As waves approach shore, they tend to refract so that Question 42 options: waves become more parallel to the shore waves approach at a greater angle to the shore wave height decreases wavelength increases

As waves approach shore, they tend to refract so that Question 42 options: waves become more parallel to the shore waves approach at a greater angle to the shore wave height decreases wavelength increases

As waves approach shore, they tend to refract so that … Read More...
MAE 214 – Fall 2015 Homework 3 Due: October 1, 2015 – Thursday by 1:00 p.m. Total Problems: 4 (including Extra Credit), Total Points: 105 1. Make a solid works part model from the given figure below. All dimensions are in millimeters. All sketches must be fully defined. Also create a drawing sheet and dimension it as shown. You can use a hole call out option under annotation to dimension a counter bore hole. (30 points) Save your part files as follows: My Documents/Homework 3 Folder/Prob1_LastName.SLDPRT My Documents/ Homework 3 Folder/Prob1_LastName.SLDDRW 2. Make a solid works part of the given figure below and also make a drawing sheet – front, top and right side views using 3rd angle projection method. Dimension the views with appropriate dimension technique. All dimensions are in mm. (30 points) Save your part file and drawing sheet as follows: Documents/Homework 3 folder/Problem 2_Last Name.SLDPRT Documents/Homework 3 folder/Problem 2_Last Name.SLDDRW 3. Make a solid works part file for the given figure below. All sketches must be fully defined. Your design tree menu must have advanced features i.e. plane, mirror, and fillet. The spot facing (SF) must be defined in a problem. The inclined cut must be created with an offset sketch and extrude cut or a suitable sketch that uses “up to surface” option. (40 points) Your part model must stick to the isometric view as it is shown here. Save your part file into: My documents/Homework 3 Folder/Problem3_Last Name.SLDPRT Given: A = 76 B = 127 Unit: MMGS ALL ROUNDS (FILLET) EQUAL 6 MM 4. (Extra Credit) Make a solid works part from the given figure below. All sketches must be fully defined. Save your part file to Documents/Homework3 Folder/Prob#4_Last Name.SLDPRT All dimensions are in millimeters. (5 points)

MAE 214 – Fall 2015 Homework 3 Due: October 1, 2015 – Thursday by 1:00 p.m. Total Problems: 4 (including Extra Credit), Total Points: 105 1. Make a solid works part model from the given figure below. All dimensions are in millimeters. All sketches must be fully defined. Also create a drawing sheet and dimension it as shown. You can use a hole call out option under annotation to dimension a counter bore hole. (30 points) Save your part files as follows: My Documents/Homework 3 Folder/Prob1_LastName.SLDPRT My Documents/ Homework 3 Folder/Prob1_LastName.SLDDRW 2. Make a solid works part of the given figure below and also make a drawing sheet – front, top and right side views using 3rd angle projection method. Dimension the views with appropriate dimension technique. All dimensions are in mm. (30 points) Save your part file and drawing sheet as follows: Documents/Homework 3 folder/Problem 2_Last Name.SLDPRT Documents/Homework 3 folder/Problem 2_Last Name.SLDDRW 3. Make a solid works part file for the given figure below. All sketches must be fully defined. Your design tree menu must have advanced features i.e. plane, mirror, and fillet. The spot facing (SF) must be defined in a problem. The inclined cut must be created with an offset sketch and extrude cut or a suitable sketch that uses “up to surface” option. (40 points) Your part model must stick to the isometric view as it is shown here. Save your part file into: My documents/Homework 3 Folder/Problem3_Last Name.SLDPRT Given: A = 76 B = 127 Unit: MMGS ALL ROUNDS (FILLET) EQUAL 6 MM 4. (Extra Credit) Make a solid works part from the given figure below. All sketches must be fully defined. Save your part file to Documents/Homework3 Folder/Prob#4_Last Name.SLDPRT All dimensions are in millimeters. (5 points)

The ghost of Banquo in Macbeth was represented for the first time in 1954, for a production in Calcutta, by using a clear glass placed at 45 degrees angle on the stage. Discuss how they used the reflection and transparency of glass to achieve this in terms of rules of reflection.

The ghost of Banquo in Macbeth was represented for the first time in 1954, for a production in Calcutta, by using a clear glass placed at 45 degrees angle on the stage. Discuss how they used the reflection and transparency of glass to achieve this in terms of rules of reflection.

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

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

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Index of refraction of glass is 1.27.Find the speed of light in glass.If you shine a light at an angle of incidence of 70 degrees, find the angle of refraction and draw the diagram

Index of refraction of glass is 1.27.Find the speed of light in glass.If you shine a light at an angle of incidence of 70 degrees, find the angle of refraction and draw the diagram

