A At what speed does a 1800 compact car have the same kinetic energy as a 1.80×104 truck going 21.0 ? Express your answer with the appropriate units. ANSWER: Correct Problem 10.5 A boy reaches out of a window and tosses a ball straight up with a speed of 13 . The ball is 21 above the ground as he releases it. 85m m kg U = 3.5×106 J kg kg km/hr vc = 66.4 km hr m/s m Part A Use energy to find the ball’s maximum height above the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Use energy to find the ball’s speed as it passes the window on its way down. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C Use energy to find the speed of impact on the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Hmax = 30 m v = 13 ms v = 24 ms Problem 10.8 A 59.0 skateboarder wants to just make it to the upper edge of a “quarter pipe,” a track that is one-quarter of a circle with a radius of 2.30 . Part A What speed does he need at the bottom? Express your answer with the appropriate units. ANSWER: Correct Problem 10.12 A 1500 car traveling at 12 suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll. Part A kg m 6.71 ms kg m/s What will be the car’s speed as it coasts into the gas station on the other side of the valley? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Ups and Downs Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy transformations that take place involve the object’s kinetic energy and its gravitational potential energy . The law of conservation of energy for such cases implies that the sum of the object’s kinetic energy and potential energy does not change with time. This idea can be expressed by the equation , where “i” denotes the “initial” moment and “f” denotes the “final” moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. First, let us consider an object launched vertically upward with an initial speed . Neglect air resistance. Part A As the projectile goes upward, what energy changes take place? ANSWER: v = 6.8 ms K = (1/2)mv2 U = mgh Ki + Ui = Kf + Uf v Correct Part B At the top point of the flight, what can be said about the projectile’s kinetic and potential energy? ANSWER: Correct Strictly speaking, it is not the ball that possesses potential energy; rather, it is the system “Earth-ball.” Although we will often talk about “the gravitational potential energy of an elevated object,” it is useful to keep in mind that the energy, in fact, is associated with the interactions between the earth and the elevated object. Part C The potential energy of the object at the moment of launch __________. ANSWER: Both kinetic and potential energy decrease. Both kinetic and potential energy increase. Kinetic energy decreases; potential energy increases. Kinetic energy increases; potential energy decreases. Both kinetic and potential energy are at their maximum values. Both kinetic and potential energy are at their minimum values. Kinetic energy is at a maximum; potential energy is at a minimum. Kinetic energy is at a minimum; potential energy is at a maximum. Correct Usually, the zero level is chosen so as to make the relevant calculations simpler. In this case, it makes good sense to assume that at the ground level–but this is not, by any means, the only choice! Part D Using conservation of energy, find the maximum height to which the object will rise. Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: Correct You may remember this result from kinematics. It is comforting to know that our new approach yields the same answer. Part E At what height above the ground does the projectile have a speed of ? Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: is negative is positive is zero depends on the choice of the “zero level” of potential energy U = 0 hmax v g hmax = v2 2g h 0.5v v g h = 3 v2 8g Correct Part F What is the speed of the object at the height of ? Express your answer in terms of and . Use three significant figures in the numeric coefficient. Hint 1. How to approach the problem You are being asked for the speed at half of the maximum height. You know that at the initial height ( ), the speed is . All of the energy is kinetic energy, and so, the total energy is . At the maximum height, all of the energy is potential energy. Since the gravitational potential energy is proportional to , half of the initial kinetic energy must have been converted to potential energy when the projectile is at . Thus, the kinetic energy must be half of its original value (i.e., when ). You need to determine the speed, as a multiple of , that corresponds to such a kinetic energy. ANSWER: Correct Let us now consider objects launched at an angle. For such situations, using conservation of energy leads to a quicker solution than can be produced by kinematics. Part G A ball is launched as a projectile with initial speed at an angle above the horizontal. Using conservation of energy, find the maximum height of the ball’s flight. Express your answer in terms of , , and . Hint 1. Find the final kinetic energy Find the final kinetic energy of the ball. Here, the best choice of “final” moment is the point at which the ball reaches its maximum height, since this is the point we are interested in. u (1/2)hmax v g h = 0 v (1/2)mv2 h (1/2)hmax (1/4)mv2 h = (1/2)hmax v u = 0.707v v hmax v g Kf Express your answer in terms of , , and . Hint 1. Find the speed at the maximum height The speed of the ball at the maximum height is __________. ANSWER: ANSWER: ANSWER: Correct Part H A ball is launched with initial speed from ground level up a frictionless slope. The slope makes an angle with the horizontal. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of , , and . You may or may not use all of these quantities. v m 0 v v cos v sin v tan Kf = 0.5m(vcos( ))2 hmax = (vsin( ))2 2g v hmax v g ANSWER: Correct Interestingly, the answer does not depend on . The difference between this situation and the projectile case is that the ball moving up a slope has no kinetic energy at the top of its trajectory whereas the projectile launched at an angle does. Part I A ball is launched with initial speed from the ground level up a frictionless hill. The hill becomes steeper as the ball slides up; however, the ball remains in contact with the hill at all times. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of and . ANSWER: Correct The profile of the hill does not matter; the equation would have the same terms regardless of the steepness of the hill. Problem 10.14 A 12- -long spring is attached to the ceiling. When a 2.2 mass is hung from it, the spring stretches to a length of 17 . Part A What is the spring constant ? Express your answer to two significant figures and include the appropriate units. hmax = v2 2g v hmax v g hmax = v2 2g Ki + Ui = Kf + Uf cm kg cm k ANSWER: Correct Part B How long is the spring when a 3.0 mass is suspended from it? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.17 A 6.2 mass hanging from a spring scale is slowly lowered onto a vertical spring, as shown in . You may want to review ( pages 255 – 257) . For help with math skills, you may want to review: Solving Algebraic Equations = 430 k Nm kg y = 19 cm kg Part A What does the spring scale read just before the mass touches the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass before it touches the scale. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? ANSWER: Correct Part B The scale reads 22 when the lower spring has been compressed by 2.7 . What is the value of the spring constant for the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? Use these to determine the force on the mass by the spring, taking note of the directions from your picture. How is the spring constant related to the force by the spring and the compression of the spring? Check your units. ANSWER: F = 61 N N cm k = 1400 k Nm Correct Part C At what compression length will the scale read zero? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces on the mass. When the scale reads zero, what is the force on the mass due to the scale? What is the gravitational force on the mass? What is the force on the mass by the spring? How is the compression length related to the force by the spring and the spring constant? Check your units. ANSWER: Correct Problem 10.18 Part A How far must you stretch a spring with = 800 to store 180 of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: y = 4.2 cm k N/m J Correct Problem 10.22 A 15 runaway grocery cart runs into a spring with spring constant 230 and compresses it by 57 . Part A What was the speed of the cart just before it hit the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Spring Gun A spring-loaded toy gun is used to shoot a ball straight up in the air. The ball reaches a maximum height , measured from the equilibrium position of the spring. s = 0.67 m kg N/m cm v = 2.2 ms H Part A The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to . Hint 1. Potential energy of the spring The potential energy of a spring is proportional to the square of the distance the spring is compressed. The spring was compressed half the distance, so the mass, when launched, has one quarter of the energy as in the first trial. Hint 2. Potential energy of the ball At the highest point in the ball’s trajectory, all of the spring’s potential energy has been converted into gravitational potential energy of the ball. ANSWER: Correct A Bullet Is Fired into a Wooden Block A bullet of mass is fired horizontally with speed at a wooden block of mass resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed . H height = H 4 mb vi mw vf Part A Which of the following best describes this collision? Hint 1. Types of collisions An inelastic collision is a collision in which kinetic energy is not conserved. In a partially inelastic collision, kinetic energy is lost, but the objects colliding do not stick together. From this information, you can infer what completely inelastic and elastic collisions are. ANSWER: Correct Part B Which of the following quantities, if any, are conserved during this collision? Hint 1. When is kinetic energy conserved? Kinetic energy is conserved only in perfectly elastic collisions. ANSWER: perfectly elastic partially inelastic perfectly inelastic Correct Part C What is the speed of the block/bullet system after the collision? Express your answer in terms of , , and . Hint 1. Find the momentum after the collision What is the total momentum of the block/bullet system after the collision? Express your answer in terms of and other given quantities. ANSWER: Hint 2. Use conservation of momentum The momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for , this time expressed as the total momentum of the system before the collision. Express your answer in terms of and other given quantities. ANSWER: kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy vi mw mb ptotal vf ptotal = (mw + mb)vf ptotal vi ptotal = mbvi ANSWER: Correct Problem 10.31 Ball 1, with a mass of 150 and traveling at 15.0 , collides head on with ball 2, which has a mass of 340 and is initially at rest. Part A What are the final velocities of each ball if the collision is perfectly elastic? Express your answer with the appropriate units. ANSWER: Correct Part B Express your answer with the appropriate units. ANSWER: Correct Part C vf = mb vi mb+mw g m/s g (vfx) = -5.82 1 ms (vfx) = 9.18 2 ms What are the final velocities of each ball if the collision is perfectly inelastic? Express your answer with the appropriate units. ANSWER: Correct Part D Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.43 A package of mass is released from rest at a warehouse loading dock and slides down the = 2.2 – high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass , from the bottom of the chute. You may want to review ( pages 265 – 269) . For help with math skills, you may want to review: Solving Algebraic Equations (vfx) = 4.59 1 ms (vfx) = 4.59 2 ms m h m 2m Part A Suppose the packages stick together. What is their common speed after the collision? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are two parts to this problem: the block sliding down the frictionless incline and the collision. What conservation laws are valid in each part? In terms of , what are the kinetic and potential energies of the block at the top of the incline? What is the potential energy of the same block at the bottom just before the collision? What are the kinetic energy and velocity of block just before the collision? What is conserved during the collision? What is the total momentum of the two blocks before the collision? What is the momentum of the two blocks stuck together after the collision? What is the velocity of the two blocks after the collision? ANSWER: Correct Part B Suppose the collision between the packages is perfectly elastic. To what height does the package of mass rebound? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are three parts to this problem: the block sliding down the incline, the collision, and mass going back up the incline. What conservation laws are valid in each part? m m v = 2.2 ms m m What is an elastic collision? For an elastic collision, how are the initial and final velocities related when one of the masses is initially at rest? Using the velocity of just before the collision from Part A, what is the velocity of just after the collision in this case? What are the kinetic and potential energies of mass just after the collision? What is the kinetic energy of mass at its maximum rebound height? Using conservation of energy, what is the potential energy of mass at its maximum height? What is the maximum height? ANSWER: Correct Problem 10.35 A cannon tilted up at a 35.0 angle fires a cannon ball at 79.0 from atop a 21.0 -high fortress wall. Part A What is the ball’s impact speed on the ground below? Express your answer with the appropriate units. ANSWER: Correct Problem 10.45 A 1000 safe is 2.5 above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 48 . m m m m m h = 24 cm $ m/s m vf = 81.6 ms kg m cm Part A What is the spring constant of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.49 A 100 block on a frictionless table is firmly attached to one end of a spring with = 21 . The other end of the spring is anchored to the wall. A 30 ball is thrown horizontally toward the block with a speed of 6.0 . Part A If the collision is perfectly elastic, what is the ball’s speed immediately after the collision? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the maximum compression of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: = 2.5×105 k Nm g k N/m g m/s v = 3.2 ms Correct Part C Repeat part A for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D Repeat part B for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.4%. You received 120.28 out of a possible total of 121 points. x = 0.19 m v = 1.4 ms x = 0.11 m

