A dipole consisting of two oppositely charged balls connected by a wooden stick is located as shown in the diagram at right. A block of plastic is located nearby, as shown. LocationsB, C, and Dall lie on a line perpendicular to the axis of the dipole, passing through the midpoint of the dipole. Which of the following statements is correct? A. The polarization of a molecule located at D would be the same as the polarization of a molecule located at C B. A molecule located at D would be polarized less than a molecule located at C C. A molecule located at C would be not be polarized at all D. A molecule located at D would be polarized more than a molecule located at C Correct + E. Information given is insufficient to fully answer the question

A dipole consisting of two oppositely charged balls connected by a wooden stick is located as shown in the diagram at right. A block of plastic is located nearby, as shown. LocationsB, C, and Dall lie on a line perpendicular to the axis of the dipole, passing through the midpoint of the dipole. Which of the following statements is correct? A. The polarization of a molecule located at D would be the same as the polarization of a molecule located at C B. A molecule located at D would be polarized less than a molecule located at C C. A molecule located at C would be not be polarized at all D. A molecule located at D would be polarized more than a molecule located at C Correct + E. Information given is insufficient to fully answer the question

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MIS 3000 – Introduction to Management Information Systems Excel Tutorial #5 (Spring 2014) Healthy Cooking Grading Criteria Do “Case Problem 3” on pages EX 322-323 of your Excel book. See details below. When you are done, turn in your Excel spreadsheet: Item Points Download the Cooking workbook (available on Pilot). Save your Workbook as HealthyCookingXXX (where XXX are your initials). Complete the Documentation Sheet: 1 – Fill in the Documentation Sheet, as directed Correct the Order Amount Filter Worksheet: ……..………………………………………………………………………….. 2 – Correct the errors in the filter (Step 3) Create the Sort Worksheet: 3 – Create a custom sort on the specified criteria – Use conditional formatting as directed in Step 8 Correct the Customer Type Subtotal Worksheet: 5 – Correct the Customer Type Subtotal Worksheet (step 11) – Complete step 12 to insert a count of orders for customer type Create the Pivot Table per Step 13: 4 – Use a slicer to filter the Pivot Table per Step 14 – Format the slicer to match the Pivot Table style Create the Pivot Chart as directed in Step 16: ……..…………………………………………………………………………..4 – Move the Pivot Chart to row 3 – Change the Chart Title as directed – Change the axis and fill colors as directed

MIS 3000 – Introduction to Management Information Systems Excel Tutorial #5 (Spring 2014) Healthy Cooking Grading Criteria Do “Case Problem 3” on pages EX 322-323 of your Excel book. See details below. When you are done, turn in your Excel spreadsheet: Item Points Download the Cooking workbook (available on Pilot). Save your Workbook as HealthyCookingXXX (where XXX are your initials). Complete the Documentation Sheet: 1 – Fill in the Documentation Sheet, as directed Correct the Order Amount Filter Worksheet: ……..………………………………………………………………………….. 2 – Correct the errors in the filter (Step 3) Create the Sort Worksheet: 3 – Create a custom sort on the specified criteria – Use conditional formatting as directed in Step 8 Correct the Customer Type Subtotal Worksheet: 5 – Correct the Customer Type Subtotal Worksheet (step 11) – Complete step 12 to insert a count of orders for customer type Create the Pivot Table per Step 13: 4 – Use a slicer to filter the Pivot Table per Step 14 – Format the slicer to match the Pivot Table style Create the Pivot Chart as directed in Step 16: ……..…………………………………………………………………………..4 – Move the Pivot Chart to row 3 – Change the Chart Title as directed – Change the axis and fill colors as directed

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PHSX 220 Homework 12 D2L – Due Thursday April 13 – 5:00 pm Exam 3 MC Review Problem 1: A 1.0-kg with a velocity of 2.0m/s perpendicular towards a wall rebounds from the wall at 1.5m/s perpendicularlly away from the wall. The change in the momentum of the ball is: A. zero B. 0.5 N s away from wall C. 0.5 N s toward wall D. 3.5 N s away from wall E. 3.5 N s toward wall Problem 2: A 64 kg man stands on a frictionless surface with a 0.10 kg stone at his feet. Both the man and the person are initially at rest. He kicks the stone with his foot so that his end velocity is 0.0017m/s in the forward direction. The velocity of the stone is now: A. 1.1m/s forward B. 1.1m/s backward C. 0.0017m/s forward D. 0.0017m/s backward E. none of these Problem 3: A 2-kg cart, traveling on a rctionless surface with a speed of 3m/s, collides with a stationary 4-kg cart. The carts then stick together. Calculate the magnitude of the impulse exerted by one cart on the other: A. 0 B. 4N s C. 6N s D. 9N s E. 12N s Problem 4: A disc has an initial angular velocity of 18 radians per second. It has a constant angular acceleration of 2.0 radians per second every second and is slowing at rst. How much time elapses before its angular velocity is 18 rad/s in the direction opposite to its initial angular velocity? A. 3.0 s B. 6.0 s C. 9.0 s D. 18 s E. 36 s Problem 5: Three point masses of M, 2M, and 3M, are fastened to a massless rod of length L as shown. The rotational inertia about the rotational axis shown is: A. (ML2=2) B. (ML2) C. (3ML2)=2 D. (6ML2) E. (3ML2)=4 Problem 6: A board is allowed to pivot about its center. A 5-N force is applied 2m from the pivot and another 5-N force is applied 4m from the pivot. These forces are applied at the angles shown in the gure. The magnitude of the net torque about the pivot is: A. 0 Nm B. 5 Nm C. 8.7 Nm D. 15 Nm E. 26 Nm Problem 7: A solid disk (r=0.03 m) and a rotational inertia of 4:5×10􀀀3kgm2 hangs from the ceiling. A string passes over it with a 2.0-kg block and a 4.0-kg block hanging on either end of the string and does not slip as the system starts to move. When the speed of the 4 kg block is 2.0m/s the kinetic energy of the pulley is: A. 0.15 J B. 0.30 J C. 1.0J D. 10 J E. 20 J Problem 8: A merry go round (r= 3.0m, I =600 kgm2) is initially spinning with an angular velocity of 0.80 radians per second when a 20 kg point mass moves from the center to the rim. Calculate the nal angular velocity of the system: A. 0.62 rad/s B. 0.73 rad/s C. 0.80 rad/s D. 0.89 rad/s E. 1.1 rad/s

PHSX 220 Homework 12 D2L – Due Thursday April 13 – 5:00 pm Exam 3 MC Review Problem 1: A 1.0-kg with a velocity of 2.0m/s perpendicular towards a wall rebounds from the wall at 1.5m/s perpendicularlly away from the wall. The change in the momentum of the ball is: A. zero B. 0.5 N s away from wall C. 0.5 N s toward wall D. 3.5 N s away from wall E. 3.5 N s toward wall Problem 2: A 64 kg man stands on a frictionless surface with a 0.10 kg stone at his feet. Both the man and the person are initially at rest. He kicks the stone with his foot so that his end velocity is 0.0017m/s in the forward direction. The velocity of the stone is now: A. 1.1m/s forward B. 1.1m/s backward C. 0.0017m/s forward D. 0.0017m/s backward E. none of these Problem 3: A 2-kg cart, traveling on a rctionless surface with a speed of 3m/s, collides with a stationary 4-kg cart. The carts then stick together. Calculate the magnitude of the impulse exerted by one cart on the other: A. 0 B. 4N s C. 6N s D. 9N s E. 12N s Problem 4: A disc has an initial angular velocity of 18 radians per second. It has a constant angular acceleration of 2.0 radians per second every second and is slowing at rst. How much time elapses before its angular velocity is 18 rad/s in the direction opposite to its initial angular velocity? A. 3.0 s B. 6.0 s C. 9.0 s D. 18 s E. 36 s Problem 5: Three point masses of M, 2M, and 3M, are fastened to a massless rod of length L as shown. The rotational inertia about the rotational axis shown is: A. (ML2=2) B. (ML2) C. (3ML2)=2 D. (6ML2) E. (3ML2)=4 Problem 6: A board is allowed to pivot about its center. A 5-N force is applied 2m from the pivot and another 5-N force is applied 4m from the pivot. These forces are applied at the angles shown in the gure. The magnitude of the net torque about the pivot is: A. 0 Nm B. 5 Nm C. 8.7 Nm D. 15 Nm E. 26 Nm Problem 7: A solid disk (r=0.03 m) and a rotational inertia of 4:5×10􀀀3kgm2 hangs from the ceiling. A string passes over it with a 2.0-kg block and a 4.0-kg block hanging on either end of the string and does not slip as the system starts to move. When the speed of the 4 kg block is 2.0m/s the kinetic energy of the pulley is: A. 0.15 J B. 0.30 J C. 1.0J D. 10 J E. 20 J Problem 8: A merry go round (r= 3.0m, I =600 kgm2) is initially spinning with an angular velocity of 0.80 radians per second when a 20 kg point mass moves from the center to the rim. Calculate the nal angular velocity of the system: A. 0.62 rad/s B. 0.73 rad/s C. 0.80 rad/s D. 0.89 rad/s E. 1.1 rad/s

