According to previous research, women are generally higher than men in affect intensity. This finding may provide evidence for why ________. it is more adaptive for women to express their anger than it is for men women are more emotionally intelligent compared to men women have a higher threshold for pain compared to men women are more prone to depression compared to men

According to previous research, women are generally higher than men in affect intensity. This finding may provide evidence for why ________. it is more adaptive for women to express their anger than it is for men women are more emotionally intelligent compared to men women have a higher threshold for pain compared to men women are more prone to depression compared to men

According to previous research, women are generally higher than men … Read More...
Chapter 7 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Book on a Table A book weighing 5 N rests on top of a table. Part A A downward force of magnitude 5 N is exerted on the book by the force of ANSWER: Part B An upward force of magnitude _____ is exerted on the _____ by the table. the table gravity inertia . ANSWER: Part C Do the downward force in Part A and the upward force in Part B constitute a 3rd law pair? You did not open hints for this part. ANSWER: Part D The reaction to the force in Part A is a force of magnitude _____, exerted on the _____ by the _____. Its direction is _____ . You did not open hints for this part. ANSWER: 6 N / table 5 N / table 5 N / book 6 N / book yes no Part E The reaction to the force in Part B is a force of magnitude _____, exerted on the _____ by the _____. Its direction is _____. ANSWER: Part F Which of Newton’s laws dictates that the forces in Parts A and B are equal and opposite? ANSWER: Part G Which of Newton’s laws dictates that the forces in Parts B and E are equal and opposite? ANSWER: 5 N / earth / book / upward 5 N / book / table / upward 5 N / book / earth / upward 5 N / earth / book / downward 5 N / table / book / upward 5 N / table / earth / upward 5 N / book / table / upward 5 N / table / book / downward 5 N / earth / book / downward Newton’s 1st or 2nd law Newton’s 3rd law Blocks in an Elevator Ranking Task Three blocks are stacked on top of each other inside an elevator as shown in the figure. Answer the following questions with reference to the eight forces defined as follows. the force of the 3 block on the 2 block, , the force of the 2 block on the 3 block, , the force of the 3 block on the 1 block, , the force of the 1 block on the 3 block, , the force of the 2 block on the 1 block, , the force of the 1 block on the 2 block, , the force of the 1 block on the floor, , and the force of the floor on the 1 block, . Part A Assume the elevator is at rest. Rank the magnitude of the forces. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: Newton’s 1st or 2nd law Newton’s 3rd law kg kg F3 on 2 kg kg F2 on 3 kg kg F3 on 1 kg kg F1 on 3 kg kg F2 on 1 kg kg F1 on 2 kg F1 on floor kg Ffloor on 1 Part B This question will be shown after you complete previous question(s). Newton’s 3rd Law Discussed Learning Goal: To understand Newton’s 3rd law, which states that a physical interaction always generates a pair of forces on the two interacting bodies. In Principia, Newton wrote: To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts. (translation by Cajori) The phrase after the colon (often omitted from textbooks) makes it clear that this is a statement about the nature of force. The central idea is that physical interactions (e.g., due to gravity, bodies touching, or electric forces) cause forces to arise between pairs of bodies. Each pairwise interaction produces a pair of opposite forces, one acting on each body. In summary, each physical interaction between two bodies generates a pair of forces. Whatever the physical cause of the interaction, the force on body A from body B is equal in magnitude and opposite in direction to the force on body B from body A. Incidentally, Newton states that the word “action” denotes both (a) the force due to an interaction and (b) the changes in momentum that it imparts to the two interacting bodies. If you haven’t learned about momentum, don’t worry; for now this is just a statement about the origin of forces. Mark each of the following statements as true or false. If a statement refers to “two bodies” interacting via some force, you are not to assume that these two bodies have the same mass. Part A Every force has one and only one 3rd law pair force. ANSWER: Part B The two forces in each pair act in opposite directions. ANSWER: Part C The two forces in each pair can either both act on the same body or they can act on different bodies. ANSWER: true false true false Part D The two forces in each pair may have different physical origins (for instance, one of the forces could be due to gravity, and its pair force could be due to friction or electric charge). ANSWER: Part E The two forces of a 3rd law pair always act on different bodies. ANSWER: Part F Given that two bodies interact via some force, the accelerations of these two bodies have the same magnitude but opposite directions. (Assume no other forces act on either body.) You did not open hints for this part. ANSWER: true false true false true false Part G According to Newton’s 3rd law, the force on the (smaller) moon due to the (larger) earth is ANSWER: Pulling Three Blocks Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force . The magnitude of the tension in the string between blocks B and C is = 3.00 . Assume that each block has mass = 0.400 . true false greater in magnitude and antiparallel to the force on the earth due to the moon. greater in magnitude and