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Chapter 5 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 5.1 Drawing Force Vectors Learning Goal: To practice Tactics Box 5.1 Drawing Force Vectors. To visualize how forces are exerted on objects, we can use simple diagrams such as vectors. This Tactics Box illustrates the process of drawing a force vector by using the particle model, in which objects are treated as points. TACTICS BOX 5.1 Drawing force vectors Represent the object 1. as a particle. 2. Place the tail of the force vector on the particle. 3. Draw the force vector as an arrow pointing in the proper direction and with a length proportional to the size of the force. 4. Give the vector an appropriate label. The resulting diagram for a force exerted on an object is shown in the drawing. Note that the object is represented as a black dot. Part A A book lies on a table. A pushing force parallel to the table top and directed to the right is exerted on the book. Follow the steps above to draw the force vector . Use the black dot as the particle representing the book. F  F push F push

Chapter 5 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 5.1 Drawing Force Vectors Learning Goal: To practice Tactics Box 5.1 Drawing Force Vectors. To visualize how forces are exerted on objects, we can use simple diagrams such as vectors. This Tactics Box illustrates the process of drawing a force vector by using the particle model, in which objects are treated as points. TACTICS BOX 5.1 Drawing force vectors Represent the object 1. as a particle. 2. Place the tail of the force vector on the particle. 3. Draw the force vector as an arrow pointing in the proper direction and with a length proportional to the size of the force. 4. Give the vector an appropriate label. The resulting diagram for a force exerted on an object is shown in the drawing. Note that the object is represented as a black dot. Part A A book lies on a table. A pushing force parallel to the table top and directed to the right is exerted on the book. Follow the steps above to draw the force vector . Use the black dot as the particle representing the book. F  F push F push

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Design of Electrical Systems Name: ______________________________ Note: All problems weighted equally. Show your work on all problems to receive partial credit. Resources: a) The Fundamental Logic Gate Family, Author Unknown b) Electric Devices and Circuit Theory 7th Edition, Boylestad c) Introductory Circuit Analysis 10th Edition, Boylestad d) Power Supplies (Voltage Regulators) Chapter 19, Boylestad e) Electronic Devices and Circuit Theory Chapter 5, Boylestad f) Operational Amplifiers Handout, Self g) Switch Mode Power Supplies, Philips Semiconductor h) NI Tutorial 13714-en October 6, 2013 i) NI Tutorial 13714-en V2.0 October 6, 2013 j) National Instruments Circuit Design Applications http://www.ni.com/multisim/applications/pro/ k) ENERGY STAR https://www.energystar.gov/index.cfm?c=most_efficient.me_comp_monitor_under_23_inches l) Manufactures Device Data Sheets 1) For the VDB shown below, please find the following quantities and plot the load line (Saturation / Cutoff), Q pt (Quiescent Point) and sketch input waveform and output wave form. Remember to test for Exact vs. Approximate Method. Given Bdc = hfe = 150 and RL of 10KΩ. Efficiency _ Class _____ Degrees ___ VR2_______ VE_______ VC _______ VCE ______ IC _______ IE _______ IB _______ PD _______ re’ _______ Av _______ mpp ______ Vout______ What is the effect of reducing RL to 500Ω ________________________________ What is the effect of reducing the Source Frequency to 50 Hz ________________ | | | | | | |____________________________________________ 2) For the following Networks, please complete the Truth Tables, Logic Gate Type, provide the Boolean Logic Expression. A | Vout 0 | 1 | Logic Gate Type _______ Boolean Logic Expression _________ A B| Vout 0 0| 0 1| 1 0| 1 1| Logic Gate Type _______ Boolean Logic Expression _________ A B C| Vout 0 0 0| 0 0 1| 0 1 0| 0 1 1| 1 0 0| 1 0 1| 1 1 0| 1 1 1| Logic Gate Type _______ Boolean Logic Expression _________ Operation of Transistors ____________ 3) For the Network shown below, please refer to Electronic Devices and Circuit Theory Chapter 5, Boylestad to solve for the following values: Given: Bdc1 = hfe1 = 55 Bdc2 = hfe2 = 70 Bdc Total ______ IB1 _________ IB2 _________ VC1 __________ VC2 __________ VE1 __________ VE2 __________ What is this Transistor Configuration? _______________________ What are the advantages of this Transistor Configuration? _________________________________________ _________________________________________ _________________________________________ _________________________________________ 4) Design a Four (4) output Power Supply with the following Specifications, Provide a clean schematic sketch of circuit (Please provide the schematic sketch on a separate piece of graph paper). Use a straight edge and label everything. Refer to Data Sheets as necessary. Specifications: 120 VAC rms 60 Hz Source Positive + 15 VDC Driving a 15Ω 20 Watt Resistive Load Positive +8 VDC Driving a 10Ω 2 Watt Resistive Load Negative – 12 VDC Driving a 10Ω 2 Watt Resistive Load Negative – 5 VDC Driving a 4Ω 2 Watt Resistive Load Parts available (Must use parts): 1x 120 VAC 40 Volt 3.5 Amp Center Tap Transformer 1x Fuse 1x Bridge Rectifier 12 Amp 1x LM7808 1x LM7815 1x LM7905 1x LM7912 Psource _____________ Fuse size with 25% Service Factor, 1-10 Amps increments of 1A, 10 – 50 Amps increments of 5 Amps ______ Are we exceeding Power Dissipation of any components? If so please identify and provide a brief explanation: _________________________________________________________________ _________________________________________________________________ 5) For the circuit shown below please calculate the following quantities, and Plot the Trans-Conductance Curve (Transfer Curve), (Please provide the plot on a separate piece of graph paper): You will need to refer to the 2N3819 N-Channel JFET ON Semiconductor Data Sheet Posted on Bb. VDS _________ VP ___________ VGS(off) ______ VS __________ VD __________ VG __________ PDD _________ PSource ______ VGSQ ________ IDQ __________ 6) Determine both the Upper and Lower Cutoff frequencies. Sketch Bode plot and label everything including dB Role-Off. Construct Network in Multisim and perform AC Analysis verifying frequency response and Upper and Lower Cutoff Frequencies in support of your calculations. Attach Screen shot of your Multisim Model and AC Analysis. Repeat the above for a 2nd Order Active BP Filter. You will need to research this configuration. Make sure that you use the same values for R and C. Upper and Lower Cutoff Frequencies are determined by for the 2nd Order Active BP Filter fc = 1/(2(3.14)SQRT(R1R2C1C2)). Demonstrate a change in Roll-Off from 1st Order to 2nd Order. First Order: Lower Cutoff Frequency ________ Upper Cutoff Frequency ________ Roll-Off ______________________ | | | | | | | |_____________________________________________________________ Second Order: Lower Cutoff Frequency ________ Upper Cutoff Frequency ________ Roll-Off ______________________ | | | | | | |_____________________________________________________________ 7) The following questions relate to LED Backlight LCD Monitors. (Please feel free to use more paper if need be). See Resources. Please explain the differences between LED Backlight LCD Monitor, LCD and CCFL Monitors (Cold Cathode Fluorescent Lamp) Monitors. What are some advantages of LED Backlight LCD Monitors when compared with LCD and CCFL Monitors? What color LEDs are used in the creation of an LED Backlight LCD Monitor? Does a Black Background use less energy than a White Background? If you can believe the hype, how and why are LED Backlight LCD Monitors among the most energy efficient, higher than heirs apparent? 8) In this problem the goal is to verify the Transfer Characteristics of the 2N7000G Enhancement Mode N-Channel MOSFET against the manufactures Data Sheets. Please create in Multisim a Model as exampled below. First Plot by hand on Graph Paper various VGS Voltages vs ID. Second simulate using the DC Sweep Analysis. From these results verify against the 2N7000G ON Semiconductor Data Sheet Posted on Bb, remembering that the 2N7000G ON Semiconductor Data Sheet includes both Tabulated Data and Figure 2. Transfer Characteristics. Attach all results, screen shots and write a brief description of your work. • I estimate that my mark for this exam will be: ________ % • Time spent on this exam: __________ Hours • Average of time spent per week on work for EGR-330 (outside class sessions): ______________ Hours

Design of Electrical Systems Name: ______________________________ Note: All problems weighted equally. Show your work on all problems to receive partial credit. Resources: a) The Fundamental Logic Gate Family, Author Unknown b) Electric Devices and Circuit Theory 7th Edition, Boylestad c) Introductory Circuit Analysis 10th Edition, Boylestad d) Power Supplies (Voltage Regulators) Chapter 19, Boylestad e) Electronic Devices and Circuit Theory Chapter 5, Boylestad f) Operational Amplifiers Handout, Self g) Switch Mode Power Supplies, Philips Semiconductor h) NI Tutorial 13714-en October 6, 2013 i) NI Tutorial 13714-en V2.0 October 6, 2013 j) National Instruments Circuit Design Applications http://www.ni.com/multisim/applications/pro/ k) ENERGY STAR https://www.energystar.gov/index.cfm?c=most_efficient.me_comp_monitor_under_23_inches l) Manufactures Device Data Sheets 1) For the VDB shown below, please find the following quantities and plot the load line (Saturation / Cutoff), Q pt (Quiescent Point) and sketch input waveform and output wave form. Remember to test for Exact vs. Approximate Method. Given Bdc = hfe = 150 and RL of 10KΩ. Efficiency _ Class _____ Degrees ___ VR2_______ VE_______ VC _______ VCE ______ IC _______ IE _______ IB _______ PD _______ re’ _______ Av _______ mpp ______ Vout______ What is the effect of reducing RL to 500Ω ________________________________ What is the effect of reducing the Source Frequency to 50 Hz ________________ | | | | | | |____________________________________________ 2) For the following Networks, please complete the Truth Tables, Logic Gate Type, provide the Boolean Logic Expression. A | Vout 0 | 1 | Logic Gate Type _______ Boolean Logic Expression _________ A B| Vout 0 0| 0 1| 1 0| 1 1| Logic Gate Type _______ Boolean Logic Expression _________ A B C| Vout 0 0 0| 0 0 1| 0 1 0| 0 1 1| 1 0 0| 1 0 1| 1 1 0| 1 1 1| Logic Gate Type _______ Boolean Logic Expression _________ Operation of Transistors ____________ 3) For the Network shown below, please refer to Electronic Devices and Circuit Theory Chapter 5, Boylestad to solve for the following values: Given: Bdc1 = hfe1 = 55 Bdc2 = hfe2 = 70 Bdc Total ______ IB1 _________ IB2 _________ VC1 __________ VC2 __________ VE1 __________ VE2 __________ What is this Transistor Configuration? _______________________ What are the advantages of this Transistor Configuration? _________________________________________ _________________________________________ _________________________________________ _________________________________________ 4) Design a Four (4) output Power Supply with the following Specifications, Provide a clean schematic sketch of circuit (Please provide the schematic sketch on a separate piece of graph paper). Use a straight edge and label everything. Refer to Data Sheets as necessary. Specifications: 120 VAC rms 60 Hz Source Positive + 15 VDC Driving a 15Ω 20 Watt Resistive Load Positive +8 VDC Driving a 10Ω 2 Watt Resistive Load Negative – 12 VDC Driving a 10Ω 2 Watt Resistive Load Negative – 5 VDC Driving a 4Ω 2 Watt Resistive Load Parts available (Must use parts): 1x 120 VAC 40 Volt 3.5 Amp Center Tap Transformer 1x Fuse 1x Bridge Rectifier 12 Amp 1x LM7808 1x LM7815 1x LM7905 1x LM7912 Psource _____________ Fuse size with 25% Service Factor, 1-10 Amps increments of 1A, 10 – 50 Amps increments of 5 Amps ______ Are we exceeding Power Dissipation of any components? If so please identify and provide a brief explanation: _________________________________________________________________ _________________________________________________________________ 5) For the circuit shown below please calculate the following quantities, and Plot the Trans-Conductance Curve (Transfer Curve), (Please provide the plot on a separate piece of graph paper): You will need to refer to the 2N3819 N-Channel JFET ON Semiconductor Data Sheet Posted on Bb. VDS _________ VP ___________ VGS(off) ______ VS __________ VD __________ VG __________ PDD _________ PSource ______ VGSQ ________ IDQ __________ 6) Determine both the Upper and Lower Cutoff frequencies. Sketch Bode plot and label everything including dB Role-Off. Construct Network in Multisim and perform AC Analysis verifying frequency response and Upper and Lower Cutoff Frequencies in support of your calculations. Attach Screen shot of your Multisim Model and AC Analysis. Repeat the above for a 2nd Order Active BP Filter. You will need to research this configuration. Make sure that you use the same values for R and C. Upper and Lower Cutoff Frequencies are determined by for the 2nd Order Active BP Filter fc = 1/(2(3.14)SQRT(R1R2C1C2)). Demonstrate a change in Roll-Off from 1st Order to 2nd Order. First Order: Lower Cutoff Frequency ________ Upper Cutoff Frequency ________ Roll-Off ______________________ | | | | | | | |_____________________________________________________________ Second Order: Lower Cutoff Frequency ________ Upper Cutoff Frequency ________ Roll-Off ______________________ | | | | | | |_____________________________________________________________ 7) The following questions relate to LED Backlight LCD Monitors. (Please feel free to use more paper if need be). See Resources. Please explain the differences between LED Backlight LCD Monitor, LCD and CCFL Monitors (Cold Cathode Fluorescent Lamp) Monitors. What are some advantages of LED Backlight LCD Monitors when compared with LCD and CCFL Monitors? What color LEDs are used in the creation of an LED Backlight LCD Monitor? Does a Black Background use less energy than a White Background? If you can believe the hype, how and why are LED Backlight LCD Monitors among the most energy efficient, higher than heirs apparent? 8) In this problem the goal is to verify the Transfer Characteristics of the 2N7000G Enhancement Mode N-Channel MOSFET against the manufactures Data Sheets. Please create in Multisim a Model as exampled below. First Plot by hand on Graph Paper various VGS Voltages vs ID. Second simulate using the DC Sweep Analysis. From these results verify against the 2N7000G ON Semiconductor Data Sheet Posted on Bb, remembering that the 2N7000G ON Semiconductor Data Sheet includes both Tabulated Data and Figure 2. Transfer Characteristics. Attach all results, screen shots and write a brief description of your work. • I estimate that my mark for this exam will be: ________ % • Time spent on this exam: __________ Hours • Average of time spent per week on work for EGR-330 (outside class sessions): ______________ Hours

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Describe and discuss: how your study of special education has informed your professional identity

Describe and discuss: how your study of special education has informed your professional identity

The force on culture variety and linguistic diversity in special … Read More...