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Assignment 5 Due: 11:59pm on Wednesday, March 5, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 6.13 A hand presses down on the book in the figure. Part A Is the normal force of the table on the book larger than, smaller than, or equal to ? ANSWER: Correct mg Equal to Larger than Smaller than mg mg mg Problem 6.2 The three ropes in the figure are tied to a small, very light ring. Two of these ropes are anchored to walls at right angles with the tensions shown in the figure. Part A What is the magnitude of the tension in the third rope? Express your answer using two significant figures. ANSWER: Correct Part B What is the direction of the tension in the third rope? Express your answer using two significant figures. T  3 T3 = 94 N T  3 Typesetting math: 100% ANSWER: Correct The Normal Force When an object rests on a surface, there is always a force perpendicular to the surface; we call this the normal force, denoted by . The two questions to the right will explore the normal force. Part A A man attempts to pick up his suitcase of weight by pulling straight up on the handle. However, he is unable to lift the suitcase from the floor. Which statement about the magnitude of the normal force acting on the suitcase is true during the time that the man pulls upward on the suitcase? Hint 1. How to approach this problem First, identify the forces that act on the suitcase and draw a free-body diagram. Then use the fact that the suitcase is in equilibrium, , to examine how the forces acting on the suitcase relate to each other. Hint 2. Identify the correct free-body diagram Which of the figures represents the free-body diagram of the suitcase while the man is pulling on the handle with a force of magnitude ? = 58   below horizontal n ws n F = 0 fpull Typesetting math: 100% ANSWER: ANSWER: Correct Part B A B C D The magnitude of the normal force is equal to the magnitude of the weight of the suitcase. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull. The magnitude of the normal force is equal to the sum of the magnitude of the force of the pull and the magnitude of the suitcase’s weight. The magnitude of the normal force is greater than the magnitude of the weight of the suitcase. Typesetting math: 100% Now assume that the man of weight is tired and decides to sit on his suitcase. Which statement about the magnitude of the normal force acting on the suitcase is true during the time that the man is sitting on the suitcase? Hint 1. Identify the correct free-body diagram. Which of the figures represents the free-body diagram while the man is sitting atop the suitcase? Here the vector labeled is a force that has the same magnitude as the man’s weight. ANSWER: wm n wm Typesetting math: 100% ANSWER: Correct Recognize that the normal force acting on an object is not always equal to the weight of that object. This is an important point to understand. Problem 6.5 A construction worker with a weight of 880 stands on a roof that is sloped at 18 . Part A What is the magnitude of the normal force of the roof on the worker? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct A B C D The magnitude of the normal force is equal to the magnitude of the suitcase’s weight. The magnitude of the normal force is equal to the magnitude of the suitcase’s weight minus the magnitude of the man’s weight. The magnitude of the normal force is equal to the sum of the magnitude of the man’s weight and the magnitude of the suitcase’s weight. The magnitude of the normal force is less than the magnitude of the suitcase’s weight. N  n = 840 N Typesetting math: 100% Problem 6.6 In each of the two free-body diagrams, the forces are acting on a 3.0 object. Part A For diagram , find the value of , the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B For diagram the part A, find the value of the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: kg ax x ax = -0.67 m s2 ay, y Typesetting math: 100% Correct Part C For diagram , find the value of , the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D For diagram the part C, find the value of , the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: ay = 0 m s2 ax x ax = 0.67 m s2 ay y Typesetting math: 100% Correct Problem 6.7 In each of the two free-body diagrams, the forces are acting on a 3.0 object. Part A Find the value of , the component of the acceleration in diagram (a). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct ay = 0 m s2 kg ax x ax = 0.99 m s2 Typesetting math: 100% Part B Find the value of , the component of the acceleration in diagram (a). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C Find the value of , the component of the acceleration in diagram (b). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D Find the value of , the component of the acceleration in diagram (b). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct ay y ay = 0 m s2 ax x ax = -0.18 m s2 ay y ay = 0 m s2 Typesetting math: 100% Problem 6.10 A horizontal rope is tied to a 53.0 box on frictionless ice. What is the tension in the rope if: Part A The box is at rest? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part B The box moves at a steady = 4.80 ? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part C The box = 4.80 and = 4.60 ? Express your answer to three significant figures and include the appropriate units. ANSWER: kg T = 0 N vx m/s T = 0 N vx m/s ax m/s2 Typesetting math: 100% Correct Problem 6.14 It takes the elevator in a skyscraper 4.5 to reach its cruising speed of 11 . A 60 passenger gets aboard on the ground floor. Part A What is the passenger’s weight before the elevator starts moving? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the passenger’s weight while the elevator is speeding up? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the passenger’s weight after the elevator reaches its cruising speed? T = 244 N s m/s kg w = 590 N w = 730 N Typesetting math: 100% Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Block on an Incline A block lies on a plane raised an angle from the horizontal. Three forces act upon the block: , the force of gravity; , the normal force; and , the force of friction. The coefficient of friction is large enough to prevent the block from sliding . Part A Consider coordinate system a, with the x axis along the plane. Which forces lie along the axes? ANSWER: w = 590 N  F  w F n F  f Typesetting math: 100% Correct Part B Which forces lie along the axes of the coordinate system b, in which the y axis is vertical? ANSWER: Correct only only only and and and and and F  f F  n F  w F  f F  n F  f F  w F  n F w F  f F  n F w only only only and and and and and F  f F  n F  w F  f F  n F  f F  w F  n F w F  f F  n F w Typesetting math: 100% Usually the best advice is to choose coordinate system so that the acceleration of the system is directly along one of the coordinate axes. If the system isn’t accelerating, then you are better off choosing the coordinate system with the most vectors along the coordinate axes. But now you are going to ignore that advice. You will find the normal force, , using vertical coordinate system b. In these coordinates you will find the magnitude appearing in both the x and y equations, each multiplied by a trigonometric function. Part C Because the block is not moving, the sum of the y components of the forces acting on the block must be zero. Find an expression for the sum of the y components of the forces acting on the block, using coordinate system b. Express your answer in terms of some or all of the variables , , , and . Hint 1. Find the y component of Write an expression for , the y component of the force , using coordinate system b. Express your answer in terms of and . Hint 1. Some geometry help – a useful angle The smaller angle between and the y-axis is also , as shown in the figure. ANSWER: F  n Fn Fn Ff Fw  F n Fny F  n Fn  F  n  Typesetting math: 100% Hint 2. Find the y component of Write an expression for , the y component of the force , using coordinate system b. Express your answer in terms of and . Hint 1. Some geometry help – a useful angle The smaller angle between and the x-axis is also , as shown in the figure. ANSWER: ANSWER: Fny = Fncos() F f Ffy F f Ff  F  f  Ffy = Ffsin() Fy = 0 = Fncos() + Ffsin() − Fw Typesetting math: 100% Correct Part D Because the block is not moving, the sum of the x components of the forces acting on the block must be zero. Find an expression for the sum of the x components of the forces acting on the block, using coordinate system b. Express your answer in terms of some or all of the variables , , , and . Hint 1. Find the x component of Write an expression for , the x component of the force , using coordinate system b. Express your answer in terms of and . ANSWER: ANSWER: Correct Part E To find the magnitude of the normal force, you must express in terms of since is an unknown. Using the equations you found in the two previous parts, find an expression for involving and but not . Hint 1. How to approach the problem From your answers to the previous two parts you should have two force equations ( and ). Combine these equations to eliminate . The key is to multiply the Fn Ff Fw  F n Fnx F  n Fn  Fnx = −Fnsin() Fx = 0 = −Fnsin() + Ffcos() Fn Fw Ff Fn Fw  Ff Typesetting math: 100% Fy = 0 Fx = 0 Ff equation for the y components by and the equation for the x components by , then add or subtract the two equations to eliminate the term . An alternative motivation for the algebra is to eliminate the trig functions in front of by using the trig identity . At the very least this would result in an equation that is simple to solve for . ANSWER: Correct Congratulations on working this through. Now realize that in coordinate system a, which is aligned with the plane, the y-coordinate equation is , which leads immediately to the result obtained here for . CONCLUSION: A thoughtful examination of which coordinate system to choose can save a lot of algebra. Contact Forces Introduced Learning Goal: To introduce contact forces (normal and friction forces) and to understand that, except for friction forces under certain circumstances, these forces must be determined from: net Force = ma. Two solid objects cannot occupy the same space at the same time. Indeed, when the objects touch, they exert repulsive normal forces on each other, as well as frictional forces that resist their slipping relative to each other. These contact forces arise from a complex interplay between the electrostatic forces between the electrons and ions in the objects and the laws of quantum mechanics. As two surfaces are pushed together these forces increase exponentially over an atomic distance scale, easily becoming strong enough to distort the bulk material in the objects if they approach too close. In everyday experience, contact forces are limited by the deformation or acceleration of the objects, rather than by the fundamental interatomic forces. Hence, we can conclude the following: The magnitude of contact forces is determined by , that is, by the other forces on, and acceleration of, the contacting bodies. The only exception is that the frictional forces cannot exceed (although they can be smaller than this or even zero). Normal and friction forces Two types of contact forces operate in typical mechanics problems, the normal and frictional forces, usually designated by and (or , or something similar) respectively. These are the components of the overall contact force: perpendicular to and parallel to the plane of contact. Kinetic friction when surfaces slide cos  sin  Ff cos() sin() Fn sin2() + cos2 () = 1 Fn Fn = Fwcos() Fy = Fn − FW cos() = 0 Fn F = ma μn n f Ffric n f Typesetting math: 100% When one surface is sliding past the other, experiments show three things about the friction force (denoted ): The frictional force opposes the relative motion at the 1. point of contact, 2. is proportional to the normal force, and 3. the ratio of the magnitude of the frictional force to that of the normal force is fairly constant over a wide range of speeds. The constant of proportionality is called the coefficient of kinetic friction, often designated . As long as the sliding continues, the frictional force is then (valid when the surfaces slide by each other). Static friction when surfaces don’t slide When there is no relative motion of the surfaces, the frictional force can assume any value from zero up to a maximum , where is the coefficient of static friction. Invariably, is larger than , in agreement with the observation that when a force is large enough that something breaks loose and starts to slide, it often accelerates. The frictional force for surfaces with no relative motion is therefore (valid when the contacting surfaces have no relative motion). The actual magnitude and direction of the static friction force are such that it (together with other forces on the object) causes the object to remain motionless with respect to the contacting surface as long as the static friction force required does not exceed . The equation is valid only when the surfaces are on the verge of sliding. Part A When two objects slide by one another, which of the following statements about the force of friction between them, is true? ANSWER: Correct Part B fk fk μk fk = μkn μsn μs μs μk fs ! μsn μsn fs = μsn The frictional force is always equal to . The frictional force is always less than . The frictional force is determined by other forces on the objects so it can be either equal to or less than . μkn μkn μkn Typesetting math: 100% When two objects are in contact with no relative motion, which of the following statements about the frictional force between them, is true? ANSWER: Correct For static friction, the actual magnitude and direction of the friction force are such that it, together with any other forces present, will cause the object to have the observed acceleration. The magnitude of the force cannot exceed . If the magnitude of static friction needed to keep acceleration equal to zero exceeds , then the object will slide subject to the resistance of kinetic friction. Do not automatically assume that unless you are considering a situation in which the magnitude of the static friction force is as large as possible (i.e., when determining at what point an object will just begin to slip). Whether the actual magnitude of the friction force is 0, less than , or equal to depends on the magnitude of the other forces (if any) as well as the acceleration of the object through . Part C When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some point “break loose” and start to accelerate down the board. The box begins to slide once the component of gravity acting parallel to the board just begins to exceeds the maximum force of static friction. Which of the following is the most general explanation for why the box accelerates down the board? ANSWER: The frictional force is always equal to . The frictional force is always less than . The frictional force is determined by other forces on the objects so it can be either equal to or less than . μsn μsn μsn μsn μsn fs = μsn μsn μsn F = ma Fg The force of kinetic friction is smaller than that of maximum static friction, but remains the same. Once the box is moving, is smaller than the force of maximum static friction but larger than the force of kinetic friction. Once the box is moving, is larger than the force of maximum static friction. When the box is stationary, equals the force of static friction, but once the box starts moving, the sliding reduces the normal force, which in turn reduces the friction. Fg Fg Fg Fg Typesetting math: 100% Correct At the point when the box finally does “break loose,” you know that the component of the box’s weight that is parallel to the board just exceeds (i.e., this component of gravitational force on the box has just reached a magnitude such that the force of static friction, which has a maximum value of , can no longer oppose it.) For the box to then accelerate, there must be a net force on the box along the board. Thus, the component of the box’s weight parallel to the board must be greater than the force of kinetic friction. Therefore the force of kinetic friction must be less than the force of static friction which implies , as expected. Part D Consider a problem in which a car of mass is on a road tilted at an angle . The normal force Select the best answer. ANSWER: Correct The key point is that contact forces must be determined from Newton’s equation. In the problem described above, there is not enough information given to determine the normal force (e.g., the acceleration is unknown). Each of the answer options is valid under some conditions ( , the car is sliding down an icy incline, or the car is going around a banked turn), but in fact none is likely to be correct if there are other forces on the car or if the car is accelerating. Do not memorize values for the normal force valid in different problems–you must determine from . Problem 6.17 Bonnie and Clyde are sliding a 323 bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 375 of force while Bonnie pulls forward on a rope with 335 of force. μsn μsn μkn μsn μk < μs M  is found using n = Mg n = Mg cos() n = Mg cos() F  = Ma  = 0 n F = ma kg N N Typesetting math: 100% Part A What is the safe's coefficient of kinetic friction on the bank floor? ANSWER: Correct Problem 6.19 A crate is placed on a horizontal conveyor belt. The materials are such that and . Part A Draw a free-body diagram showing all the forces on the crate if the conveyer belt runs at constant speed. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: 0.224 10 kg μs = 0.5 μk = 0.3 Typesetting math: 100% Correct Part B Draw a free-body diagram showing all the forces on the crate if the conveyer belt is speeding up. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Typesetting math: 100% Correct Part C What is the maximum acceleration the belt can have without the crate slipping? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct amax = 4.9 m s2 Typesetting math: 100% Problem 6.28 A 1100 steel beam is supported by two ropes. Part A What is the tension in rope 1? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the tension in rope 2? Express your answer to two significant figures and include the appropriate units. ANSWER: kg T1 = 7000 N Typesetting math: 100% Correct Problem 6.35 The position of a 1.4 mass is given by , where is in seconds. Part A What is the net horizontal force on the mass at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the net horizontal force on the mass at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 6.39 T2 = 4800 N kg x = (2t3 − 3t2 )m t t = 0 s F = -8.4 N t = 1 s F = 8.4 N Typesetting math: 100% A rifle with a barrel length of 61 fires a 8 bullet with a horizontal speed of 400 . The bullet strikes a block of wood and penetrates to a depth of 11 . Part A What resistive force (assumed to be constant) does the wood exert on the bullet? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How long does it take the bullet to come to rest after entering the wood? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 6.45 You and your friend Peter are putting new shingles on a roof pitched at 21 . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.0 away, asks you for the box of nails. Rather than carry the 2.0 box of nails down to Peter, you decide to give the box a push and have it slide down to him. Part A If the coefficient of kinetic friction between the box and the roof is 0.55, with what speed should you push the box to have it gently come to rest right at the edge of the roof? Express your answer to two significant figures and include the appropriate units. cm g m/s cm fk = 5800 N = 5.5×10−4 t s  m kg Typesetting math: 100% ANSWER: Correct Problem 6.54 The 2.0 wood box in the figure slides down a vertical wood wall while you push on it at a 45 angle. Part A What magnitude of force should you apply to cause the box to slide down at a constant speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct v = 3.9 ms kg  F = 23 N Typesetting math: 100% Score Summary: Your score on this assignment is 98.8%. You received 114.57 out of a possible total of 116 points. Typesetting math: 100%