parallel to the force on the earth due to the moon. equal in magnitude but antiparallel to the force on the earth due to the moon. equal in magnitude and parallel to the force on the earth due to the moon. smaller in magnitude and antiparallel to the force on the earth due to the moon. smaller in magnitude and parallel to the force on the earth due to the moon. F T N m kg Part A What is the magnitude of the force? Express your answer numerically in newtons. You did not open hints for this part. ANSWER: Part B What is the tension in the string between block A and block B? Express your answer numerically in newtons You did not open hints for this part. ANSWER: Pulling Two Blocks In the situation shown in the figure, a person is pulling with a constant, nonzero force on string 1, which is attached to block A. Block A is also attached to block B via string 2, as shown. For this problem, assume that neither string stretches and that friction is negligible. Both blocks have finite (nonzero) mass. F F = N TAB TAB = N F Part A Which one of the following statements correctly descibes the relationship between the accelerations of blocks A and B? You did not open hints for this part. ANSWER: Part B How does the magnitude of the tension in string 1, , compare with the tension in string 2, ? You did not open hints for this part. Block A has a larger acceleration than block B. Block B has a larger acceleration than block A. Both blocks have the same acceleration. More information is needed to determine the relationship between the accelerations. T1 T2 ANSWER: Tension in a Massless Rope Learning Goal: To understand the concept of tension and the relationship between tension and force. This problem introduces the concept of tension. The example is a rope, oriented vertically, that is being pulled from both ends. Let and (with u for up and d for down) represent the magnitude of the forces acting on the top and bottom of the rope, respectively. Assume that the rope is massless, so that its weight is negligible compared with the tension. (This is not a ridiculous approximation–modern rope materials such as Kevlar can carry tensions thousands of times greater than the weight of tens of meters of such rope.) Consider the three sections of rope labeled a, b, and c in the figure. At point 1, a downward force of magnitude acts on section a. At point 1, an upward force of magnitude acts on section b. At point 1, the tension in the rope is . At point 2, a downward force of magnitude acts on section b. At point 2, an upward force of magnitude acts on section c. At point 2, the tension in the rope is . Assume, too, that the rope is at equilibrium. Part A What is the magnitude of the downward force on section a? Express your answer in terms of the tension . ANSWER: More information is needed to determine the relationship between and . T1 > T2 T1 = T2 T1 < T2 T1 T2 Fu Fd Fad Fbu T1 Fbd Fcu T2 Fad T1 Part B What is the magnitude of the upward force on section b? Express your answer in terms of the tension . ANSWER: Part C The magnitude of the upward force on c, , and the magnitude of the downward force on b, , are equal because of which of Newton's laws? ANSWER: Part D The magnitude of the force is ____ . ANSWER: Fad = Fbu T1 Fbu = Fcu Fbd 1st 2nd 3rd Fbu Fbd Part E Now consider the forces on the ends of the rope. What is the relationship between the magnitudes of these two forces? You did not open hints for this part. ANSWER: Part F The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straight line. For this situation, which of the following statements are true? Check all that apply. ANSWER: less than greater than equal to Fu > Fd Fu = Fd Fu < Fd The tension in the rope is everywhere the same. The magnitudes of the forces exerted on the two objects by the rope are the same. The forces exerted on the two objects by the rope must be in opposite directions. The forces exerted on the two objects by the rope must be in the direction of the rope. Two Hanging Masses Two blocks with masses and hang one under the other. For this problem, take the positive direction to be upward, and use for the magnitude of the acceleration due to gravity. Case 1: Blocks at rest For Parts A and B assume the blocks are at rest. Part A Find , the tension in the lower rope. Express your answer in terms of some or all of the variables , , and . You did not open hints for this part. ANSWER: M1 M2 g T2 M1 M2 g Part B Find , the tension in the upper rope. Express your answer in terms of some or all of the variables , , and . You did not open hints for this part. ANSWER: Case 2: Accelerating blocks For Parts C and D the blocks are now accelerating upward (due to the tension in the strings) with acceleration of magnitude . Part C Find , the tension in the lower rope. Express your answer in terms of some or all of the variables , , , and . You did not open hints for this part. ANSWER: T2 = T1 M1 M2 g T1 = a T2 M1 M2 a g Part D Find , the tension in the upper rope. Express your answer in terms of some or all of the variables , , , and . You did not open hints for this part. ANSWER: Video Tutor: Suspended Balls: Which String Breaks? First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point. T2 = T1 M1 M2 a g T1 = Part A A heavy crate is attached to the wall by a light rope, as shown in the figure. Another rope hangs off the opposite edge of the box. If you slowly increase the force on the free rope by pulling on it in a horizontal direction, which rope will break? Ignore friction and the mass of the ropes. 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. The rope attached to the wall will break. The rope that you are pulling on will break. Both ropes are equally likely to break.