ENG 100 – Critique Assignment Sheet Rough Draft Due for Peer Response: Tuesday, September 29 First Draft Due (submit for feedback): Thursday, October 1 Final Draft with Outline Due: Thursday, October 8 Highlighting, Labeling, and Reflection: Thursday, October 8 Submit hard copies in class and upload to turnitin.com (Password: English, Class ID: 10423941) What is a Critique? A critique is a “formal evaluation [that offers your] judgment of a text—whether the reading was effective, ineffective, valuable, or trivial.” In a critique, “your goal is to convince readers to accept your judgments concerning the quality of the reading” based on specific criteria you have established (Wilhoit 87). Additionally, a critique is comprised of many integrated parts: introduction to the text, introduction to and brief background on the general topic, brief summary properly placed in the essay, a discussion of the criteria chosen for evaluation, a discussion of the criteria using specific examples/information from the text (this discussion should be the largest section of your essay by far!!), instances of personal response, and a conclusion. All of these items should relate to your overall evaluation/thesis of the text. The Assignment: Instead of a written essay, your “text” will be either a movie or a documentary. You will follow the same standards that you would use for a critique based off of an essay but you will adapt the integrated parts to fit a film critique. In order to effectively address this assignment, complete the following steps: STEP I: Choose either a movie or documentary • Base your choice on the strength of your feelings, whether hate, love, respect, etc., because you do not have to like the film in order to write a solid and coherent critique. You might have more to say about a film you dislike. Also choose a genre of film that you understand (i.e. romantic comedy, drama, indie-film, comedy, documentary). • Think about the important components for this specific genre. STEP II: Watch and Annotate the film • Note the major points within the film, how you felt while watching it, and what made you feel that way. • Keep in mind the film’s genre and whether or not your chosen film fits any of those criteria. STEP III: Analyze (break the film into parts) • Break the film down into your genre-driven criteria. • Choose 4-5 criteria and then determine what sections/components of the film either represent effectiveness or ineffectiveness. STEP IV: Evaluate the film (using the criteria and your personal standards) • Evaluate the film according to the criteria list we will generate in class. • To help create your thesis claim, determine whether the film, based on your criteria and standards, is an excellent, mediocre, terrible, etc. representation of your chosen genre. • For example: While the costume and design are fantastic and interesting, the film 300 is a mediocre example of historical drama because the history of Greece and Asia is inaccurate and the female characters are weak. STEP V: Find outside sources—one should agree with you and one should disagree • Check out a review website, such as imdb.com, and locate a few reviews of your film. In your critique, you will be expected to reference other film reviewers to develop and support your own arguments. Please note that those reviews must be cited properly, both in-text citations and the Works Cited page entries. The basic structure of the critique is as follows: • An introduction that o Introduces the film and provides an adequate amount of background information, including the intended audience, to give the reader context (i.e. a cartoon might not be meant for college-age viewers) o Includes a thesis statement that presents the film as either an excellent, mediocre, or terrible representation of your chosen genre o Explains at least three-four different criteria as the basis for your thesis/argument • A summary that is o Brief, neutral and comprehensive o No more than one paragraph in length • Body Paragraphs including o Support of your thesis using specific examples from the film o More than one example to support your argument o Either direct quotes or paraphrased information from the source text, reviews, outside information (websites, blogs, credible sources) or a combination of all three to support your argument • A counter-claim o Based on an outside review/blog/article disagreeing with your opinion or one criteria o Includes either a refutation or concession of the reviewer’s opinion • A conclusion including o A restatement of your main points and thesis o A final recommendation • A Work Cited page that o Includes all referenced materials including the source text The bulk of your critique should consist of your qualified opinion of the film – unlike the summary, your opinion matters here. In the body of your paper, you will need about three to five main points to support your thesis statement. You will develop each of these points in a section of your essay, each section consisting of about three paragraphs. You will make claims in your topic sentences, provide examples from the text, and then explain your reasons, using source support where possible. Evaluation A successful critique will contain all of the following: • Creative and clearly stated criteria • A debatable thesis statement • A brief background and summary of the film • 80% of the essay is located within the body paragraphs • Topic sentences that transition from one criteria to the next • Body paragraphs clearly and accurately reflecting your criteria and opinion • Body paragraphs that include more than one example as support • Conclusion including a summation and thoughtful recommendation • Correct MLA documentation including signal phrases and in-text citations • A Work Cited page including all sources referenced • Correct grammar and mechanics • Effective and meaningful transitions • Meaningful and descriptive word choices • Literary present tense and grammatical 3rd person • Length of 3-5 pages • Follows the basic structure for a critique Possible Points (25 % of final grade): • Outline 5 % • Peer Response Workshop with Rough Draft 5 % • Highlighted Revisions, & Reflection 10 % • Final Draft: 80 % Upload to Turnitin.com, using Password: English and Class ID: 10423941. Your grade will not be finalized until you have done this.

ENG 100 – Critique Assignment Sheet Rough Draft Due for Peer Response: Tuesday, September 29 First Draft Due (submit for feedback): Thursday, October 1 Final Draft with Outline Due: Thursday, October 8 Highlighting, Labeling, and Reflection: Thursday, October 8 Submit hard copies in class and upload to turnitin.com (Password: English, Class ID: 10423941) What is a Critique? A critique is a “formal evaluation [that offers your] judgment of a text—whether the reading was effective, ineffective, valuable, or trivial.” In a critique, “your goal is to convince readers to accept your judgments concerning the quality of the reading” based on specific criteria you have established (Wilhoit 87). Additionally, a critique is comprised of many integrated parts: introduction to the text, introduction to and brief background on the general topic, brief summary properly placed in the essay, a discussion of the criteria chosen for evaluation, a discussion of the criteria using specific examples/information from the text (this discussion should be the largest section of your essay by far!!), instances of personal response, and a conclusion. All of these items should relate to your overall evaluation/thesis of the text. The Assignment: Instead of a written essay, your “text” will be either a movie or a documentary. You will follow the same standards that you would use for a critique based off of an essay but you will adapt the integrated parts to fit a film critique. In order to effectively address this assignment, complete the following steps: STEP I: Choose either a movie or documentary • Base your choice on the strength of your feelings, whether hate, love, respect, etc., because you do not have to like the film in order to write a solid and coherent critique. You might have more to say about a film you dislike. Also choose a genre of film that you understand (i.e. romantic comedy, drama, indie-film, comedy, documentary). • Think about the important components for this specific genre. STEP II: Watch and Annotate the film • Note the major points within the film, how you felt while watching it, and what made you feel that way. • Keep in mind the film’s genre and whether or not your chosen film fits any of those criteria. STEP III: Analyze (break the film into parts) • Break the film down into your genre-driven criteria. • Choose 4-5 criteria and then determine what sections/components of the film either represent effectiveness or ineffectiveness. STEP IV: Evaluate the film (using the criteria and your personal standards) • Evaluate the film according to the criteria list we will generate in class. • To help create your thesis claim, determine whether the film, based on your criteria and standards, is an excellent, mediocre, terrible, etc. representation of your chosen genre. • For example: While the costume and design are fantastic and interesting, the film 300 is a mediocre example of historical drama because the history of Greece and Asia is inaccurate and the female characters are weak. STEP V: Find outside sources—one should agree with you and one should disagree • Check out a review website, such as imdb.com, and locate a few reviews of your film. In your critique, you will be expected to reference other film reviewers to develop and support your own arguments. Please note that those reviews must be cited properly, both in-text citations and the Works Cited page entries. The basic structure of the critique is as follows: • An introduction that o Introduces the film and provides an adequate amount of background information, including the intended audience, to give the reader context (i.e. a cartoon might not be meant for college-age viewers) o Includes a thesis statement that presents the film as either an excellent, mediocre, or terrible representation of your chosen genre o Explains at least three-four different criteria as the basis for your thesis/argument • A summary that is o Brief, neutral and comprehensive o No more than one paragraph in length • Body Paragraphs including o Support of your thesis using specific examples from the film o More than one example to support your argument o Either direct quotes or paraphrased information from the source text, reviews, outside information (websites, blogs, credible sources) or a combination of all three to support your argument • A counter-claim o Based on an outside review/blog/article disagreeing with your opinion or one criteria o Includes either a refutation or concession of the reviewer’s opinion • A conclusion including o A restatement of your main points and thesis o A final recommendation • A Work Cited page that o Includes all referenced materials including the source text The bulk of your critique should consist of your qualified opinion of the film – unlike the summary, your opinion matters here. In the body of your paper, you will need about three to five main points to support your thesis statement. You will develop each of these points in a section of your essay, each section consisting of about three paragraphs. You will make claims in your topic sentences, provide examples from the text, and then explain your reasons, using source support where possible. Evaluation A successful critique will contain all of the following: • Creative and clearly stated criteria • A debatable thesis statement • A brief background and summary of the film • 80% of the essay is located within the body paragraphs • Topic sentences that transition from one criteria to the next • Body paragraphs clearly and accurately reflecting your criteria and opinion • Body paragraphs that include more than one example as support • Conclusion including a summation and thoughtful recommendation • Correct MLA documentation including signal phrases and in-text citations • A Work Cited page including all sources referenced • Correct grammar and mechanics • Effective and meaningful transitions • Meaningful and descriptive word choices • Literary present tense and grammatical 3rd person • Length of 3-5 pages • Follows the basic structure for a critique Possible Points (25 % of final grade): • Outline 5 % • Peer Response Workshop with Rough Draft 5 % • Highlighted Revisions, & Reflection 10 % • Final Draft: 80 % Upload to Turnitin.com, using Password: English and Class ID: 10423941. 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Assignment 9 Due: 11:59pm on Friday, April 11, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 11.2 Part A Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Part B Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Problem 11.4  A B = 5 − 6 A i ^ j ^ = −9 − 5 B i ^ j ^ A  B  = -15  A B = −5 + 9 A i ^ j ^ = 5 + 6 B i ^ j ^ A  B  = 29 Part A What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part B What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct ± All Work and No Play Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement If an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the force can be calculated as , where is the work done by force on the object that undergoes displacement directed at angle relative to .  