Assignment 5 Due: 11:59pm on Wednesday, March 5, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 6.13 A hand presses down on the book in the figure. Part A Is the normal force of the table on the book larger than, smaller than, or equal to ? ANSWER: Correct mg Equal to Larger than Smaller than mg mg mg Problem 6.2 The three ropes in the figure are tied to a small, very light ring. Two of these ropes are anchored to walls at right angles with the tensions shown in the figure. Part A What is the magnitude of the tension in the third rope? Express your answer using two significant figures. ANSWER: Correct Part B What is the direction of the tension in the third rope? Express your answer using two significant figures. T  3 T3 = 94 N T  3 Typesetting math: 100% ANSWER: Correct The Normal Force When an object rests on a surface, there is always a force perpendicular to the surface; we call this the normal force, denoted by . The two questions to the right will explore the normal force. Part A A man attempts to pick up his suitcase of weight by pulling straight up on the handle. However, he is unable to lift the suitcase from the floor. Which statement about the magnitude of the normal force acting on the suitcase is true during the time that the man pulls upward on the suitcase? Hint 1. How to approach this problem First, identify the forces that act on the suitcase and draw a free-body diagram. Then use the fact that the suitcase is in equilibrium, , to examine how the forces acting on the suitcase relate to each other. Hint 2. Identify the correct free-body diagram Which of the figures represents the free-body diagram of the suitcase while the man is pulling on the handle with a force of magnitude ? = 58   below horizontal n ws n F = 0 fpull Typesetting math: 100% ANSWER: ANSWER: Correct Part B A B C D The magnitude of the normal force is equal to the magnitude of the weight of the suitcase. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull. The magnitude of the normal force is equal to the sum of the magnitude of the force of the pull and the magnitude of the suitcase’s weight. The magnitude of the normal force is greater than the magnitude of the weight of the suitcase. Typesetting math: 100% Now assume that the man of weight is tired and decides to sit on his suitcase. Which statement about the magnitude of the normal force acting on the suitcase is true during the time that the man is sitting on the suitcase? Hint 1. Identify the correct free-body diagram. Which of the figures represents the free-body diagram while the man is sitting atop the suitcase? Here the vector labeled is a force that has the same magnitude as the man’s weight. ANSWER: wm n wm Typesetting math: 100% ANSWER: Correct Recognize that the normal force acting on an object is not always equal to the weight of that object. This is an important point to understand. Problem 6.5 A construction worker with a weight of 880 stands on a roof that is sloped at 18 . Part A What is the magnitude of the normal force of the roof on the worker? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct A B C D The magnitude of the normal force is equal to the magnitude of the suitcase’s weight. The magnitude of the normal force is equal to the magnitude of the suitcase’s weight minus the magnitude of the man’s weight. The magnitude of the normal force is equal to the sum of the magnitude of the man’s weight and the magnitude of the suitcase’s weight. The magnitude of the normal force is less than the magnitude of the suitcase’s weight. N  n = 840 N Typesetting math: 100% Problem 6.6 In each of the two free-body diagrams, the forces are acting on a 3.0 object. Part A For diagram , find the value of , the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B For diagram the part A, find the value of the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: kg ax x ax = -0.67 m s2 ay, y Typesetting math: 100% Correct Part C For diagram , find the value of , the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D For diagram the part C, find the value of , the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: ay = 0 m s2 ax x ax = 0.67 m s2 ay y Typesetting math: 100% Correct Problem 6.7 In each of the two free-body diagrams, the forces are acting on a 3.0 object. Part A Find the value of , the component of the acceleration in diagram (a). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct ay = 0 m s2 kg ax x ax = 0.99 m s2 Typesetting math: 100% Part B Find the value of , the component of the acceleration in diagram (a). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C Find the value of , the component of the acceleration in diagram (b). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D Find the value of , the component of the acceleration in diagram (b). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct ay y ay = 0 m s2 ax x ax = -0.18 m s2 ay y ay = 0 m s2 Typesetting math: 100% Problem 6.10 A horizontal rope is tied to a 53.0 box on frictionless ice. What is the tension in the rope if: Part A The box is at rest? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part B The box moves at a steady = 4.80 ? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part C The box = 4.80 and = 4.60 ? Express your answer to three significant figures and include the appropriate units. ANSWER: kg T = 0 N vx m/s T = 0 N vx m/s ax m/s2 Typesetting math: 100% Correct Problem 6.14 It takes the elevator in a skyscraper 4.5 to reach its cruising speed of 11 . A 60 passenger gets aboard on the ground floor. Part A What is the passenger’s weight before the elevator starts moving? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the passenger’s weight while the elevator is speeding up? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the passenger’s weight after the elevator reaches its cruising speed? T = 244 N s m/s kg w = 590 N w = 730 N Typesetting math: 100% Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Block on an Incline A block lies on a plane raised an angle from the horizontal. Three forces act upon the block: , the force of gravity; , the normal force; and , the force of friction. The coefficient of friction is large enough to prevent the block from sliding . Part A Consider coordinate system a, with the x axis along the plane. Which forces lie along the axes? ANSWER: w = 590 N  F  w F n F  f Typesetting math: 100% Correct Part B Which forces lie along the axes of the coordinate system b, in which the y axis is vertical? ANSWER: Correct only only only and and and and and F  f F  n F  w F  f F  n F  f F  w F  n F w F  f F  n F w only only only and and and and and F  f F  n F  w F  f F  n F  f F  w F  n F w F  f F  n F w Typesetting math: 100% Usually the best advice is to choose coordinate system so that the acceleration of the system is directly along one of the coordinate axes. If the system isn’t accelerating, then you are better off choosing the coordinate system with the most vectors along the coordinate axes. But now you are going to ignore that advice. You will find the normal force, , using vertical coordinate system b. In these coordinates you will find the magnitude appearing in both the x and y equations, each multiplied by a trigonometric function. Part C Because the block is not moving, the sum of the y components of the forces acting on the block must be zero. Find an expression for the sum of the y components of the forces acting on the block, using coordinate system b. Express your answer in terms of some or all of the variables , , , and . Hint 1. Find the y component of Write an expression for , the y component of the force , using coordinate system b. Express your answer in terms of and . Hint 1. Some geometry help – a useful angle The smaller angle between and the y-axis is also , as shown in the figure. ANSWER: F  n Fn Fn Ff Fw  F n Fny F  n Fn  F  n  Typesetting math: 100% Hint 2. Find the y component of Write an expression for , the y component of the force , using coordinate system b. Express your answer in terms of and . Hint 1. Some geometry help – a useful angle The smaller angle between and the x-axis is also , as shown in the figure. ANSWER: ANSWER: Fny = Fncos() F f Ffy F f Ff  F  f  Ffy = Ffsin() Fy = 0 = Fncos() + Ffsin() − Fw Typesetting math: 100% Correct Part D Because the block is not moving, the sum of the x components of the forces acting on the block must be zero. Find an expression for the sum of the x components of the forces acting on the block, using coordinate system b. Express your answer in terms of some or all of the variables , , , and . Hint 1. Find the x component of Write an expression for , the x component of the force , using coordinate system b. Express your answer in terms of and . ANSWER: ANSWER: Correct Part E To find the magnitude of the normal force, you must express in terms of since is an unknown. Using the equations you found in the two previous parts, find an expression for involving and but not . Hint 1. How to approach the problem From your answers to the previous two parts you should have two force equations ( and ). Combine these equations to eliminate . The key is to multiply the Fn Ff Fw  F n Fnx F  n Fn  Fnx = −Fnsin() Fx = 0 = −Fnsin() + Ffcos() Fn Fw Ff Fn Fw  Ff Typesetting math: 100% Fy = 0 Fx = 0 Ff equation for the y components by and the equation for the x components by , then add or subtract the two equations to eliminate the term . An alternative motivation for the algebra is to eliminate the trig functions in front of by using the trig identity . At the very least this would result in an equation that is simple to solve for . ANSWER: Correct Congratulations on working this through. Now realize that in coordinate system a, which is aligned with the plane, the y-coordinate equation is , which leads immediately to the result obtained here for . CONCLUSION: A thoughtful examination of which coordinate system to choose can save a lot of algebra. Contact Forces Introduced Learning Goal: To introduce contact forces (normal and friction forces) and to understand that, except for friction forces under certain circumstances, these forces must be determined from: net Force = ma. Two solid objects cannot occupy the same space at the same time. Indeed, when the objects touch, they exert repulsive normal forces on each other, as well as frictional forces that resist their slipping relative to each other. These contact forces arise from a complex interplay between the electrostatic forces between the electrons and ions in the objects and the laws of quantum mechanics. As two surfaces are pushed together these forces increase exponentially over an atomic distance scale, easily becoming strong enough to distort the bulk material in the objects if they approach too close. In everyday experience, contact forces are limited by the deformation or acceleration of the objects, rather than by the fundamental interatomic forces. Hence, we can conclude the following: The magnitude of contact forces is determined by , that is, by the other forces on, and acceleration of, the contacting bodies. The only exception is that the frictional forces cannot exceed (although they can be smaller than this or even zero). Normal and friction forces Two types of contact forces operate in typical mechanics problems, the normal and frictional forces, usually designated by and (or , or something similar) respectively. These are the components of the overall contact force: perpendicular to and parallel to the plane of contact. Kinetic friction when surfaces slide cos  sin  Ff cos() sin() Fn sin2() + cos2 () = 1 Fn Fn = Fwcos() Fy = Fn − FW cos() = 0 Fn F = ma μn n f Ffric n f Typesetting math: 100% When one surface is sliding past the other, experiments show three things about the friction force (denoted ): The frictional force opposes the relative motion at the 1. point of contact, 2. is proportional to the normal force, and 3. the ratio of the magnitude of the frictional force to that of the normal force is fairly constant over a wide range of speeds. The constant of proportionality is called the coefficient of kinetic friction, often designated . As long as the sliding continues, the frictional force is then (valid when the surfaces slide by each other). Static friction when surfaces don’t slide When there is no relative motion of the surfaces, the frictional force can assume any value from zero up to a maximum , where is the coefficient of static friction. Invariably, is larger than , in agreement with the observation that when a force is large enough that something breaks loose and starts to slide, it often accelerates. The frictional force for surfaces with no relative motion is therefore (valid when the contacting surfaces have no relative motion). The actual magnitude and direction of the static friction force are such that it (together with other forces on the object) causes the object to remain motionless with respect to the contacting surface as long as the static friction force required does not exceed . The equation is valid only when the surfaces are on the verge of sliding. Part A When two objects slide by one another, which of the following statements about the force of friction between them, is true? ANSWER: Correct Part B fk fk μk fk = μkn μsn μs μs μk fs ! μsn μsn fs = μsn The frictional force is always equal to . The frictional force is always less than . The frictional force is determined by other forces on the objects so it can be either equal to or less than . μkn μkn μkn Typesetting math: 100% When two objects are in contact with no relative motion, which of the following statements about the frictional force between them, is true? ANSWER: Correct For static friction, the actual magnitude and direction of the friction force are such that it, together with any other forces present, will cause the object to have the observed acceleration. The magnitude of the force cannot exceed . If the magnitude of static friction needed to keep acceleration equal to zero exceeds , then the object will slide subject to the resistance of kinetic friction. Do not automatically assume that unless you are considering a situation in which the magnitude of the static friction force is as large as possible (i.e., when determining at what point an object will just begin to slip). Whether the actual magnitude of the friction force is 0, less than , or equal to depends on the magnitude of the other forces (if any) as well as the acceleration of the object through . Part C When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some point “break loose” and start to accelerate down the board. The box begins to slide once the component of gravity acting parallel to the board just begins to exceeds the maximum force of static friction. Which of the following is the most general explanation for why the box accelerates down the board? ANSWER: The frictional force is always equal to . The frictional force is always less than . The frictional force is determined by other forces on the objects so it can be either equal to or less than . μsn μsn μsn μsn μsn fs = μsn μsn μsn F = ma Fg The force of kinetic friction is smaller than that of maximum static friction, but remains the same. Once the box is moving, is smaller than the force of maximum static friction but larger than the force of kinetic friction. Once the box is moving, is larger than the force of maximum static friction. When the box is stationary, equals the force of static friction, but once the box starts moving, the sliding reduces the normal force, which in turn reduces the friction. Fg Fg Fg Fg Typesetting math: 100% Correct At the point when the box finally does “break loose,” you know that the component of the box’s weight that is parallel to the board just exceeds (i.e., this component of gravitational force on the box has just reached a magnitude such that the force of static friction, which has a maximum value of , can no longer oppose it.) For the box to then accelerate, there must be a net force on the box along the board. Thus, the component of the box’s weight parallel to the board must be greater than the force of kinetic friction. Therefore the force of kinetic friction must be less than the force of static friction which implies , as expected. Part D Consider a problem in which a car of mass is on a road tilted at an angle . The normal force Select the best answer. ANSWER: Correct The key point is that contact forces must be determined from Newton’s equation. In the problem described above, there is not enough information given to determine the normal force (e.g., the acceleration is unknown). Each of the answer options is valid under some conditions ( , the car is sliding down an icy incline, or the car is going around a banked turn), but in fact none is likely to be correct if there are other forces on the car or if the car is accelerating. Do not memorize values for the normal force valid in different problems–you must determine from . Problem 6.17 Bonnie and Clyde are sliding a 323 bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 375 of force while Bonnie pulls forward on a rope with 335 of force. μsn μsn μkn μsn μk < μs M  is found using n = Mg n = Mg cos() n = Mg cos() F  = Ma  = 0 n F = ma kg N N Typesetting math: 100% Part A What is the safe's coefficient of kinetic friction on the bank floor? ANSWER: Correct Problem 6.19 A crate is placed on a horizontal conveyor belt. The materials are such that and . Part A Draw a free-body diagram showing all the forces on the crate if the conveyer belt runs at constant speed. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: 0.224 10 kg μs = 0.5 μk = 0.3 Typesetting math: 100% Correct Part B Draw a free-body diagram showing all the forces on the crate if the conveyer belt is speeding up. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Typesetting math: 100% Correct Part C What is the maximum acceleration the belt can have without the crate slipping? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct amax = 4.9 m s2 Typesetting math: 100% Problem 6.28 A 1100 steel beam is supported by two ropes. Part A What is the tension in rope 1? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the tension in rope 2? Express your answer to two significant figures and include the appropriate units. ANSWER: kg T1 = 7000 N Typesetting math: 100% Correct Problem 6.35 The position of a 1.4 mass is given by , where is in seconds. Part A What is the net horizontal force on the mass at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the net horizontal force on the mass at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 6.39 T2 = 4800 N kg x = (2t3 − 3t2 )m t t = 0 s F = -8.4 N t = 1 s F = 8.4 N Typesetting math: 100% A rifle with a barrel length of 61 fires a 8 bullet with a horizontal speed of 400 . The bullet strikes a block of wood and penetrates to a depth of 11 . Part A What resistive force (assumed to be constant) does the wood exert on the bullet? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How long does it take the bullet to come to rest after entering the wood? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 6.45 You and your friend Peter are putting new shingles on a roof pitched at 21 . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.0 away, asks you for the box of nails. Rather than carry the 2.0 box of nails down to Peter, you decide to give the box a push and have it slide down to him. Part A If the coefficient of kinetic friction between the box and the roof is 0.55, with what speed should you push the box to have it gently come to rest right at the edge of the roof? Express your answer to two significant figures and include the appropriate units. cm g m/s cm fk = 5800 N = 5.5×10−4 t s  m kg Typesetting math: 100% ANSWER: Correct Problem 6.54 The 2.0 wood box in the figure slides down a vertical wood wall while you push on it at a 45 angle. Part A What magnitude of force should you apply to cause the box to slide down at a constant speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct v = 3.9 ms kg  F = 23 N Typesetting math: 100% Score Summary: Your score on this assignment is 98.8%. You received 114.57 out of a possible total of 116 points. Typesetting math: 100%