Index of refraction of glass is 1.27. A) Find the … Read More...
Geometric versus Componentwise Vector Addition Learning Goal: To understand that adding vectors geometrically or using components yields the same result. Vectors may be manipulated using either geometry or components. In this tutorial, we consider the addition of two vectors using these methods. Vectors A and B have lengths A and B, respectively, and B makes an angle θ from the direction of A. Vector addition using geometry Vector addition using geometry is accomplished by placing the tail of one vector, in this case B, at the tip of the other vector, A (Figure 1) and using the laws of plane geometry to find C=A2+B2−2ABcos(c)−−−−−−−−−−−−−−−−−√ and b=sin−1(Bsin(c)C), where the length C and angle b are those of the resultant (or sum) vector, C=A+B. Vector addition using components Vector addition using components requires that a coordinate system be chosen. Here, the x axis is chosen along the direction of A (Figure 2) . Given the coordinate system, the x and y components of B are Bcos(θ) and Bsin(θ), respectively. Therefore, the x and y components of C are given by the equations Cx=A+Bcos(θ) and Cy=Bsin(θ). Part A Which of the following sets of conditions, if true, would show that Equations 1 and 2 above define the same vector C as Equations 3 and 4? The two pairs of equations give the same Check all that apply. length and direction for C. length and x component for C. direction and x component for C. length and y component for C. direction and y component for C. x and y components for C.

Geometric versus Componentwise Vector Addition Learning Goal: To understand that adding vectors geometrically or using components yields the same result. Vectors may be manipulated using either geometry or components. In this tutorial, we consider the addition of two vectors using these methods. Vectors A and B have lengths A and B, respectively, and B makes an angle θ from the direction of A. Vector addition using geometry Vector addition using geometry is accomplished by placing the tail of one vector, in this case B, at the tip of the other vector, A (Figure 1) and using the laws of plane geometry to find C=A2+B2−2ABcos(c)−−−−−−−−−−−−−−−−−√ and b=sin−1(Bsin(c)C), where the length C and angle b are those of the resultant (or sum) vector, C=A+B. Vector addition using components Vector addition using components requires that a coordinate system be chosen. Here, the x axis is chosen along the direction of A (Figure 2) . Given the coordinate system, the x and y components of B are Bcos(θ) and Bsin(θ), respectively. Therefore, the x and y components of C are given by the equations Cx=A+Bcos(θ) and Cy=Bsin(θ). Part A Which of the following sets of conditions, if true, would show that Equations 1 and 2 above define the same vector C as Equations 3 and 4? The two pairs of equations give the same Check all that apply. length and direction for C. length and x component for C. direction and x component for C. length and y component for C. direction and y component for C. x and y components for C.

A factory receives power at 480 Vrms @ 60 Hz. from the electric utility company. The factory’s electrical load can be simply represented by 2 loads. LOAD1 describes the manufacturing equipment on the assembly line. LOAD2 describes the power used in office rooms. From time to time, the assembly line shuts down thereby removing LOAD1 from the grid. SWITCH1 accounts for this effect in the equivalent circuit model shown above. Note that the 2 dependent sources represent a device called a “transformer” that steps the 480 Vrms down to 120 Vrms for use in the offices. (But don’t take my word for it; circuit analysis calculations will confirm this.) Given: Receiving End Voltage (with SWITCH1 closed): RV = 480 Vrms Wiring parameters: RW = 0.005 Ω, LW = 0.52052 mH Find: a) With SWITCH1 closed, find the value of C (in Farads) so that the total LOADt at the Receiving End has unity pf. Find the magnitude of the Sending End Voltage SV , and the magnitude of the “Office” load voltage, 2V. Note that RMS480VRV= for this case. b) With SWITCH1 open, using the value of C and SV found in part a), find the new values of the magnitudes of the Receiving End Voltage RV and Office Voltage 2V. Why will this be a problem for the office? How could you change the capacitor connection to avoid this problem? Hints: Note that no phase angles were given, and only magnitudes were asked for. You can choose one voltage or current to have 0 degree phase angle and then allow the calculations of any other voltages and currents be relative to that. In part b) RMS480VRV≠.

A factory receives power at 480 Vrms @ 60 Hz. from the electric utility company. The factory’s electrical load can be simply represented by 2 loads. LOAD1 describes the manufacturing equipment on the assembly line. LOAD2 describes the power used in office rooms. From time to time, the assembly line shuts down thereby removing LOAD1 from the grid. SWITCH1 accounts for this effect in the equivalent circuit model shown above. Note that the 2 dependent sources represent a device called a “transformer” that steps the 480 Vrms down to 120 Vrms for use in the offices. (But don’t take my word for it; circuit analysis calculations will confirm this.) Given: Receiving End Voltage (with SWITCH1 closed): RV = 480 Vrms Wiring parameters: RW = 0.005 Ω, LW = 0.52052 mH Find: a) With SWITCH1 closed, find the value of C (in Farads) so that the total LOADt at the Receiving End has unity pf. Find the magnitude of the Sending End Voltage SV , and the magnitude of the “Office” load voltage, 2V. Note that RMS480VRV= for this case. b) With SWITCH1 open, using the value of C and SV found in part a), find the new values of the magnitudes of the Receiving End Voltage RV and Office Voltage 2V. Why will this be a problem for the office? How could you change the capacitor connection to avoid this problem? Hints: Note that no phase angles were given, and only magnitudes were asked for. You can choose one voltage or current to have 0 degree phase angle and then allow the calculations of any other voltages and currents be relative to that. In part b) RMS480VRV≠.

A factory receives power at 480 Vrms @ 60 Hz. … Read More...