A B A = 2 + 5 ı ^  ^ B = −2 − 4 ı ^  ^  = 175  A B A = −6 + 2 ı ^  ^ B = − − 3 ı ^  ^  = 90 W =  = cos  F  s  F   s  W F  s  F  Note that depending on the value of , the work done can be positive, negative, or zero. In this problem, you will practice calculating work done on an object moving in a straight line. The first series of questions is related to the accompanying figure. Part A What can be said about the sign of the work done by the force ? ANSWER: Correct When , the cosine of is zero, and therefore the work done is zero. Part B cos  F  1 It is positive. It is negative. It is zero. There is not enough information to answer the question.  = 90  What can be said about the work done by force ? ANSWER: Correct When , is positive, and so the work done is positive. Part C The work done by force is ANSWER: Correct When , is negative, and so the work done is negative. Part D The work done by force is ANSWER: F  2 It is positive. It is negative. It is zero. 0 <  < 90 cos  F  3 positive negative zero 90 <  < 180 cos  F  4 Correct Part E The work done by force is ANSWER: Correct positive negative zero F  5 positive negative zero Part F The work done by force is ANSWER: Correct Part G The work done by force is ANSWER: Correct In the next series of questions, you will use the formula to calculate the work done by various forces on an object that moves 160 meters to the right. F  6 positive negative zero F  7 positive negative zero W =  = cos  F  s  F   s  Part H Find the work done by the 18-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part I Find the work done by the 30-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part J Find the work done by the 12-newton force. Use two significant figures in your answer. Express your answer in joules. W W = 2900 J W W = 4200 J W ANSWER: Correct Part K Find the work done by the 15-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Introduction to Potential Energy Learning Goal: Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that must be added to the kinetic energy to get the total mechanical energy. The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part we introduce the concept of potential energy. But for now, please answer in terms of the work-energy theorem. Work-Energy Theorem The work-energy theorem states , where is the work done by all forces that act on the object, and and are the initial and final kinetic energies, respectively. Part A The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion. W = -1900 J W W = -1800 J Kf = Ki + Wall Wall Ki Kf Choose the best answer to fill in the blanks above: ANSWER: Correct It is important that the force have a component acting in the direction of motion. For example, if a ball is attached to a string and whirled in uniform circular motion, the string does apply a force to the ball, but since the string's force is always perpendicular to the motion it does no work and cannot change the kinetic energy of the ball. Part B To calculate the change in energy, you must know the force as a function of _______. The work done by the force causes the energy change. Choose the best answer to fill in the blank above: ANSWER: Correct Part C To illustrate the work-energy concept, consider the case of a stone falling from to under the influence of gravity. Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone. Choose the best answer to fill in the blanks above: distance / potential distance / kinetic vertical displacement / potential none of the above acceleration work distance potential energy xi xf ANSWER: Correct Potential Energy You should read about potential energy in your text before answering the following questions. Potential energy is a concept that builds on the work-energy theorem, enlarging the concept of energy in the most physically useful way. The key aspect that allows for potential energy is the existence of conservative forces, forces for which the work done on an object does not depend on the path of the object, only the initial and final positions of the object. The gravitational force is conservative; the frictional force is not. The change in potential energy is the negative of the work done by conservative forces. Hence considering the initial and final potential energies is equivalent to calculating the work done by the conservative forces. When potential energy is used, it replaces the work done by the associated conservative force. Then only the work due to nonconservative forces needs to be calculated. In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which now changes the total energy: , where and are the final and initial potential energies, and is the work due only to nonconservative forces. Now, we will revisit the falling stone example using the concept of potential energy. Part D Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy) say that the increased kinetic energy comes from the ______ of the _______ energy. Choose the best answer to fill in the blanks above: ANSWER: force / kinetic potential energy / potential force / potential potential energy / kinetic Kf + Uf = Ef = Wnc + Ei = Wnc + Ki + Ui Uf Ui Wnc Correct Part E This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potential energies, is _______. Choose the best answer to fill in the blanks above: ANSWER: Correct Problem 11.7 Part A How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: work / potential force / kinetic change / potential sum / conserved sum / zero sum / not conserved difference / conserved F  = (− + i ^ ) N j ^ ! = r m i ^ Correct Part B How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.10 A 1.8 book is lying on a 0.80- -high table. You pick it up and place it on a bookshelf 2.27 above the floor. Part A How much work does gravity do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B W = -8.6 J F  = (− + i ^ ) N j ^ ! = r m? j ^ W = 26 J kg m m Wg = -26 J How much work does your hand do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.12 The three ropes shown in the bird's-eye view of the figure are used to drag a crate 3.3 across the floor. Part A How much work is done by each of the three forces? Express your answers using two significant figures. Enter your answers numerically separated by commas. ANSWER: WH = 26 J m W1 , W2 , W3 = 1.9,1.2,-2.1 kJ Correct Enhanced EOC: Problem 11.16 A 1.2 particle moving along the x-axis experiences the force shown in the figure. The particle's velocity is 4.6 at . You may want to review ( pages 286 - 287) . For help with math skills, you may want to review: The Definite Integral Part A What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: kg m/s x = 0m x = 2m x = 0 m x = 0 m x = 2 m x = 2 m x = 2 m Correct Part B What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Can the work be negative? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: Correct Work on a Sliding Block A block of weight sits on a frictionless inclined plane, which makes an angle with respect to the horizontal, as shown. A force of magnitude , applied parallel to the incline, pulls the block up the plane at constant speed. v = 6.2 ms x = 4m x = 0 m x = 0 m x = 4 m x = 4 m x = 4 m v = 4.6 ms w  F Part A The block moves a distance up the incline. The block does not stop after moving this distance but continues to move with constant speed. What is the total work done on the block by all forces? (Include only the work done after the block has started moving, not the work needed to start the block moving from rest.) Express your answer in terms of given quantities. Hint 1. What physical principle to use To find the total work done on the block, use the work-energy theorem: . Hint 2. Find the change in kinetic energy What is the change in the kinetic energy of the block, from the moment it starts moving until it has been pulled a distance ? Remember that the block is pulled at constant speed. Hint 1. Consider kinetic energy If the block's speed does not change, its kinetic energy cannot change. ANSWER: ANSWER: L Wtot Wtot = Kf − Ki L Kf − Ki = 0 Wtot = 0 Correct Part B What is , the work done on the block by the force of gravity as the block moves a distance up the incline? Express the work done by gravity in terms of the weight and any other quantities given in the problem introduction. Hint 1. Force diagram Hint 2. Force of gravity component What is the component of the force of gravity in the direction of the block's displacement (along the inclined plane)? Express your answer in terms of and . Hint 1. Relative direction of the force and the motion Remember that the force of gravity acts down the plane, whereas the block's displacement is directed up the plane. ANSWER: Wg L w w  ANSWER: Correct Part C What is , the work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of and other given quantities. Hint 1. How to find the work done by a constant force Remember that the work done on an object by a particular force is the integral of the dot product of the force and the instantaneous displacement of the object, over the path followed by the object. In this case, since the force is constant and the path is a straight segment of length up the inclined plane, the dot product becomes simple multiplication. ANSWER: Correct Part D What is , the work done on the block by the normal force as the block moves a distance up the inclined plane? Express your answer in terms of given quantities. Hint 1. First step in computing the work Fg|| = −wsin() Wg = −wLsin() WF F L F L WF = FL Wnormal L The work done by the normal force is equal to the dot product of the force vector and the block's displacement vector. The normal force and the block's displacement vector are perpendicular. Therefore, what is their dot product? ANSWER: ANSWER: Correct Problem 11.20 A particle moving along the -axis has the potential energy , where is in . Part A What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. N  L = 0 Wnormal = 0 y U = 3.2y3 J y m y y = 0 m Fy = 0 N y y = 1 m ANSWER: Correct Part C What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.28 A cable with 25.0 of tension pulls straight up on a 1.08 block that is initially at rest. Part A What is the block's speed after being lifted 2.40 ? Solve this problem using work and energy. Express your answer with the appropriate units. ANSWER: Correct Fy = -9.6 N y y = 2 m Fy = -38 N N kg m vf = 8.00 ms Problem 11.29 Part A How much work does an elevator motor do to lift a 1500 elevator a height of 110 ? Express your answer with the appropriate units. ANSWER: Correct Part B How much power must the motor supply to do this in 50 at constant speed? Express your answer with the appropriate units. ANSWER: Correct Problem 11.32 How many energy is consumed by a 1.20 hair dryer used for 10.0 and a 11.0 night light left on for 16.0 ? Part A Hair dryer: Express your answer with the appropriate units. kg m Wext = 1.62×106 J s = 3.23×104 P W kW min W hr ANSWER: Correct Part B Night light: Express your answer with the appropriate units. ANSWER: Correct Problem 11.42 A 2500 elevator accelerates upward at 1.20 for 10.0 , starting from rest. Part A How much work does gravity do on the elevator? Express your answer with the appropriate units. ANSWER: Correct W = 7.20×105 J = 6.34×105 W J kg m/s2 m −2.45×105 J Part B How much work does the tension in the elevator cable do on the elevator? Express your answer with the appropriate units. ANSWER: Correct Part C Use the work-kinetic energy theorem to find the kinetic energy of the elevator as it reaches 10.0 . Express your answer with the appropriate units. ANSWER: Correct Part D What is the speed of the elevator as it reaches 10.0 ? Express your answer with the appropriate units. ANSWER: Correct 2.75×105 J m 3.00×104 J m 4.90 ms Problem 11.47 A horizontal spring with spring constant 130 is compressed 17 and used to launch a 2.4 box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.15. Part A Use work and energy to find how far the box slides across the rough surface before stopping. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.49 Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0 and the coefficient of rolling friction is 0.45. Part A Use work and energy to find the length of a ramp that will stop a 15,000 truck that enters the ramp at 30 . Express your answer to two significant figures and include the appropriate units. ANSWER: Correct N/m cm kg l = 53 cm kg m/s l = 83 m Problem 11.51 Use work and energy to find an expression for the speed of the block in the following figure just before it hits the floor. Part A Find an expression for the speed of the block if the coefficient of kinetic friction for the block on the table is . Express your answer in terms of the variables , , , , and free fall acceleration . ANSWER: Part B Find an expression for the speed of the block if the table is frictionless. Express your answer in terms of the variables , , , and free fall acceleration . ANSWER: μk M m h μk g v = M m h g Problem 11.57 The spring shown in the figure is compressed 60 and used to launch a 100 physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the incline is 0.12 . Part A What is the student's speed just after losing contact with the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How far up the incline does the student go? Express your answer to two significant figures and include the appropriate units. ANSWER: v = cm kg 30 v = 17 ms Correct Score Summary: Your score on this assignment is 93.6%. You received 112.37 out of a possible total of 120 points. !s = 41 m