Assignment 5 Due: 11:59pm on Wednesday, March 5, 2014 You … Read More...
EGT 300 – Statics and Strength of Materials 1. Two structural mebers B and C are bolted to the brackedt A. Knowing that the tension in member B is 2000 N and the tension in C is 1500 N, determine the magnitude (in N.) of the resultant force acting on the bracket. 2. A hoist trolley is subjected to the three forces shown. Knowing that P = 250 lb, determine the magnitude of the resultant (in lb.). 3. The collar A may slide freely on the horizontal smooth rod. Determine the magnitude of the force P (in kN.) required to maintain equilibrium when the dimension c is 8 m. 4. Determine the volume (in mm3) of the solid obtained by rotating the area of the figure show, about the x axis. 5. In the previous problem (6), determine the volume (in dm3) of the solid, by rotating the area about the y axis. 6. Using the method of joints, determine the force (in kN) in the member CA of the truss shown. 7. The coefficient of friction between the block and the rail are µs =0.30 and µk = 0.25. Find the magnitude of the smallest force P (in N) required to start the block up to the rail. 8. Determine the momentum at the beam support B (in lb.in) for the given loading.

EGT 300 – Statics and Strength of Materials 1. Two structural mebers B and C are bolted to the brackedt A. Knowing that the tension in member B is 2000 N and the tension in C is 1500 N, determine the magnitude (in N.) of the resultant force acting on the bracket. 2. A hoist trolley is subjected to the three forces shown. Knowing that P = 250 lb, determine the magnitude of the resultant (in lb.). 3. The collar A may slide freely on the horizontal smooth rod. Determine the magnitude of the force P (in kN.) required to maintain equilibrium when the dimension c is 8 m. 4. Determine the volume (in mm3) of the solid obtained by rotating the area of the figure show, about the x axis. 5. In the previous problem (6), determine the volume (in dm3) of the solid, by rotating the area about the y axis. 6. Using the method of joints, determine the force (in kN) in the member CA of the truss shown. 7. The coefficient of friction between the block and the rail are µs =0.30 and µk = 0.25. Find the magnitude of the smallest force P (in N) required to start the block up to the rail. 8. Determine the momentum at the beam support B (in lb.in) for the given loading.

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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) Part A Find , the x component of the total momentum of the system at time . Express your answer in terms of , , , and . ANSWER: Part B Find the time derivative of the x component of the system’s total momentum. Express your answer in terms of , , , and . You did not open hints for this part. ANSWER: Why did we bother with all this math? The expression for the derivative of momentum that we just obtained will be useful in reaching our desired conclusion, if only for this very special case. Part C The quantity (mass times acceleration) is dimensionally equivalent to which of the following? ANSWER: p(t) t m1 m2 v1 (t) v2 (t) p(t) = dp(t)/dt a1 (t) a2 (t) m1 m2 dp(t)/dt = ma Part D Acceleration is due to which of the following physical quantities? ANSWER: Part E Since we have assumed that the system composed of blocks 1 and 2 is closed, what could be the reason for the acceleration of block 1? You did not open hints for this part. ANSWER: momentum energy force acceleration inertia velocity speed energy momentum force Part F This question will be shown after you complete previous question(s). Part G Let us denote the x component of the force exerted by block 1 on block 2 by , and the x component of the force exerted by block 2 on block 1 by . Which of the following pairs equalities is a direct consequence of Newton’s second law? ANSWER: Part H Let us recall that we have denoted the force exerted by block 1 on block 2 by , and the force exerted by block 2 on block 1 by . If we suppose that is greater than , which of the following statements about forces is true? You did not open hints for this part. the large mass of block 1 air resistance Earth’s gravitational attraction a force exerted by block 2 on block 1 a force exerted by block 1 on block 2 F12 F21 and and and and F12 = m2a2 F21 = m1a1 F12 = m1a1 F21 = m2a2 F12 = m1a2 F21 = m2a1 F12 = m2a1 F21 = m1a2 F12 F21 m1 m2 ANSWER: Part I Now recall the expression for the time derivative of the x component of the system’s total momentum: . Considering the information that you now have, choose the best alternative for an equivalent expression to . You did not open hints for this part. ANSWER: Impulse and Momentum Ranking Task Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest. Part A Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. ANSWER: Both forces have equal magnitudes. |F12 | > |F21| |F21 | > |F12| dpx(t)/dt = Fx dpx(t)/dt 0 nonzero constant kt kt2 Part B Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: Part C Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: A Game of Frictionless Catch Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his cart, , is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest. Chuck then picks up a ball of mass and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is . The speed of the thrown ball relative to the ground is . Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie’s speed relative to the ground after she catches the ball is . When answering the questions in this problem, keep the following in mind: The original mass of Chuck and his cart does not include the 1. mass of the ball. 2. The speed of an object is the magnitude of its velocity. An object’s speed will always be a nonnegative quantity. mcart mball vc vb vj mcart Part A Find the relative speed between Chuck and the ball after Chuck has thrown the ball. Express the speed in terms of and . You did not open hints for this part. ANSWER: Part B What is the speed of the ball (relative to the ground) while it is in the air? Express your answer in terms of , , and . You did not open hints for this part. ANSWER: Part C What is Chuck’s speed (relative to the ground) after he throws the ball? Express your answer in terms of , , and . u vc vb u = vb mball mcart u vb = vc mball mcart u You did not open hints for this part. ANSWER: Part D Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: Part E Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: vc = vj vb vj mball mcart vb vj = vj u vj mball mcart u Momentum in an Explosion A giant “egg” explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with the masses indicated in the diagram, traveling in opposite directions. Part A What is the momentum of piece A before the explosion? Express your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: vj = pA,i Part B During the explosion, is the force of piece A on piece B greater than, less than, or equal to the force of piece B on piece A? You did not open hints for this part. ANSWER: Part C The momentum of piece B is measured to be 500 after the explosion. Find the momentum of piece A after the explosion. Enter your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: pA,i = kg  m/s greater than less than equal to cannot be determined kg  m/s pA,f pA,f = kg  m/s ± PSS 9.1 Conservation of Momentum Learning Goal: To practice Problem-Solving Strategy 9.1 for conservation of momentum problems. An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be moving backward just after releasing the ball? PROBLEM-SOLVING STRATEGY 9.1 Conservation of momentum MODEL: Clearly define the system. If possible, choose a system that is isolated ( ) or within which the interactions are sufficiently short and intense that you can ignore external forces for the duration of the interaction (the impulse approximation). Momentum is conserved. If it is not possible to choose an isolated system, try to divide the problem into parts such that momentum is conserved during one segment of the motion. Other segments of the motion can be analyzed using Newton’s laws or, as you will learn later, conservation of energy. VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you are trying to find. SOLVE: The mathematical representation is based on the law of conservation of momentum: . In component form, this is ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model The interaction at study in this problem is the action of throwing the ball, performed by the quarterback while being off the ground. To apply conservation of momentum to this interaction, you will need to clearly define a system that is isolated or within which the impulse approximation can be applied. Part A Sort the following objects as part of the system or not. Drag the appropriate objects to their respective bins. ANSWER: kg kg m/s F = net 0 P = f P  i (pfx + ( + ( += ( + ( + ( + )1 pfx)2 pfx)3 pix)1 pix)2 pix)3 (pfy + ( + ( += ( + ( + ( + )1 pfy)2 pfy)3 piy)1 piy)2 piy)3 Part B This question will be shown after you complete previous question(s). Visualize Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). Conservation of Momentum in Inelastic Collisions Learning Goal: To understand the vector nature of momentum in the case in which two objects collide and stick together. In this problem we will consider a collision of two moving objects such that after the collision, the objects stick together and travel off as a single unit. The collision is therefore completely inelastic. You have probably learned that “momentum is conserved” in an inelastic collision. But how does this fact help you to solve collision problems? The following questions should help you to clarify the meaning and implications of the statement “momentum is conserved.” Part A What physical quantities are conserved in this collision? ANSWER: Part B Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speeds are and . What is the speed of the two-car system after the collision? the magnitude of the momentum only the net momentum (considered as a vector) only the momentum of each object considered individually v1 v2 You did not open hints for this part. ANSWER: Part C Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, what is the magnitude of their combined momentum? You did not open hints for this part. ANSWER: The answer depends on the directions in which the cars were moving before the collision. v1 + v2 v1 − v2 v2 − v1 v1v2 −−−− ” v1+v2 2 v1 + 2 v2 2 −−−−−−−  p1 p2 Part D Two cars collide inelastically and stick together after the collision. Before the collision, their momenta are and . After the collision, their combined momentum is . Of what can one be certain? You did not open hints for this part. ANSWER: Part E Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, the magnitude of their combined momentum is . Of what can one be certain? The answer depends on the directions in which the cars were moving before the collision. p1 + p2 p1 − p2 p2 − p1 p1p2 −−−− ” p1+p2 2 p1 + 2 p2 2 −−−−−−−  p 1 p 2 p p = p1 + # p2 # p = p1 − # p2 # p = p2 − # p1 # p1 p2 p You did not open hints for this part. ANSWER: Colliding Cars In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses and collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of , and car 2 was traveling northward at a speed of . After the collision, the two cars stick together and travel off in the direction shown. Part A p1 + p2 $ p $ p1p2 −−−− ” p1 +p2 $ p $ p1+p2 2 p1 + p2 $ p $ |p1 − p2 | p1 + p2 $ p $ p1 + 2 p2 2 −−−−−−−  m1 m2 v1 v2 First, find the magnitude of , that is, the speed of the two-car unit after the collision. Express in terms of , , and the cars’ initial speeds and . You did not open hints for this part. ANSWER: Part B Find the tangent of the angle . Express your answer in terms of the momenta of the two cars, and . ANSWER: Part C Suppose that after the collision, ; in other words, is . This means that before the collision: ANSWER: v v v m1 m2 v1 v2 v = p1 p2 tan( ) = tan = 1 45′ The magnitudes of the momenta of the cars were equal. The masses of the cars were equal. The velocities of the cars were equal. ± Catching a Ball on Ice Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 that is traveling horizontally at 11.2 . Olaf’s mass is 67.1 . Part A If Olaf catches the ball, with what speed do Olaf and the ball move afterward? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: Part B kg m/s kg vf vf = m/s If the ball hits Olaf and bounces off his chest horizontally at 8.00 in the opposite direction, what is his speed after the collision? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: A One-Dimensional Inelastic Collision Block 1, of mass = 2.90 , moves along a frictionless air track with speed = 25.0 . It collides with block 2, of mass = 17.0 , which was initially at rest. The blocks stick together after the collision. Part A Find the magnitude of the total initial momentum of the two-block system. Express your answer numerically. m/s vf vf = m/s m1 kg v1 m/s m2 kg pi You did not open hints for this part. ANSWER: Part B Find , the magnitude of the final velocity of the two-block system. Express your answer numerically. You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. pi = kg  m/s vf vf = m/s