Assignment 9 Due: 11:59pm on Friday, April 11, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 11.2 Part A Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Part B Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Problem 11.4  A B = 5 − 6 A i ^ j ^ = −9 − 5 B i ^ j ^ A  B  = -15  A B = −5 + 9 A i ^ j ^ = 5 + 6 B i ^ j ^ A  B  = 29 Part A What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part B What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct ± All Work and No Play Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement If an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the force can be calculated as , where is the work done by force on the object that undergoes displacement directed at angle relative to .  A B A = 2 + 5 ı ^  ^ B = −2 − 4 ı ^  ^  = 175  A B A = −6 + 2 ı ^  ^ B = − − 3 ı ^  ^  = 90 W =  = cos  F  s  F   s  W F  s  F  Note that depending on the value of , the work done can be positive, negative, or zero. In this problem, you will practice calculating work done on an object moving in a straight line. The first series of questions is related to the accompanying figure. Part A What can be said about the sign of the work done by the force ? ANSWER: Correct When , the cosine of is zero, and therefore the work done is zero. Part B cos  F  1 It is positive. It is negative. It is zero. There is not enough information to answer the question.  = 90  What can be said about the work done by force ? ANSWER: Correct When , is positive, and so the work done is positive. Part C The work done by force is ANSWER: Correct When , is negative, and so the work done is negative. Part D The work done by force is ANSWER: F  2 It is positive. It is negative. It is zero. 0 <  < 90 cos  F  3 positive negative zero 90 <  < 180 cos  F  4 Correct Part E The work done by force is ANSWER: Correct positive negative zero F  5 positive negative zero Part F The work done by force is ANSWER: Correct Part G The work done by force is ANSWER: Correct In the next series of questions, you will use the formula to calculate the work done by various forces on an object that moves 160 meters to the right. F  6 positive negative zero F  7 positive negative zero W =  = cos  F  s  F   s  Part H Find the work done by the 18-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part I Find the work done by the 30-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part J Find the work done by the 12-newton force. Use two significant figures in your answer. Express your answer in joules. W W = 2900 J W W = 4200 J W ANSWER: Correct Part K Find the work done by the 15-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Introduction to Potential Energy Learning Goal: Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that must be added to the kinetic energy to get the total mechanical energy. The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part we introduce the concept of potential energy. But for now, please answer in terms of the work-energy theorem. Work-Energy Theorem The work-energy theorem states , where is the work done by all forces that act on the object, and and are the initial and final kinetic energies, respectively. Part A The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion. W = -1900 J W W = -1800 J Kf = Ki + Wall Wall Ki Kf Choose the best answer to fill in the blanks above: ANSWER: Correct It is important that the force have a component acting in the direction of motion. For example, if a ball is attached to a string and whirled in uniform circular motion, the string does apply a force to the ball, but since the string's force is always perpendicular to the motion it does no work and cannot change the kinetic energy of the ball. Part B To calculate the change in energy, you must know the force as a function of _______. The work done by the force causes the energy change. Choose the best answer to fill in the blank above: ANSWER: Correct Part C To illustrate the work-energy concept, consider the case of a stone falling from to under the influence of gravity. Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone. Choose the best answer to fill in the blanks above: distance / potential distance / kinetic vertical displacement / potential none of the above acceleration work distance potential energy xi xf ANSWER: Correct Potential Energy You should read about potential energy in your text before answering the following questions. Potential energy is a concept that builds on the work-energy theorem, enlarging the concept of energy in the most physically useful way. The key aspect that allows for potential energy is the existence of conservative forces, forces for which the work done on an object does not depend on the path of the object, only the initial and final positions of the object. The gravitational force is conservative; the frictional force is not. The change in potential energy is the negative of the work done by conservative forces. Hence considering the initial and final potential energies is equivalent to calculating the work done by the conservative forces. When potential energy is used, it replaces the work done by the associated conservative force. Then only the work due to nonconservative forces needs to be calculated. In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which now changes the total energy: , where and are the final and initial potential energies, and is the work due only to nonconservative forces. Now, we will revisit the falling stone example using the concept of potential energy. Part D Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy) say that the increased kinetic energy comes from the ______ of the _______ energy. Choose the best answer to fill in the blanks above: ANSWER: force / kinetic potential energy / potential force / potential potential energy / kinetic Kf + Uf = Ef = Wnc + Ei = Wnc + Ki + Ui Uf Ui Wnc Correct Part E This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potential energies, is _______. Choose the best answer to fill in the blanks above: ANSWER: Correct Problem 11.7 Part A How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: work / potential force / kinetic change / potential sum / conserved sum / zero sum / not conserved difference / conserved F  = (− + i ^ ) N j ^ ! = r m i ^ Correct Part B How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.10 A 1.8 book is lying on a 0.80- -high table. You pick it up and place it on a bookshelf 2.27 above the floor. Part A How much work does gravity do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B W = -8.6 J F  = (− + i ^ ) N j ^ ! = r m? j ^ W = 26 J kg m m Wg = -26 J How much work does your hand do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.12 The three ropes shown in the bird's-eye view of the figure are used to drag a crate 3.3 across the floor. Part A How much work is done by each of the three forces? Express your answers using two significant figures. Enter your answers numerically separated by commas. ANSWER: WH = 26 J m W1 , W2 , W3 = 1.9,1.2,-2.1 kJ Correct Enhanced EOC: Problem 11.16 A 1.2 particle moving along the x-axis experiences the force shown in the figure. The particle's velocity is 4.6 at . You may want to review ( pages 286 - 287) . For help with math skills, you may want to review: The Definite Integral Part A What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: kg m/s x = 0m x = 2m x = 0 m x = 0 m x = 2 m x = 2 m x = 2 m Correct Part B What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Can the work be negative? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: Correct Work on a Sliding Block A block of weight sits on a frictionless inclined plane, which makes an angle with respect to the horizontal, as shown. A force of magnitude , applied parallel to the incline, pulls the block up the plane at constant speed. v = 6.2 ms x = 4m x = 0 m x = 0 m x = 4 m x = 4 m x = 4 m v = 4.6 ms w  F Part A The block moves a distance up the incline. The block does not stop after moving this distance but continues to move with constant speed. What is the total work done on the block by all forces? (Include only the work done after the block has started moving, not the work needed to start the block moving from rest.) Express your answer in terms of given quantities. Hint 1. What physical principle to use To find the total work done on the block, use the work-energy theorem: . Hint 2. Find the change in kinetic energy What is the change in the kinetic energy of the block, from the moment it starts moving until it has been pulled a distance ? Remember that the block is pulled at constant speed. Hint 1. Consider kinetic energy If the block's speed does not change, its kinetic energy cannot change. ANSWER: ANSWER: L Wtot Wtot = Kf − Ki L Kf − Ki = 0 Wtot = 0 Correct Part B What is , the work done on the block by the force of gravity as the block moves a distance up the incline? Express the work done by gravity in terms of the weight and any other quantities given in the problem introduction. Hint 1. Force diagram Hint 2. Force of gravity component What is the component of the force of gravity in the direction of the block's displacement (along the inclined plane)? Express your answer in terms of and . Hint 1. Relative direction of the force and the motion Remember that the force of gravity acts down the plane, whereas the block's displacement is directed up the plane. ANSWER: Wg L w w  ANSWER: Correct Part C What is , the work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of and other given quantities. Hint 1. How to find the work done by a constant force Remember that the work done on an object by a particular force is the integral of the dot product of the force and the instantaneous displacement of the object, over the path followed by the object. In this case, since the force is constant and the path is a straight segment of length up the inclined plane, the dot product becomes simple multiplication. ANSWER: Correct Part D What is , the work done on the block by the normal force as the block moves a distance up the inclined plane? Express your answer in terms of given quantities. Hint 1. First step in computing the work Fg|| = −wsin() Wg = −wLsin() WF F L F L WF = FL Wnormal L The work done by the normal force is equal to the dot product of the force vector and the block's displacement vector. The normal force and the block's displacement vector are perpendicular. Therefore, what is their dot product? ANSWER: ANSWER: Correct Problem 11.20 A particle moving along the -axis has the potential energy , where is in . Part A What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. N  L = 0 Wnormal = 0 y U = 3.2y3 J y m y y = 0 m Fy = 0 N y y = 1 m ANSWER: Correct Part C What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.28 A cable with 25.0 of tension pulls straight up on a 1.08 block that is initially at rest. Part A What is the block's speed after being lifted 2.40 ? Solve this problem using work and energy. Express your answer with the appropriate units. ANSWER: Correct Fy = -9.6 N y y = 2 m Fy = -38 N N kg m vf = 8.00 ms Problem 11.29 Part A How much work does an elevator motor do to lift a 1500 elevator a height of 110 ? Express your answer with the appropriate units. ANSWER: Correct Part B How much power must the motor supply to do this in 50 at constant speed? Express your answer with the appropriate units. ANSWER: Correct Problem 11.32 How many energy is consumed by a 1.20 hair dryer used for 10.0 and a 11.0 night light left on for 16.0 ? Part A Hair dryer: Express your answer with the appropriate units. kg m Wext = 1.62×106 J s = 3.23×104 P W kW min W hr ANSWER: Correct Part B Night light: Express your answer with the appropriate units. ANSWER: Correct Problem 11.42 A 2500 elevator accelerates upward at 1.20 for 10.0 , starting from rest. Part A How much work does gravity do on the elevator? Express your answer with the appropriate units. ANSWER: Correct W = 7.20×105 J = 6.34×105 W J kg m/s2 m −2.45×105 J Part B How much work does the tension in the elevator cable do on the elevator? Express your answer with the appropriate units. ANSWER: Correct Part C Use the work-kinetic energy theorem to find the kinetic energy of the elevator as it reaches 10.0 . Express your answer with the appropriate units. ANSWER: Correct Part D What is the speed of the elevator as it reaches 10.0 ? Express your answer with the appropriate units. ANSWER: Correct 2.75×105 J m 3.00×104 J m 4.90 ms Problem 11.47 A horizontal spring with spring constant 130 is compressed 17 and used to launch a 2.4 box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.15. Part A Use work and energy to find how far the box slides across the rough surface before stopping. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.49 Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0 and the coefficient of rolling friction is 0.45. Part A Use work and energy to find the length of a ramp that will stop a 15,000 truck that enters the ramp at 30 . Express your answer to two significant figures and include the appropriate units. ANSWER: Correct N/m cm kg l = 53 cm kg m/s l = 83 m Problem 11.51 Use work and energy to find an expression for the speed of the block in the following figure just before it hits the floor. Part A Find an expression for the speed of the block if the coefficient of kinetic friction for the block on the table is . Express your answer in terms of the variables , , , , and free fall acceleration . ANSWER: Part B Find an expression for the speed of the block if the table is frictionless. Express your answer in terms of the variables , , , and free fall acceleration . ANSWER: μk M m h μk g v = M m h g Problem 11.57 The spring shown in the figure is compressed 60 and used to launch a 100 physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the incline is 0.12 . Part A What is the student's speed just after losing contact with the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How far up the incline does the student go? Express your answer to two significant figures and include the appropriate units. ANSWER: v = cm kg 30 v = 17 ms Correct Score Summary: Your score on this assignment is 93.6%. You received 112.37 out of a possible total of 120 points. !s = 41 m