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) Part A Find , the x component of the total momentum of the system at time . Express your answer in terms of , , , and . ANSWER: Part B Find the time derivative of the x component of the system’s total momentum. Express your answer in terms of , , , and . You did not open hints for this part. ANSWER: Why did we bother with all this math? The expression for the derivative of momentum that we just obtained will be useful in reaching our desired conclusion, if only for this very special case. Part C The quantity (mass times acceleration) is dimensionally equivalent to which of the following? ANSWER: p(t) t m1 m2 v1 (t) v2 (t) p(t) = dp(t)/dt a1 (t) a2 (t) m1 m2 dp(t)/dt = ma Part D Acceleration is due to which of the following physical quantities? ANSWER: Part E Since we have assumed that the system composed of blocks 1 and 2 is closed, what could be the reason for the acceleration of block 1? You did not open hints for this part. ANSWER: momentum energy force acceleration inertia velocity speed energy momentum force Part F This question will be shown after you complete previous question(s). Part G Let us denote the x component of the force exerted by block 1 on block 2 by , and the x component of the force exerted by block 2 on block 1 by . Which of the following pairs equalities is a direct consequence of Newton’s second law? ANSWER: Part H Let us recall that we have denoted the force exerted by block 1 on block 2 by , and the force exerted by block 2 on block 1 by . If we suppose that is greater than , which of the following statements about forces is true? You did not open hints for this part. the large mass of block 1 air resistance Earth’s gravitational attraction a force exerted by block 2 on block 1 a force exerted by block 1 on block 2 F12 F21 and and and and F12 = m2a2 F21 = m1a1 F12 = m1a1 F21 = m2a2 F12 = m1a2 F21 = m2a1 F12 = m2a1 F21 = m1a2 F12 F21 m1 m2 ANSWER: Part I Now recall the expression for the time derivative of the x component of the system’s total momentum: . Considering the information that you now have, choose the best alternative for an equivalent expression to . You did not open hints for this part. ANSWER: Impulse and Momentum Ranking Task Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest. Part A Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. ANSWER: Both forces have equal magnitudes. |F12 | > |F21| |F21 | > |F12| dpx(t)/dt = Fx dpx(t)/dt 0 nonzero constant kt kt2 Part B Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: Part C Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: A Game of Frictionless Catch Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his cart, , is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest. Chuck then picks up a ball of mass and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is . The speed of the thrown ball relative to the ground is . Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie’s speed relative to the ground after she catches the ball is . When answering the questions in this problem, keep the following in mind: The original mass of Chuck and his cart does not include the 1. mass of the ball. 2. The speed of an object is the magnitude of its velocity. An object’s speed will always be a nonnegative quantity. mcart mball vc vb vj mcart Part A Find the relative speed between Chuck and the ball after Chuck has thrown the ball. Express the speed in terms of and . You did not open hints for this part. ANSWER: Part B What is the speed of the ball (relative to the ground) while it is in the air? Express your answer in terms of , , and . You did not open hints for this part. ANSWER: Part C What is Chuck’s speed (relative to the ground) after he throws the ball? Express your answer in terms of , , and . u vc vb u = vb mball mcart u vb = vc mball mcart u You did not open hints for this part. ANSWER: Part D Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: Part E Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: vc = vj vb vj mball mcart vb vj = vj u vj mball mcart u Momentum in an Explosion A giant “egg” explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with the masses indicated in the diagram, traveling in opposite directions. Part A What is the momentum of piece A before the explosion? Express your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: vj = pA,i Part B During the explosion, is the force of piece A on piece B greater than, less than, or equal to the force of piece B on piece A? You did not open hints for this part. ANSWER: Part C The momentum of piece B is measured to be 500 after the explosion. Find the momentum of piece A after the explosion. Enter your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: pA,i = kg  m/s greater than less than equal to cannot be determined kg  m/s pA,f pA,f = kg  m/s ± PSS 9.1 Conservation of Momentum Learning Goal: To practice Problem-Solving Strategy 9.1 for conservation of momentum problems. An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be moving backward just after releasing the ball? PROBLEM-SOLVING STRATEGY 9.1 Conservation of momentum MODEL: Clearly define the system. If possible, choose a system that is isolated ( ) or within which the interactions are sufficiently short and intense that you can ignore external forces for the duration of the interaction (the impulse approximation). Momentum is conserved. If it is not possible to choose an isolated system, try to divide the problem into parts such that momentum is conserved during one segment of the motion. Other segments of the motion can be analyzed using Newton’s laws or, as you will learn later, conservation of energy. VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you are trying to find. SOLVE: The mathematical representation is based on the law of conservation of momentum: . In component form, this is ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model The interaction at study in this problem is the action of throwing the ball, performed by the quarterback while being off the ground. To apply conservation of momentum to this interaction, you will need to clearly define a system that is isolated or within which the impulse approximation can be applied. Part A Sort the following objects as part of the system or not. Drag the appropriate objects to their respective bins. ANSWER: kg kg m/s F = net 0 P = f P  i (pfx + ( + ( += ( + ( + ( + )1 pfx)2 pfx)3 pix)1 pix)2 pix)3 (pfy + ( + ( += ( + ( + ( + )1 pfy)2 pfy)3 piy)1 piy)2 piy)3 Part B This question will be shown after you complete previous question(s). Visualize Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). Conservation of Momentum in Inelastic Collisions Learning Goal: To understand the vector nature of momentum in the case in which two objects collide and stick together. In this problem we will consider a collision of two moving objects such that after the collision, the objects stick together and travel off as a single unit. The collision is therefore completely inelastic. You have probably learned that “momentum is conserved” in an inelastic collision. But how does this fact help you to solve collision problems? The following questions should help you to clarify the meaning and implications of the statement “momentum is conserved.” Part A What physical quantities are conserved in this collision? ANSWER: Part B Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speeds are and . What is the speed of the two-car system after the collision? the magnitude of the momentum only the net momentum (considered as a vector) only the momentum of each object considered individually v1 v2 You did not open hints for this part. ANSWER: Part C Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, what is the magnitude of their combined momentum? You did not open hints for this part. ANSWER: The answer depends on the directions in which the cars were moving before the collision. v1 + v2 v1 − v2 v2 − v1 v1v2 −−−− ” v1+v2 2 v1 + 2 v2 2 −−−−−−−  p1 p2 Part D Two cars collide inelastically and stick together after the collision. Before the collision, their momenta are and . After the collision, their combined momentum is . Of what can one be certain? You did not open hints for this part. ANSWER: Part E Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, the magnitude of their combined momentum is . Of what can one be certain? The answer depends on the directions in which the cars were moving before the collision. p1 + p2 p1 − p2 p2 − p1 p1p2 −−−− ” p1+p2 2 p1 + 2 p2 2 −−−−−−−  p 1 p 2 p p = p1 + # p2 # p = p1 − # p2 # p = p2 − # p1 # p1 p2 p You did not open hints for this part. ANSWER: Colliding Cars In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses and collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of , and car 2 was traveling northward at a speed of . After the collision, the two cars stick together and travel off in the direction shown. Part A p1 + p2 $ p $ p1p2 −−−− ” p1 +p2 $ p $ p1+p2 2 p1 + p2 $ p $ |p1 − p2 | p1 + p2 $ p $ p1 + 2 p2 2 −−−−−−−  m1 m2 v1 v2 First, find the magnitude of , that is, the speed of the two-car unit after the collision. Express in terms of , , and the cars’ initial speeds and . You did not open hints for this part. ANSWER: Part B Find the tangent of the angle . Express your answer in terms of the momenta of the two cars, and . ANSWER: Part C Suppose that after the collision, ; in other words, is . This means that before the collision: ANSWER: v v v m1 m2 v1 v2 v = p1 p2 tan( ) = tan = 1 45′ The magnitudes of the momenta of the cars were equal. The masses of the cars were equal. The velocities of the cars were equal. ± Catching a Ball on Ice Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 that is traveling horizontally at 11.2 . Olaf’s mass is 67.1 . Part A If Olaf catches the ball, with what speed do Olaf and the ball move afterward? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: Part B kg m/s kg vf vf = m/s If the ball hits Olaf and bounces off his chest horizontally at 8.00 in the opposite direction, what is his speed after the collision? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: A One-Dimensional Inelastic Collision Block 1, of mass = 2.90 , moves along a frictionless air track with speed = 25.0 . It collides with block 2, of mass = 17.0 , which was initially at rest. The blocks stick together after the collision. Part A Find the magnitude of the total initial momentum of the two-block system. Express your answer numerically. m/s vf vf = m/s m1 kg v1 m/s m2 kg pi You did not open hints for this part. ANSWER: Part B Find , the magnitude of the final velocity of the two-block system. Express your answer numerically. You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. pi = kg  m/s vf vf = m/s