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Chapter 9 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Momentum and Internal Forces Learning Goal: To understand the concept of total momentum for a system of objects and the effect of the internal forces on the total momentum. We begin by introducing the following terms: System: Any collection of objects, either pointlike or extended. In many momentum-related problems, you have a certain freedom in choosing the objects to be considered as your system. Making a wise choice is often a crucial step in solving the problem. Internal force: Any force interaction between two objects belonging to the chosen system. Let us stress that both interacting objects must belong to the system. External force: Any force interaction between objects at least one of which does not belong to the chosen system; in other words, at least one of the objects is external to the system. Closed system: a system that is not subject to any external forces. Total momentum: The vector sum of the individual momenta of all objects constituting the system. In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses and . To simplify the analysis, we will make several assumptions: The blocks can move in only one dimension, namely, 1. along the x axis. 2. The masses of the blocks remain constant. 3. The system is closed. At time , the x components of the velocity and the acceleration of block 1 are denoted by and . Similarly, the x components of the velocity and acceleration of block 2 are denoted by and . In this problem, you will show that the total momentum of the system is not changed by the presence of internal forces. m1 m2 t v1(t) a1 (t) v2 (t) a2 (t)

Chapter 9 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Momentum and Internal Forces Learning Goal: To understand the concept of total momentum for a system of objects and the effect of the internal forces on the total momentum. We begin by introducing the following terms: System: Any collection of objects, either pointlike or extended. In many momentum-related problems, you have a certain freedom in choosing the objects to be considered as your system. Making a wise choice is often a crucial step in solving the problem. Internal force: Any force interaction between two objects belonging to the chosen system. Let us stress that both interacting objects must belong to the system. External force: Any force interaction between objects at least one of which does not belong to the chosen system; in other words, at least one of the objects is external to the system. Closed system: a system that is not subject to any external forces. Total momentum: The vector sum of the individual momenta of all objects constituting the system. In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses and . To simplify the analysis, we will make several assumptions: The blocks can move in only one dimension, namely, 1. along the x axis. 2. The masses of the blocks remain constant. 3. The system is closed. At time , the x components of the velocity and the acceleration of block 1 are denoted by and . Similarly, the x components of the velocity and acceleration of block 2 are denoted by and . In this problem, you will show that the total momentum of the system is not changed by the presence of internal forces. m1 m2 t v1(t) a1 (t) v2 (t) a2 (t)

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Morgan Extra Pages Graphing with Excel to be carried out in a computer lab, 3rd floor Calloway Hall or elsewhere The Excel spreadsheet consists of vertical columns and horizontal rows; a column and row intersect at a cell. A cell can contain data for use in calculations of all sorts. The Name Box shows the currently selected cell (Fig. 1). In the Excel 2007 and 2010 versions the drop-down menus familiar in most software screens have been replaced by tabs with horizontally-arranged command buttons of various categories (Fig. 2) ___________________________________________________________________ Open Excel, click on the Microsoft circle, upper left, and Save As your surname. xlsx on the desktop. Before leaving the lab e-mail the file to yourself and/or save to a flash drive. Also e-mail it to your instructor. Figure 1. Parts of an Excel spreadsheet. Name Box Figure 2. Tabs. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 1: BASIC OPERATIONS Click Save often as you work. 1. Type the heading “Edge Length” in Cell A1 and double click the crack between the A and B column heading for automatic widening of column A. Similarly, write headings for columns B and C and enter numbers in Cells A2 and A3 as in Fig. 3. Highlight Cells A2 and A3 by dragging the cursor (chunky plus-shape) over the two of them and letting go. 2. Note that there are three types of cursor crosses: chunky for selecting, barbed for moving entries or blocks of entries from cell to cell, and tiny (appearing only at the little square in the lower-right corner of a cell). Obtain a tiny arrow for Cell A3 and perform a plus-drag down Column A until the cells are filled up to 40 (in Cell A8). Note that the two highlighted cells set both the starting value of the fill and the intervals. 3. Click on Cell B2 and enter a formula for face area of a cube as follows: type =, click on Cell A2, type ^2, and press Enter (note the formula bar in Fig. 4). 4. Enter the formula for cube volume in Cell C2 (same procedure, but “=, click on A2, ^3, Enter”). 5. Highlight Cells B2 and C2; plus-drag down to Row 8 (Fig. 5). Do the numbers look correct? Click on some cells in the newly filled area and notice how Excel steps the row designations as it moves down the column (it can do it for horizontal plusdrags along rows also). This is the major programming development that has led to the popularity of spreadsheets. Figure 3. Entries. Figure 4. A formula. Figure 5. Plus-dragging formulas. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 6. Now let’s graph the Face Area versus Edge Length: select Cells A1 through B8, choose the Insert tab, and click the Scatter drop-down menu and select “Scatter with only Markers” (Fig. 6). 7. Move the graph (Excel calls it a “chart”) that appears up alongside your number table and dress it up as follows: a. Note that some Chart Layouts have appeared above. Click Layout 1 and alter each title to read Face Area for the vertical axis, Edge Length for the horizontal and Face Area vs. Edge Length for the Graph Title. b. Activate the Excel Least squares routine, called “fitting a trendline” in the program: right click any of the data markers and click Add Trendline. Choose Power and also check “Display equation on chart” and “Display R-squared value on chart.” Fig. 7 shows what the graph will look like at this point. c. The titles are explicit, so the legend is unnecessary. Click on it and press the delete button to remove it. Figure 6. Creating a scatter graph. Figure 7. A graph with a fitted curve. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 8. Now let’s overlay the Volume vs. Edge Length curve onto the same graph (optional for 203L/205L): Make a copy of your graph by clicking on the outer white area, clicking ctrl-c (or right click, copy), and pasting the copy somewhere else (ctrl-v). If you wish, delete the trendline as in Fig. 8. a. Right click on the outer white space, choose Select Data and click the Add button. b. You can type in the cell ranges by hand in the dialog box that comes up, but it is easier to click the red, white, and blue button on the right of each space and highlight what you want to go in. Click the red, white, and blue of the bar that has appeared, and you will bounce back to the Add dialog box. Use the Edge Length column for the x’s and Volume for the y’s. c. Right-click on any volume data point and choose Format Data Series. Clicking Secondary Axis will place its scale on the right of the graph as in Fig. 8. d. Dress up your graph with two axis titles (Layout-Labels-Axis Titles), etc. Figure 8. Adding a second curve and y-axis to the graph Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2: INTERPRETING A LINEAR GRAPH Introduction: Many experiments are repeated a number of times with one of the parameters involved varied from run to run. Often the goal is to measure the rate of change of a dependent variable, rather than a particular value. If the dependent variable can be expressed as a linear function of the independent parameter, then the slope and yintercept of an appropriate graph will give the rate of change and a particular value, respectively. An example of such an experiment in PHYS.203L/205L is the first part of Lab 20, in which weights are added to the bottom of a suspended spring (Figure 9). This experiment shows that a spring exerts a force Fs proportional to the distance stretched y = (y-yo), a relationship known as Hooke’s Law: Fs = – k(y – yo) (Eq. 1) where k is called the Hooke’s Law constant. The minus sign shows that the spring opposes any push or pull on it. In Lab 20 Fs is equal to (- Mg) and y is given by the reading on a meter stick. Masses were added to the bottom of the spring in 50-g increments giving weights in newtons of 0.49, 0.98, etc. The weight pan was used as the pointer for reading y and had a mass of 50 g, so yo could not be directly measured. For convenient graphing Equation 1 can be rewritten: -(Mg) = – ky + kyo Or (Mg) = ky – kyo (Eq. 1′) Procedure 1. On your spreadsheet note the tabs at the bottom left and double-click Sheet1. Type in “Basics,” and then click the Sheet2 tab to bring up a fresh worksheet. Change the sheet name to “Linear Fit” and fill in data as in this table. Hooke’s Law Experiment y (m) -Fs = Mg (N) 0.337 0.49 0.388 0.98 0.446 1.47 0.498 1.96 0.550 2.45 2. Highlight the cells with the numbers, and graph (Mg) versus y as in Steps 6 and 7 of the Basics section. Your Trendline this time will be Linear of course. If you are having trouble remembering what’s versus what, “y” looks like “v”, so what comes before the “v” of “versus” goes on the y (vertical) axis. Yes, this graph is confusing: the horizontal (“x”) axis is distance y, and the “y” axis is something else. 3. Click on the Equation/R2 box on the graph and highlight just the slope, that is, only the number that comes before the “x.” Copy it (control-c is a fast way to Figure 9. A spring with a weight stretching it Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com do it) and paste it (control-v) into an empty cell. Do likewise for the intercept (including the minus sign). SAVE YOUR FILE! 5. The next steps use the standard procedure for obtaining information from linear data. Write the general equation for a straight line immediately below a hand-written copy of Equation 1′ then circle matching items: (Mg) = k y + (- k yo) (Eq. 1′) y = m x + b Note the parentheses around the intercept term of Equation 1′ to emphasize that the minus sign is part of it. Equating above and below, you can create two useful new equations: slope m = k (Eq. 2) y-intercept b = -kyo (Eq. 3) 6. Solve Equation 2 for k, that is, rewrite left to right. Then substitute the value for slope m from your graph, and you have an experimental value for the Hooke’s Law constant k. Next solve Equation 3 for yo, substitute the value for intercept b from your graph and the value of k that you just found, and calculate yo. 7. Examine your linear graph for clues to finding the units of the slope and the yintercept. Use these units to find the units of k and yo. 8. Present your values of k and yo with their units neatly at the bottom of your spreadsheet. 9. R2 in Excel, like r in our lab manual and Corr. in the LoggerPro software, is a measure of how well the calculated line matches the data points. 1.00 would indicate a perfect match. State how good a match you think was made in this case? 10. Do the Homework, Further Exercises on Interpreting Linear Graphs, on the following pages. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com Eq.1 M m f M a g               , (Eq.2) M slope m g       (Eq.3) M b f        Morgan Extra Pages Homework: Graph Interpretation Exercises EXAMPLE WITH COMPLETE SOLUTION In PHYS.203L and 205L we do Lab 9 Newton’s Second Law on Atwood’s Machine using a photogate sensor (Fig. 1). The Atwood’s apparatus can slow the rate of fall enough to be measured even with primitive timing devices. In our experiment LoggerPro software automatically collects and analyzes the data giving reliable measurements of g, the acceleration of gravity. The equation governing motion for Atwood’s Machine can be written: where a is the acceleration of the masses and string, g is the acceleration of gravity, M is the total mass at both ends of the string, m is the difference between the masses, and f is the frictional force at the hub of the pulley wheel. In this exercise you are given a graph of a vs. m obtained in this experiment with the values of M and the slope and intercept (Fig. 2). The goal is to extract values for acceleration of gravity g and frictional force f from this information. To analyze the graph we write y = mx + b, the general equation for a straight line, directly under Equation 1 and match up the various parameters: Equating above and below, you can create two new equations: and y m x b M m f M a g                Figure 1. The Atwood’s Machine setup (from the LoggerPro handout). Figure 2. Graph of acceleration versus mass difference; data from a Physics I experiment. Atwood’s Machine M = 0.400 kg a = 24.4 m – 0.018 R2 = 0.998 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 0.000 0.010 0.020 0.030 0.040 0.050 0.060  m (kg) a (m/s2) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 2 2 9.76 / 0.400 24.4 /( ) m s kg m kg s g Mm      To handle Equation 2 it pays to consider what the units of the slope are. A slope is “the rise over the run,“ so its units must be the units of the vertical axis divided by those of the horizontal axis. In this case: Now let’s solve Equation 2 for g and substitute the values of total mass M and of the slope m from the graph: Using 9.80 m/s2 as the Baltimore accepted value for g, we can calculate the percent error: A similar process with Equation 3 leads to a value for f, the frictional force at the hub of the pulley wheel. Note that the units of intercept b are simply whatever the vertical axis units are, m/s2 in this case. Solving Equation 3 for f: EXERCISE 1 The Picket Fence experiment makes use of LoggerPro software to calculate velocities at regular time intervals as the striped plate passes through the photogate (Fig. 3). The theoretical equation is v = vi + at (Eq. 4) where vi = 0 (the fence is dropped from rest) and a = g. a. Write Equation 4 with y = mx + b under it and circle matching factors as in the Example. b. What is the experimental value of the acceleration of gravity? What is its percent error from the accepted value for Baltimore, 9.80 m/s2? c. Does the value of the y-intercept make sense? d. How well did the straight Trendline match the data? 2 / 2 kg s m kg m s   0.4% 100 9.80 9.76 9.80 100 . . . %        Acc Exp Acc Error kg m s mN kg m s f Mb 7.2 10 / 7.2 0.400 ( 0.018 / ) 3 2 2           Figure 3. Graph of speed versus time as calculated by LoggerPro as a picket fence falls freely through a photogate. Picket Fence Drop y = 9.8224x + 0.0007 R2 = 0.9997 0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2 t (s) v (m/s) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2 This is an electrical example from PHYS.204L/206L, potential difference, V, versus current, I (Fig. 4). The theoretical equation is V = IR (Eq. 5) and is known as “Ohm’s Law.” The unit symbols stand for volts, V, and Amperes, A. The factor R stands for resistance and is measured in units of ohms, symbol  (capital omega). The definition of the ohm is: V (Eq. 6) By coincidence the letter symbols for potential (a quantity ) and volts (its unit) are identical. Thus “voltage” has become the laboratory slang name for potential. a. Rearrange the Ohm’s Law equation to match y = mx + b.. b. What is the experimental resistance? c. Comment on the experimental intercept: is its value reasonable? EXERCISE 3 This graph (Fig. 5) also follows Ohm’s Law, but solved for current I. For this graph the experimenter held potential difference V constant at 15.0V and measured the current for resistances of 100, 50, 40, and 30  Solve Ohm’s Law for I and you will see that 1/R is the logical variable to use on the x axis. For units, someone once jokingly referred to a “reciprocal ohm” as a “mho,” and the name stuck. a. Rearrange Equation 5 solved for I to match y = mx + b. b. What is the experimental potential difference? c. Calculate the percent difference from the 15.0 V that the experimenter set on the power supply (the instrument used for such experiments). d. Comment on the experimental intercept: is its value reasonable? Figure 4. Graph of potential difference versus current; data from a Physics II experiment. The theoretical equation, V = IR, is known as “Ohm’s Law.” Ohm’s Law y = 0.628x – 0.0275 R2 = 0.9933 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 0.5 0.6 Current, I (A) Potential difference, V (V) Figure 5. Another application of Ohm’s Law: a graph of current versus the inverse of resistance, from a different electric circuit experiment. Current versus (1/Resistance) y = 14.727x – 0.2214 R2 = 0.9938 0 100 200 300 400 500 600 5 10 15 20 25 30 35 R-1 (millimhos) I (milliamperes) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 4 The Atwood’s Machine experiment (see the solved example above) can be done in another way: keep mass difference m the same and vary the total mass M (Fig. 6). a. Rewrite Equation 1 and factor out (1/M). b. Equate the coefficient of (1/M) with the experimental slope and solve for acceleration of gravity g. c. Substitute the values for slope, mass difference, and frictional force and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? EXERCISE 5 In the previous two exercises the reciprocal of a variable was used to make the graph come out linear. In this one the trick will be to use the square root of a variable (Fig. 7). In PHYS.203L and 205L Lab 19 The Pendulum the theoretical equation is where the period T is the time per cycle, L is the length of the string, and g is the acceleration of gravity. a. Rewrite Equation 7 with the square root of L factored out and placed at the end. b. Equate the coefficient of √L with the experimental slope and solve for acceleration of gravity g. c. Substitute the value for slope and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? 2 (Eq . 7) g T   L Figure 6. Graph of acceleration versus the reciprocal of total mass; data from a another Physics I experiment. Atwood’s Machine m = 0.020 kg f = 7.2 mN y = 0.1964x – 0.0735 R2 = 0.995 0.400 0.600 0.800 1.000 2.000 2.500 3.000 3.500 4.000 4.500 5.000 1/M (1/kg) a (m/s2) Effect of Pendulum Length on Period y = 2.0523x – 0.0331 R2 = 0.999 0.400 0.800 1.200 1.600 2.000 2.400 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 L1/2 (m1/2) T (s) Figure 7. Graph of period T versus the square root of pendulum length; data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 6 In Exercise 5 another approach would have been to square both sides of Equation 7 and plot T2 versus L. Lab 20 directs us to use that alternative. It involves another case of periodic or harmonic motion with a similar, but more complicated, equation for the period: where T is the period of the bobbing (Fig. 8), M is the suspended mass, ms is the mass of the spring, k is a measure of stiffness called the spring constant, and C is a dimensionless factor showing how much of the spring mass is effectively bobbing. a. Square both sides of Equation 8 and rearrange it to match y = mx + b. b. Write y = mx + b under your rearranged equation and circle matching factors as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating k and the second for finding C from the data of Fig. 9. d. A theoretical analysis has shown that for most springs C = 1/3. Find the percent error from that value. e. Derive the units of the slope and intercept; show that the units of k come out as N/m and that C is dimensionless. 2 (Eq . 8) k T M Cm s    Figure 8. In Lab 20 mass M is suspended from a spring which is set to bobbing up and down, a good approximation to simple harmonic motion (SHM), described by Equation 8. Lab 20: SHM of a Spring Mass of the spring, ms = 25.1 g y = 3.0185x + 0.0197 R2 = 0.9965 0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 0 0.05 0.1 0.15 0.2 0.25 0.3 M (kg) T 2 2 Figure 9. Graph of the square of the period T2 versus suspended mass M data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 7 This last exercise deals with an exponential equation, and the trick is to take the logarithm of both sides. In PHYS.204L/206L we do Lab 33 The RC Time Constant with theoretical equation: where V is the potential difference at time t across a circuit element called a capacitor (the  is dropped for simplicity), Vo is V at t = 0 (try it), and  (tau) is a characteristic of the circuit called the time constant. a. Take the natural log of both sides and apply the addition rule for logarithms of a product on the right-hand side. b. Noting that the graph (Fig. 10) plots lnV versus t, arrange your equation in y = mx + b order, write y = mx + b under it, and circle the parts as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating  and the second for finding lnVo and then Vo. d. Note that the units of lnV are the natural log of volts, lnV. As usual derive the units of the slope and interecept and use them to obtain the units of your experimental V and t. V V e (Eq. 9) t o    Figure 10. Graph of a logarithm versus time; data from Lab 33, a Physics II experiment. Discharge of a Capacitor y = -9.17E-03x + 2.00E+00 R2 = 9.98E-01 0.00 0.50 1.00 1.50 2.00 2.50

Morgan Extra Pages Graphing with Excel to be carried out in a computer lab, 3rd floor Calloway Hall or elsewhere The Excel spreadsheet consists of vertical columns and horizontal rows; a column and row intersect at a cell. A cell can contain data for use in calculations of all sorts. The Name Box shows the currently selected cell (Fig. 1). In the Excel 2007 and 2010 versions the drop-down menus familiar in most software screens have been replaced by tabs with horizontally-arranged command buttons of various categories (Fig. 2) ___________________________________________________________________ Open Excel, click on the Microsoft circle, upper left, and Save As your surname. xlsx on the desktop. Before leaving the lab e-mail the file to yourself and/or save to a flash drive. Also e-mail it to your instructor. Figure 1. Parts of an Excel spreadsheet. Name Box Figure 2. Tabs. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 1: BASIC OPERATIONS Click Save often as you work. 1. Type the heading “Edge Length” in Cell A1 and double click the crack between the A and B column heading for automatic widening of column A. Similarly, write headings for columns B and C and enter numbers in Cells A2 and A3 as in Fig. 3. Highlight Cells A2 and A3 by dragging the cursor (chunky plus-shape) over the two of them and letting go. 2. Note that there are three types of cursor crosses: chunky for selecting, barbed for moving entries or blocks of entries from cell to cell, and tiny (appearing only at the little square in the lower-right corner of a cell). Obtain a tiny arrow for Cell A3 and perform a plus-drag down Column A until the cells are filled up to 40 (in Cell A8). Note that the two highlighted cells set both the starting value of the fill and the intervals. 3. Click on Cell B2 and enter a formula for face area of a cube as follows: type =, click on Cell A2, type ^2, and press Enter (note the formula bar in Fig. 4). 4. Enter the formula for cube volume in Cell C2 (same procedure, but “=, click on A2, ^3, Enter”). 5. Highlight Cells B2 and C2; plus-drag down to Row 8 (Fig. 5). Do the numbers look correct? Click on some cells in the newly filled area and notice how Excel steps the row designations as it moves down the column (it can do it for horizontal plusdrags along rows also). This is the major programming development that has led to the popularity of spreadsheets. Figure 3. Entries. Figure 4. A formula. Figure 5. Plus-dragging formulas. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 6. Now let’s graph the Face Area versus Edge Length: select Cells A1 through B8, choose the Insert tab, and click the Scatter drop-down menu and select “Scatter with only Markers” (Fig. 6). 7. Move the graph (Excel calls it a “chart”) that appears up alongside your number table and dress it up as follows: a. Note that some Chart Layouts have appeared above. Click Layout 1 and alter each title to read Face Area for the vertical axis, Edge Length for the horizontal and Face Area vs. Edge Length for the Graph Title. b. Activate the Excel Least squares routine, called “fitting a trendline” in the program: right click any of the data markers and click Add Trendline. Choose Power and also check “Display equation on chart” and “Display R-squared value on chart.” Fig. 7 shows what the graph will look like at this point. c. The titles are explicit, so the legend is unnecessary. Click on it and press the delete button to remove it. Figure 6. Creating a scatter graph. Figure 7. A graph with a fitted curve. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 8. Now let’s overlay the Volume vs. Edge Length curve onto the same graph (optional for 203L/205L): Make a copy of your graph by clicking on the outer white area, clicking ctrl-c (or right click, copy), and pasting the copy somewhere else (ctrl-v). If you wish, delete the trendline as in Fig. 8. a. Right click on the outer white space, choose Select Data and click the Add button. b. You can type in the cell ranges by hand in the dialog box that comes up, but it is easier to click the red, white, and blue button on the right of each space and highlight what you want to go in. Click the red, white, and blue of the bar that has appeared, and you will bounce back to the Add dialog box. Use the Edge Length column for the x’s and Volume for the y’s. c. Right-click on any volume data point and choose Format Data Series. Clicking Secondary Axis will place its scale on the right of the graph as in Fig. 8. d. Dress up your graph with two axis titles (Layout-Labels-Axis Titles), etc. Figure 8. Adding a second curve and y-axis to the graph Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2: INTERPRETING A LINEAR GRAPH Introduction: Many experiments are repeated a number of times with one of the parameters involved varied from run to run. Often the goal is to measure the rate of change of a dependent variable, rather than a particular value. If the dependent variable can be expressed as a linear function of the independent parameter, then the slope and yintercept of an appropriate graph will give the rate of change and a particular value, respectively. An example of such an experiment in PHYS.203L/205L is the first part of Lab 20, in which weights are added to the bottom of a suspended spring (Figure 9). This experiment shows that a spring exerts a force Fs proportional to the distance stretched y = (y-yo), a relationship known as Hooke’s Law: Fs = – k(y – yo) (Eq. 1) where k is called the Hooke’s Law constant. The minus sign shows that the spring opposes any push or pull on it. In Lab 20 Fs is equal to (- Mg) and y is given by the reading on a meter stick. Masses were added to the bottom of the spring in 50-g increments giving weights in newtons of 0.49, 0.98, etc. The weight pan was used as the pointer for reading y and had a mass of 50 g, so yo could not be directly measured. For convenient graphing Equation 1 can be rewritten: -(Mg) = – ky + kyo Or (Mg) = ky – kyo (Eq. 1′) Procedure 1. On your spreadsheet note the tabs at the bottom left and double-click Sheet1. Type in “Basics,” and then click the Sheet2 tab to bring up a fresh worksheet. Change the sheet name to “Linear Fit” and fill in data as in this table. Hooke’s Law Experiment y (m) -Fs = Mg (N) 0.337 0.49 0.388 0.98 0.446 1.47 0.498 1.96 0.550 2.45 2. Highlight the cells with the numbers, and graph (Mg) versus y as in Steps 6 and 7 of the Basics section. Your Trendline this time will be Linear of course. If you are having trouble remembering what’s versus what, “y” looks like “v”, so what comes before the “v” of “versus” goes on the y (vertical) axis. Yes, this graph is confusing: the horizontal (“x”) axis is distance y, and the “y” axis is something else. 3. Click on the Equation/R2 box on the graph and highlight just the slope, that is, only the number that comes before the “x.” Copy it (control-c is a fast way to Figure 9. A spring with a weight stretching it Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com do it) and paste it (control-v) into an empty cell. Do likewise for the intercept (including the minus sign). SAVE YOUR FILE! 5. The next steps use the standard procedure for obtaining information from linear data. Write the general equation for a straight line immediately below a hand-written copy of Equation 1′ then circle matching items: (Mg) = k y + (- k yo) (Eq. 1′) y = m x + b Note the parentheses around the intercept term of Equation 1′ to emphasize that the minus sign is part of it. Equating above and below, you can create two useful new equations: slope m = k (Eq. 2) y-intercept b = -kyo (Eq. 3) 6. Solve Equation 2 for k, that is, rewrite left to right. Then substitute the value for slope m from your graph, and you have an experimental value for the Hooke’s Law constant k. Next solve Equation 3 for yo, substitute the value for intercept b from your graph and the value of k that you just found, and calculate yo. 7. Examine your linear graph for clues to finding the units of the slope and the yintercept. Use these units to find the units of k and yo. 8. Present your values of k and yo with their units neatly at the bottom of your spreadsheet. 9. R2 in Excel, like r in our lab manual and Corr. in the LoggerPro software, is a measure of how well the calculated line matches the data points. 1.00 would indicate a perfect match. State how good a match you think was made in this case? 10. Do the Homework, Further Exercises on Interpreting Linear Graphs, on the following pages. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com Eq.1 M m f M a g               , (Eq.2) M slope m g       (Eq.3) M b f        Morgan Extra Pages Homework: Graph Interpretation Exercises EXAMPLE WITH COMPLETE SOLUTION In PHYS.203L and 205L we do Lab 9 Newton’s Second Law on Atwood’s Machine using a photogate sensor (Fig. 1). The Atwood’s apparatus can slow the rate of fall enough to be measured even with primitive timing devices. In our experiment LoggerPro software automatically collects and analyzes the data giving reliable measurements of g, the acceleration of gravity. The equation governing motion for Atwood’s Machine can be written: where a is the acceleration of the masses and string, g is the acceleration of gravity, M is the total mass at both ends of the string, m is the difference between the masses, and f is the frictional force at the hub of the pulley wheel. In this exercise you are given a graph of a vs. m obtained in this experiment with the values of M and the slope and intercept (Fig. 2). The goal is to extract values for acceleration of gravity g and frictional force f from this information. To analyze the graph we write y = mx + b, the general equation for a straight line, directly under Equation 1 and match up the various parameters: Equating above and below, you can create two new equations: and y m x b M m f M a g                Figure 1. The Atwood’s Machine setup (from the LoggerPro handout). Figure 2. Graph of acceleration versus mass difference; data from a Physics I experiment. Atwood’s Machine M = 0.400 kg a = 24.4 m – 0.018 R2 = 0.998 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 0.000 0.010 0.020 0.030 0.040 0.050 0.060  m (kg) a (m/s2) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 2 2 9.76 / 0.400 24.4 /( ) m s kg m kg s g Mm      To handle Equation 2 it pays to consider what the units of the slope are. A slope is “the rise over the run,“ so its units must be the units of the vertical axis divided by those of the horizontal axis. In this case: Now let’s solve Equation 2 for g and substitute the values of total mass M and of the slope m from the graph: Using 9.80 m/s2 as the Baltimore accepted value for g, we can calculate the percent error: A similar process with Equation 3 leads to a value for f, the frictional force at the hub of the pulley wheel. Note that the units of intercept b are simply whatever the vertical axis units are, m/s2 in this case. Solving Equation 3 for f: EXERCISE 1 The Picket Fence experiment makes use of LoggerPro software to calculate velocities at regular time intervals as the striped plate passes through the photogate (Fig. 3). The theoretical equation is v = vi + at (Eq. 4) where vi = 0 (the fence is dropped from rest) and a = g. a. Write Equation 4 with y = mx + b under it and circle matching factors as in the Example. b. What is the experimental value of the acceleration of gravity? What is its percent error from the accepted value for Baltimore, 9.80 m/s2? c. Does the value of the y-intercept make sense? d. How well did the straight Trendline match the data? 2 / 2 kg s m kg m s   0.4% 100 9.80 9.76 9.80 100 . . . %        Acc Exp Acc Error kg m s mN kg m s f Mb 7.2 10 / 7.2 0.400 ( 0.018 / ) 3 2 2           Figure 3. Graph of speed versus time as calculated by LoggerPro as a picket fence falls freely through a photogate. Picket Fence Drop y = 9.8224x + 0.0007 R2 = 0.9997 0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2 t (s) v (m/s) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2 This is an electrical example from PHYS.204L/206L, potential difference, V, versus current, I (Fig. 4). The theoretical equation is V = IR (Eq. 5) and is known as “Ohm’s Law.” The unit symbols stand for volts, V, and Amperes, A. The factor R stands for resistance and is measured in units of ohms, symbol  (capital omega). The definition of the ohm is: V (Eq. 6) By coincidence the letter symbols for potential (a quantity ) and volts (its unit) are identical. Thus “voltage” has become the laboratory slang name for potential. a. Rearrange the Ohm’s Law equation to match y = mx + b.. b. What is the experimental resistance? c. Comment on the experimental intercept: is its value reasonable? EXERCISE 3 This graph (Fig. 5) also follows Ohm’s Law, but solved for current I. For this graph the experimenter held potential difference V constant at 15.0V and measured the current for resistances of 100, 50, 40, and 30  Solve Ohm’s Law for I and you will see that 1/R is the logical variable to use on the x axis. For units, someone once jokingly referred to a “reciprocal ohm” as a “mho,” and the name stuck. a. Rearrange Equation 5 solved for I to match y = mx + b. b. What is the experimental potential difference? c. Calculate the percent difference from the 15.0 V that the experimenter set on the power supply (the instrument used for such experiments). d. Comment on the experimental intercept: is its value reasonable? Figure 4. Graph of potential difference versus current; data from a Physics II experiment. The theoretical equation, V = IR, is known as “Ohm’s Law.” Ohm’s Law y = 0.628x – 0.0275 R2 = 0.9933 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 0.5 0.6 Current, I (A) Potential difference, V (V) Figure 5. Another application of Ohm’s Law: a graph of current versus the inverse of resistance, from a different electric circuit experiment. Current versus (1/Resistance) y = 14.727x – 0.2214 R2 = 0.9938 0 100 200 300 400 500 600 5 10 15 20 25 30 35 R-1 (millimhos) I (milliamperes) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 4 The Atwood’s Machine experiment (see the solved example above) can be done in another way: keep mass difference m the same and vary the total mass M (Fig. 6). a. Rewrite Equation 1 and factor out (1/M). b. Equate the coefficient of (1/M) with the experimental slope and solve for acceleration of gravity g. c. Substitute the values for slope, mass difference, and frictional force and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? EXERCISE 5 In the previous two exercises the reciprocal of a variable was used to make the graph come out linear. In this one the trick will be to use the square root of a variable (Fig. 7). In PHYS.203L and 205L Lab 19 The Pendulum the theoretical equation is where the period T is the time per cycle, L is the length of the string, and g is the acceleration of gravity. a. Rewrite Equation 7 with the square root of L factored out and placed at the end. b. Equate the coefficient of √L with the experimental slope and solve for acceleration of gravity g. c. Substitute the value for slope and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? 2 (Eq . 7) g T   L Figure 6. Graph of acceleration versus the reciprocal of total mass; data from a another Physics I experiment. Atwood’s Machine m = 0.020 kg f = 7.2 mN y = 0.1964x – 0.0735 R2 = 0.995 0.400 0.600 0.800 1.000 2.000 2.500 3.000 3.500 4.000 4.500 5.000 1/M (1/kg) a (m/s2) Effect of Pendulum Length on Period y = 2.0523x – 0.0331 R2 = 0.999 0.400 0.800 1.200 1.600 2.000 2.400 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 L1/2 (m1/2) T (s) Figure 7. Graph of period T versus the square root of pendulum length; data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 6 In Exercise 5 another approach would have been to square both sides of Equation 7 and plot T2 versus L. Lab 20 directs us to use that alternative. It involves another case of periodic or harmonic motion with a similar, but more complicated, equation for the period: where T is the period of the bobbing (Fig. 8), M is the suspended mass, ms is the mass of the spring, k is a measure of stiffness called the spring constant, and C is a dimensionless factor showing how much of the spring mass is effectively bobbing. a. Square both sides of Equation 8 and rearrange it to match y = mx + b. b. Write y = mx + b under your rearranged equation and circle matching factors as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating k and the second for finding C from the data of Fig. 9. d. A theoretical analysis has shown that for most springs C = 1/3. Find the percent error from that value. e. Derive the units of the slope and intercept; show that the units of k come out as N/m and that C is dimensionless. 2 (Eq . 8) k T M Cm s    Figure 8. In Lab 20 mass M is suspended from a spring which is set to bobbing up and down, a good approximation to simple harmonic motion (SHM), described by Equation 8. Lab 20: SHM of a Spring Mass of the spring, ms = 25.1 g y = 3.0185x + 0.0197 R2 = 0.9965 0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 0 0.05 0.1 0.15 0.2 0.25 0.3 M (kg) T 2 2 Figure 9. Graph of the square of the period T2 versus suspended mass M data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 7 This last exercise deals with an exponential equation, and the trick is to take the logarithm of both sides. In PHYS.204L/206L we do Lab 33 The RC Time Constant with theoretical equation: where V is the potential difference at time t across a circuit element called a capacitor (the  is dropped for simplicity), Vo is V at t = 0 (try it), and  (tau) is a characteristic of the circuit called the time constant. a. Take the natural log of both sides and apply the addition rule for logarithms of a product on the right-hand side. b. Noting that the graph (Fig. 10) plots lnV versus t, arrange your equation in y = mx + b order, write y = mx + b under it, and circle the parts as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating  and the second for finding lnVo and then Vo. d. Note that the units of lnV are the natural log of volts, lnV. As usual derive the units of the slope and interecept and use them to obtain the units of your experimental V and t. V V e (Eq. 9) t o    Figure 10. Graph of a logarithm versus time; data from Lab 33, a Physics II experiment. Discharge of a Capacitor y = -9.17E-03x + 2.00E+00 R2 = 9.98E-01 0.00 0.50 1.00 1.50 2.00 2.50

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