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Assignment 1 Due: 11:59pm on Wednesday, February 5, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 1.6 Part A Determine the sign (positive or negative) of the position for the particle in the figure. ANSWER: Correct Part B Determine the sign (positive or negative) of the velocity for the particle in the figure. ANSWER: Correct Positive Negative Negative Positive Part C Determine the sign (positive or negative) of the acceleration for the particle in the figure. ANSWER: Correct Conceptual Question 1.7 Part A Determine the sign (positive or negative) of the position for the particle in the figure. ANSWER: Positive Negative Correct Part B Determine the sign (positive or negative) of the velocity for the particle in the figure. ANSWER: Correct Part C Determine the sign (positive or negative) of the acceleration for the particle in the figure. ANSWER: Correct Enhanced EOC: Problem 1.18 The figure shows the motion diagram of a drag racer. The camera took one frame every 2 . Positive Negative Positive Negative Negative Positive s You may want to review ( pages 16 – 19) . For help with math skills, you may want to review: Plotting Points on a Graph Part A Make a position-versus-time graph for the drag racer. Hint 1. How to approach the problem Based on Table 1.1 in the book/e-text, what two observables are associated with each point? Which position or point of the drag racer occurs first? Which position occurs last? If you label the first point as happening at , at what time does the next point occur? At what time does the last position point occur? What is the position of a point halfway in between and ? Can you think of a way to estimate the positions of the points using a ruler? ANSWER: t = 0 s x = 0 m x = 200 m Correct Motion of Two Rockets Learning Goal: To learn to use images of an object in motion to determine velocity and acceleration. Two toy rockets are traveling in the same direction (taken to be the x axis). A diagram is shown of a time-exposure image where a stroboscope has illuminated the rockets at the uniform time intervals indicated. Part A At what time(s) do the rockets have the same velocity? Hint 1. How to determine the velocity The diagram shows position, not velocity. You can’t find instantaneous velocity from this diagram, but you can determine the average velocity between two times and : . Note that no position values are given in the diagram; you will need to estimate these based on the distance between successive positions of the rockets. ANSWER: Correct t1 t2 vavg[t1, t2] = x(t2)−x(t1) t2−t1 at time only at time only at times and at some instant in time between and at no time shown in the figure t = 1 t = 4 t = 1 t = 4 t = 1 t = 4 Part B At what time(s) do the rockets have the same x position? ANSWER: Correct Part C At what time(s) do the two rockets have the same acceleration? Hint 1. How to determine the acceleration The velocity is related to the spacing between images in a stroboscopic diagram. Since acceleration is the rate at which velocity changes, the acceleration is related to the how much this spacing changes from one interval to the next. ANSWER: at time only at time only at times and at some instant in time between and at no time shown in the figure t = 1 t = 4 t = 1 t = 4 t = 1 t = 4 at time only at time only at times and at some instant in time between and at no time shown in the figure t = 1 t = 4 t = 1 t = 4 t = 1 t = 4 Correct Part D The motion of the rocket labeled A is an example of motion with uniform (i.e., constant) __________. ANSWER: Correct Part E The motion of the rocket labeled B is an example of motion with uniform (i.e., constant) __________. ANSWER: Correct Part F At what time(s) is rocket A ahead of rocket B? and nonzero acceleration velocity displacement time and nonzero acceleration velocity displacement time Hint 1. Use the diagram You can answer this question by looking at the diagram and identifying the time(s) when rocket A is to the right of rocket B. ANSWER: Correct Dimensions of Physical Quantities Learning Goal: To introduce the idea of physical dimensions and to learn how to find them. Physical quantities are generally not purely numerical: They have a particular dimension or combination of dimensions associated with them. Thus, your height is not 74, but rather 74 inches, often expressed as 6 feet 2 inches. Although feet and inches are different units they have the same dimension–length. Part A In classical mechanics there are three base dimensions. Length is one of them. What are the other two? Hint 1. MKS system The current system of units is called the International System (abbreviated SI from the French Système International). In the past this system was called the mks system for its base units: meter, kilogram, and second. What are the dimensions of these quantities? ANSWER: before only after only before and after between and at no time(s) shown in the figure t = 1 t = 4 t = 1 t = 4 t = 1 t = 4 Correct There are three dimensions used in mechanics: length ( ), mass ( ), and time ( ). A combination of these three dimensions suffices to express any physical quantity, because when a new physical quantity is needed (e.g., velocity), it always obeys an equation that permits it to be expressed in terms of the units used for these three dimensions. One then derives a unit to measure the new physical quantity from that equation, and often its unit is given a special name. Such new dimensions are called derived dimensions and the units they are measured in are called derived units. For example, area has derived dimensions . (Note that “dimensions of variable ” is symbolized as .) You can find these dimensions by looking at the formula for the area of a square , where is the length of a side of the square. Clearly . Plugging this into the equation gives . Part B Find the dimensions of volume. Express your answer as powers of length ( ), mass ( ), and time ( ). Hint 1. Equation for volume You have likely learned many formulas for the volume of various shapes in geometry. Any of these equations will give you the dimensions for volume. You can find the dimensions most easily from the volume of a cube , where is the length of the edge of the cube. ANSWER: acceleration and mass acceleration and time acceleration and charge mass and time mass and charge time and charge l m t A [A] = l2 x [x] A = s2 s [s] = l [A] = [s] = 2 l2 [V ] l m t V = e3 e [V ] = l3 Correct Part C Find the dimensions of speed. Express your answer as powers of length ( ), mass ( ), and time ( ). Hint 1. Equation for speed Speed is defined in terms of distance and time as . Therefore, . Hint 2. Familiar units for speed You are probably accustomed to hearing speeds in miles per hour (or possibly kilometers per hour). Think about the dimensions for miles and hours. If you divide the dimensions for miles by the dimensions for hours, you will have the dimensions for speed. ANSWER: Correct The dimensions of a quantity are not changed by addition or subtraction of another quantity with the same dimensions. This means that , which comes from subtracting two speeds, has the same dimensions as speed. It does not make physical sense to add or subtract two quanitites that have different dimensions, like length plus time. You can add quantities that have different units, like miles per hour and kilometers per hour, as long as you convert both quantities to the same set of units before you actually compute the sum. You can use this rule to check your answers to any physics problem you work. If the answer involves the sum or difference of two quantities with different dimensions, then it must be incorrect. This rule also ensures that the dimensions of any physical quantity will never involve sums or differences of the base dimensions. (As in the preceeding example, is not a valid dimension for a [v] l m t v d t v = d t [v] = [d]/[t] [v] = lt−1 v l + t physical quantitiy.) A valid dimension will only involve the product or ratio of powers of the base dimensions (e.g. ). Part D Find the dimensions of acceleration. Express your answer as powers of length ( ), mass ( ), and time ( ). Hint 1. Equation for acceleration In physics, acceleration is defined as the change in velocity in a certain time. This is shown by the equation . The is a symbol that means “the change in.” ANSWER: Correct Consistency of Units In physics, every physical quantity is measured with respect to a unit. Time is measured in seconds, length is measured in meters, and mass is measured in kilograms. Knowing the units of physical quantities will help you solve problems in physics. Part A Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equation , where is the magnitude of the gravitational attraction on either body, and are the masses of the bodies, is the distance between them, and is the gravitational constant. In SI units, the units of force are , the units of mass are , and the units of distance are . For this equation to have consistent units, the units of must be which of the following? Hint 1. How to approach the problem To solve this problem, we start with the equation m2/3 l2 t−2 [a] l m t a a = v/t  [a] = lt−2 F = Gm1m2 r2 F m1 m2 r G kg  m/s2 kg m G . For each symbol whose units we know, we replace the symbol with those units. For example, we replace with . We now solve this equation for . ANSWER: Correct Part B One consequence of Einstein’s theory of special relativity is that mass is a form of energy. This mass-energy relationship is perhaps the most famous of all physics equations: , where is mass, is the speed of the light, and is the energy. In SI units, the units of speed are . For the preceding equation to have consistent units (the same units on both sides of the equation), the units of must be which of the following? Hint 1. How to approach the problem To solve this problem, we start with the equation . For each symbol whose units we know, we replace the symbol with those units. For example, we replace with . We now solve this equation for . ANSWER: F = Gm1m2 r2 m1 kg G kg3 ms2 kgs2 m3 m3 kgs2 m kgs2 E = mc2 m c E m/s E E = mc2 m kg E Correct To solve the types of problems typified by these examples, we start with the given equation. For each symbol whose units we know, we replace the symbol with those units. For example, we replace with . We now solve this equation for the units of the unknown variable. Problem 1.24 Convert the following to SI units: Part A 5.0 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B 54 Express your answer to two significant figures and include the appropriate units. kgm s kgm2 s2 kgs2 m2 kgm2 s m kg in 0.13 m ft/s ANSWER: Correct Part C 72 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D 17 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 1.55 The figure shows a motion diagram of a car traveling down a street. The camera took one frame every 10 . A distance scale is provided. 16 ms mph 32 ms in2 1.1×10−2 m2 s Part A Make a position-versus-time graph for the car. ANSWER: Incorrect; Try Again ± Moving at the Speed of Light Part A How many nanoseconds does it take light to travel a distance of 4.40 in vacuum? Express your answer numerically in nanoseconds. Hint 1. How to approach the problem Light travels at a constant speed; therefore, you can use the formula for the distance traveled in a certain amount of time by an object moving at constant speed. Before performing any calculations, it is often recommended, although it is not strictly necessary, to convert all quantities to their fundamental units rather than to multiples of the fundamental unit. km Hint 2. Find how many seconds it takes light to travel the given distance Given that the speed of light in vacuum is , how many seconds does it take light to travel a distance of 4.40 ? Express your answer numerically in seconds. Hint 1. Find the time it takes light to travel a certain distance How long does it take light to travel a distance ? Let be the speed of light. Hint 1. The speed of an object The equation that relates the distance traveled by an object with constant speed in a time is . ANSWER: Correct Hint 2. Convert the given distance to meters Convert = 4.40 to meters. Express your answer numerically in meters. Hint 1. Conversion of kilometers to meters Recall that . 3.00 × 108 m/s km r c s v t s = vt r  c r c c r d km 1 km = 103 m ANSWER: Correct ANSWER: Correct Now convert the time into nanoseconds. Recall that . ANSWER: Correct Score Summary: Your score on this assignment is 84.7%. You received 50.84 out of a possible total of 60 points. 4.40km = 4400 m 1.47×10−5 s 1 ns = 10−9 s 1.47×104 ns

Assignment 1 Due: 11:59pm on Wednesday, February 5, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 1.6 Part A Determine the sign (positive or negative) of the position for the particle in the figure. ANSWER: Correct Part B Determine the sign (positive or negative) of the velocity for the particle in the figure. ANSWER: Correct Positive Negative Negative Positive Part C Determine the sign (positive or negative) of the acceleration for the particle in the figure. ANSWER: Correct Conceptual Question 1.7 Part A Determine the sign (positive or negative) of the position for the particle in the figure. ANSWER: Positive Negative Correct Part B Determine the sign (positive or negative) of the velocity for the particle in the figure. ANSWER: Correct Part C Determine the sign (positive or negative) of the acceleration for the particle in the figure. ANSWER: Correct Enhanced EOC: Problem 1.18 The figure shows the motion diagram of a drag racer. The camera took one frame every 2 . Positive Negative Positive Negative Negative Positive s You may want to review ( pages 16 – 19) . For help with math skills, you may want to review: Plotting Points on a Graph Part A Make a position-versus-time graph for the drag racer. Hint 1. How to approach the problem Based on Table 1.1 in the book/e-text, what two observables are associated with each point? Which position or point of the drag racer occurs first? Which position occurs last? If you label the first point as happening at , at what time does the next point occur? At what time does the last position point occur? What is the position of a point halfway in between and ? Can you think of a way to estimate the positions of the points using a ruler? ANSWER: t = 0 s x = 0 m x = 200 m Correct Motion of Two Rockets Learning Goal: To learn to use images of an object in motion to determine velocity and acceleration. Two toy rockets are traveling in the same direction (taken to be the x axis). A diagram is shown of a time-exposure image where a stroboscope has illuminated the rockets at the uniform time intervals indicated. Part A At what time(s) do the rockets have the same velocity? Hint 1. How to determine the velocity The diagram shows position, not velocity. You can’t find instantaneous velocity from this diagram, but you can determine the average velocity between two times and : . Note that no position values are given in the diagram; you will need to estimate these based on the distance between successive positions of the rockets. ANSWER: Correct t1 t2 vavg[t1, t2] = x(t2)−x(t1) t2−t1 at time only at time only at times and at some instant in time between and at no time shown in the figure t = 1 t = 4 t = 1 t = 4 t = 1 t = 4 Part B At what time(s) do the rockets have the same x position? ANSWER: Correct Part C At what time(s) do the two rockets have the same acceleration? Hint 1. How to determine the acceleration The velocity is related to the spacing between images in a stroboscopic diagram. Since acceleration is the rate at which velocity changes, the acceleration is related to the how much this spacing changes from one interval to the next. ANSWER: at time only at time only at times and at some instant in time between and at no time shown in the figure t = 1 t = 4 t = 1 t = 4 t = 1 t = 4 at time only at time only at times and at some instant in time between and at no time shown in the figure t = 1 t = 4 t = 1 t = 4 t = 1 t = 4 Correct Part D The motion of the rocket labeled A is an example of motion with uniform (i.e., constant) __________. ANSWER: Correct Part E The motion of the rocket labeled B is an example of motion with uniform (i.e., constant) __________. ANSWER: Correct Part F At what time(s) is rocket A ahead of rocket B? and nonzero acceleration velocity displacement time and nonzero acceleration velocity displacement time Hint 1. Use the diagram You can answer this question by looking at the diagram and identifying the time(s) when rocket A is to the right of rocket B. ANSWER: Correct Dimensions of Physical Quantities Learning Goal: To introduce the idea of physical dimensions and to learn how to find them. Physical quantities are generally not purely numerical: They have a particular dimension or combination of dimensions associated with them. Thus, your height is not 74, but rather 74 inches, often expressed as 6 feet 2 inches. Although feet and inches are different units they have the same dimension–length. Part A In classical mechanics there are three base dimensions. Length is one of them. What are the other two? Hint 1. MKS system The current system of units is called the International System (abbreviated SI from the French Système International). In the past this system was called the mks system for its base units: meter, kilogram, and second. What are the dimensions of these quantities? ANSWER: before only after only before and after between and at no time(s) shown in the figure t = 1 t = 4 t = 1 t = 4 t = 1 t = 4 Correct There are three dimensions used in mechanics: length ( ), mass ( ), and time ( ). A combination of these three dimensions suffices to express any physical quantity, because when a new physical quantity is needed (e.g., velocity), it always obeys an equation that permits it to be expressed in terms of the units used for these three dimensions. One then derives a unit to measure the new physical quantity from that equation, and often its unit is given a special name. Such new dimensions are called derived dimensions and the units they are measured in are called derived units. For example, area has derived dimensions . (Note that “dimensions of variable ” is symbolized as .) You can find these dimensions by looking at the formula for the area of a square , where is the length of a side of the square. Clearly . Plugging this into the equation gives . Part B Find the dimensions of volume. Express your answer as powers of length ( ), mass ( ), and time ( ). Hint 1. Equation for volume You have likely learned many formulas for the volume of various shapes in geometry. Any of these equations will give you the dimensions for volume. You can find the dimensions most easily from the volume of a cube , where is the length of the edge of the cube. ANSWER: acceleration and mass acceleration and time acceleration and charge mass and time mass and charge time and charge l m t A [A] = l2 x [x] A = s2 s [s] = l [A] = [s] = 2 l2 [V ] l m t V = e3 e [V ] = l3 Correct Part C Find the dimensions of speed. Express your answer as powers of length ( ), mass ( ), and time ( ). Hint 1. Equation for speed Speed is defined in terms of distance and time as . Therefore, . Hint 2. Familiar units for speed You are probably accustomed to hearing speeds in miles per hour (or possibly kilometers per hour). Think about the dimensions for miles and hours. If you divide the dimensions for miles by the dimensions for hours, you will have the dimensions for speed. ANSWER: Correct The dimensions of a quantity are not changed by addition or subtraction of another quantity with the same dimensions. This means that , which comes from subtracting two speeds, has the same dimensions as speed. It does not make physical sense to add or subtract two quanitites that have different dimensions, like length plus time. You can add quantities that have different units, like miles per hour and kilometers per hour, as long as you convert both quantities to the same set of units before you actually compute the sum. You can use this rule to check your answers to any physics problem you work. If the answer involves the sum or difference of two quantities with different dimensions, then it must be incorrect. This rule also ensures that the dimensions of any physical quantity will never involve sums or differences of the base dimensions. (As in the preceeding example, is not a valid dimension for a [v] l m t v d t v = d t [v] = [d]/[t] [v] = lt−1 v l + t physical quantitiy.) A valid dimension will only involve the product or ratio of powers of the base dimensions (e.g. ). Part D Find the dimensions of acceleration. Express your answer as powers of length ( ), mass ( ), and time ( ). Hint 1. Equation for acceleration In physics, acceleration is defined as the change in velocity in a certain time. This is shown by the equation . The is a symbol that means “the change in.” ANSWER: Correct Consistency of Units In physics, every physical quantity is measured with respect to a unit. Time is measured in seconds, length is measured in meters, and mass is measured in kilograms. Knowing the units of physical quantities will help you solve problems in physics. Part A Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equation , where is the magnitude of the gravitational attraction on either body, and are the masses of the bodies, is the distance between them, and is the gravitational constant. In SI units, the units of force are , the units of mass are , and the units of distance are . For this equation to have consistent units, the units of must be which of the following? Hint 1. How to approach the problem To solve this problem, we start with the equation m2/3 l2 t−2 [a] l m t a a = v/t  [a] = lt−2 F = Gm1m2 r2 F m1 m2 r G kg  m/s2 kg m G . For each symbol whose units we know, we replace the symbol with those units. For example, we replace with . We now solve this equation for . ANSWER: Correct Part B One consequence of Einstein’s theory of special relativity is that mass is a form of energy. This mass-energy relationship is perhaps the most famous of all physics equations: , where is mass, is the speed of the light, and is the energy. In SI units, the units of speed are . For the preceding equation to have consistent units (the same units on both sides of the equation), the units of must be which of the following? Hint 1. How to approach the problem To solve this problem, we start with the equation . For each symbol whose units we know, we replace the symbol with those units. For example, we replace with . We now solve this equation for . ANSWER: F = Gm1m2 r2 m1 kg G kg3 ms2 kgs2 m3 m3 kgs2 m kgs2 E = mc2 m c E m/s E E = mc2 m kg E Correct To solve the types of problems typified by these examples, we start with the given equation. For each symbol whose units we know, we replace the symbol with those units. For example, we replace with . We now solve this equation for the units of the unknown variable. Problem 1.24 Convert the following to SI units: Part A 5.0 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B 54 Express your answer to two significant figures and include the appropriate units. kgm s kgm2 s2 kgs2 m2 kgm2 s m kg in 0.13 m ft/s ANSWER: Correct Part C 72 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D 17 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 1.55 The figure shows a motion diagram of a car traveling down a street. The camera took one frame every 10 . A distance scale is provided. 16 ms mph 32 ms in2 1.1×10−2 m2 s Part A Make a position-versus-time graph for the car. ANSWER: Incorrect; Try Again ± Moving at the Speed of Light Part A How many nanoseconds does it take light to travel a distance of 4.40 in vacuum? Express your answer numerically in nanoseconds. Hint 1. How to approach the problem Light travels at a constant speed; therefore, you can use the formula for the distance traveled in a certain amount of time by an object moving at constant speed. Before performing any calculations, it is often recommended, although it is not strictly necessary, to convert all quantities to their fundamental units rather than to multiples of the fundamental unit. km Hint 2. Find how many seconds it takes light to travel the given distance Given that the speed of light in vacuum is , how many seconds does it take light to travel a distance of 4.40 ? Express your answer numerically in seconds. Hint 1. Find the time it takes light to travel a certain distance How long does it take light to travel a distance ? Let be the speed of light. Hint 1. The speed of an object The equation that relates the distance traveled by an object with constant speed in a time is . ANSWER: Correct Hint 2. Convert the given distance to meters Convert = 4.40 to meters. Express your answer numerically in meters. Hint 1. Conversion of kilometers to meters Recall that . 3.00 × 108 m/s km r c s v t s = vt r  c r c c r d km 1 km = 103 m ANSWER: Correct ANSWER: Correct Now convert the time into nanoseconds. Recall that . ANSWER: Correct Score Summary: Your score on this assignment is 84.7%. You received 50.84 out of a possible total of 60 points. 4.40km = 4400 m 1.47×10−5 s 1 ns = 10−9 s 1.47×104 ns

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PHSX 220 Homework 13 Paper – Due Online April 28 – 5:00 pm SHM and Wave the Equation Problem 1: A hanging mass system with a mass of 85 kg, spring constant of k= 490 N/m is realeased from rest from a distance of 10 meters below the systems equilibrium position (similar values to the bottom of a bungee jump). Calculate the following quantities in regards to this system after being released at t=0: a) The angular frequency of the system (radians/sec) b) The frequency of oscillations for the system (Hz) c) The period of oscillations for the system (seconds) d) The time it takes to get back to the equilibrium position of the system for the rst time Problem 2: A horizontal spring-mass system (mass of 2:21×10􀀀25 kg) with no friction has an ocsillation frequency of 9,192,631,770 cycles per second. (a second is de ned by 9,192,631,770 cycles of a Cs-133 atom)). Calculate the e ective spring constant of the system Problem 3: A swinging person, such as Tarzan, can be modeled after a simple pendulum with a mass of 85 kg and a length of 10 m. Consider the mass being released from rest at t=0 at an angle of +15 degrees from the vertical. Calculate the following quantities in regards to this system. You need to be in radians mode for this problem a) The angular frequency of the system (radians/sec) b) The frequency of oscillations for the system in (Hz) c) The period of oscillations of the system (seconds) d) Sketch plots of the angular position, angular velocity and angular acceleration of the system as a function of time. Hint: These will always help you with these time to it takes to a certain point in it’s cycle questions. e) The time it takes for the mass to get half way through its rst cycle (or to the other side of the swing if you were interested in timing say a rescue e ort or something along those lines) . f) The maximum angular velocity of the mass g) The maximum angular accleration of the mass h) The magnitude of the angular momentum of the mass at 3 seconds i) The magnitude of the torque acting on the mass at 3 seconds Problem 4: A wave has a wavenumber of 1 m-1, and an angular frequency of 2 radians per second, travels in the +x direction and has a maximum transverse amplitude of 0.1 m. At t=0, and x =0 the y position is equal to 0.0 m (y(0,0) = 0.0 m). a) Calculate the wavelength of the wave b) Calculate the period of oscillations for the wave c) Calculate the wave speed along the x axis d) Calculate the magnitude and direction of the transverse position of the wave at x=0.5 m and t = 8s e) Calculate the magnitude and direction of the transverse velocity of the wave at x=0.5 m and t = 8s f) Calculate the magnitude and direction of the transverse acceleration of the wave at x=0.5 m and t = 8s Problem 5-6: Chapter 16 Problem 10, 22 Additional Suggested Problems with Solutions Provided: Chapter 16 Problems 5, 9, 15, 45

PHSX 220 Homework 13 Paper – Due Online April 28 – 5:00 pm SHM and Wave the Equation Problem 1: A hanging mass system with a mass of 85 kg, spring constant of k= 490 N/m is realeased from rest from a distance of 10 meters below the systems equilibrium position (similar values to the bottom of a bungee jump). Calculate the following quantities in regards to this system after being released at t=0: a) The angular frequency of the system (radians/sec) b) The frequency of oscillations for the system (Hz) c) The period of oscillations for the system (seconds) d) The time it takes to get back to the equilibrium position of the system for the rst time Problem 2: A horizontal spring-mass system (mass of 2:21×10􀀀25 kg) with no friction has an ocsillation frequency of 9,192,631,770 cycles per second. (a second is de ned by 9,192,631,770 cycles of a Cs-133 atom)). Calculate the e ective spring constant of the system Problem 3: A swinging person, such as Tarzan, can be modeled after a simple pendulum with a mass of 85 kg and a length of 10 m. Consider the mass being released from rest at t=0 at an angle of +15 degrees from the vertical. Calculate the following quantities in regards to this system. You need to be in radians mode for this problem a) The angular frequency of the system (radians/sec) b) The frequency of oscillations for the system in (Hz) c) The period of oscillations of the system (seconds) d) Sketch plots of the angular position, angular velocity and angular acceleration of the system as a function of time. Hint: These will always help you with these time to it takes to a certain point in it’s cycle questions. e) The time it takes for the mass to get half way through its rst cycle (or to the other side of the swing if you were interested in timing say a rescue e ort or something along those lines) . f) The maximum angular velocity of the mass g) The maximum angular accleration of the mass h) The magnitude of the angular momentum of the mass at 3 seconds i) The magnitude of the torque acting on the mass at 3 seconds Problem 4: A wave has a wavenumber of 1 m-1, and an angular frequency of 2 radians per second, travels in the +x direction and has a maximum transverse amplitude of 0.1 m. At t=0, and x =0 the y position is equal to 0.0 m (y(0,0) = 0.0 m). a) Calculate the wavelength of the wave b) Calculate the period of oscillations for the wave c) Calculate the wave speed along the x axis d) Calculate the magnitude and direction of the transverse position of the wave at x=0.5 m and t = 8s e) Calculate the magnitude and direction of the transverse velocity of the wave at x=0.5 m and t = 8s f) Calculate the magnitude and direction of the transverse acceleration of the wave at x=0.5 m and t = 8s Problem 5-6: Chapter 16 Problem 10, 22 Additional Suggested Problems with Solutions Provided: Chapter 16 Problems 5, 9, 15, 45

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