Chapter 7 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Book on a Table A book weighing 5 N rests on top of a table. Part A A downward force of magnitude 5 N is exerted on the book by the force of ANSWER: Part B An upward force of magnitude _____ is exerted on the _____ by the table. the table gravity inertia . ANSWER: Part C Do the downward force in Part A and the upward force in Part B constitute a 3rd law pair? You did not open hints for this part. ANSWER: Part D The reaction to the force in Part A is a force of magnitude _____, exerted on the _____ by the _____. Its direction is _____ . You did not open hints for this part. ANSWER: 6 N / table 5 N / table 5 N / book 6 N / book yes no Part E The reaction to the force in Part B is a force of magnitude _____, exerted on the _____ by the _____. Its direction is _____. ANSWER: Part F Which of Newton’s laws dictates that the forces in Parts A and B are equal and opposite? ANSWER: Part G Which of Newton’s laws dictates that the forces in Parts B and E are equal and opposite? ANSWER: 5 N / earth / book / upward 5 N / book / table / upward 5 N / book / earth / upward 5 N / earth / book / downward 5 N / table / book / upward 5 N / table / earth / upward 5 N / book / table / upward 5 N / table / book / downward 5 N / earth / book / downward Newton’s 1st or 2nd law Newton’s 3rd law Blocks in an Elevator Ranking Task Three blocks are stacked on top of each other inside an elevator as shown in the figure. Answer the following questions with reference to the eight forces defined as follows. the force of the 3 block on the 2 block, , the force of the 2 block on the 3 block, , the force of the 3 block on the 1 block, , the force of the 1 block on the 3 block, , the force of the 2 block on the 1 block, , the force of the 1 block on the 2 block, , the force of the 1 block on the floor, , and the force of the floor on the 1 block, . Part A Assume the elevator is at rest. Rank the magnitude of the forces. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: Newton’s 1st or 2nd law Newton’s 3rd law kg kg F3 on 2 kg kg F2 on 3 kg kg F3 on 1 kg kg F1 on 3 kg kg F2 on 1 kg kg F1 on 2 kg F1 on floor kg Ffloor on 1 Part B This question will be shown after you complete previous question(s). Newton’s 3rd Law Discussed Learning Goal: To understand Newton’s 3rd law, which states that a physical interaction always generates a pair of forces on the two interacting bodies. In Principia, Newton wrote: To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts. (translation by Cajori) The phrase after the colon (often omitted from textbooks) makes it clear that this is a statement about the nature of force. The central idea is that physical interactions (e.g., due to gravity, bodies touching, or electric forces) cause forces to arise between pairs of bodies. Each pairwise interaction produces a pair of opposite forces, one acting on each body. In summary, each physical interaction between two bodies generates a pair of forces. Whatever the physical cause of the interaction, the force on body A from body B is equal in magnitude and opposite in direction to the force on body B from body A. Incidentally, Newton states that the word “action” denotes both (a) the force due to an interaction and (b) the changes in momentum that it imparts to the two interacting bodies. If you haven’t learned about momentum, don’t worry; for now this is just a statement about the origin of forces. Mark each of the following statements as true or false. If a statement refers to “two bodies” interacting via some force, you are not to assume that these two bodies have the same mass. Part A Every force has one and only one 3rd law pair force. ANSWER: Part B The two forces in each pair act in opposite directions. ANSWER: Part C The two forces in each pair can either both act on the same body or they can act on different bodies. ANSWER: true false true false Part D The two forces in each pair may have different physical origins (for instance, one of the forces could be due to gravity, and its pair force could be due to friction or electric charge). ANSWER: Part E The two forces of a 3rd law pair always act on different bodies. ANSWER: Part F Given that two bodies interact via some force, the accelerations of these two bodies have the same magnitude but opposite directions. (Assume no other forces act on either body.) You did not open hints for this part. ANSWER: true false true false true false Part G According to Newton’s 3rd law, the force on the (smaller) moon due to the (larger) earth is ANSWER: Pulling Three Blocks Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force . The magnitude of the tension in the string between blocks B and C is = 3.00 . Assume that each block has mass = 0.400 . true false greater in magnitude and antiparallel to the force on the earth due to the moon. greater in magnitude and parallel to the force on the earth due to the moon. equal in magnitude but antiparallel to the force on the earth due to the moon. equal in magnitude and parallel to the force on the earth due to the moon. smaller in magnitude and antiparallel to the force on the earth due to the moon. smaller in magnitude and parallel to the force on the earth due to the moon. F T N m kg Part A What is the magnitude of the force? Express your answer numerically in newtons. You did not open hints for this part. ANSWER: Part B What is the tension in the string between block A and block B? Express your answer numerically in newtons You did not open hints for this part. ANSWER: Pulling Two Blocks In the situation shown in the figure, a person is pulling with a constant, nonzero force on string 1, which is attached to block A. Block A is also attached to block B via string 2, as shown. For this problem, assume that neither string stretches and that friction is negligible. Both blocks have finite (nonzero) mass. F F = N TAB TAB = N F Part A Which one of the following statements correctly descibes the relationship between the accelerations of blocks A and B? You did not open hints for this part. ANSWER: Part B How does the magnitude of the tension in string 1, , compare with the tension in string 2, ? You did not open hints for this part. Block A has a larger acceleration than block B. Block B has a larger acceleration than block A. Both blocks have the same acceleration. More information is needed to determine the relationship between the accelerations. T1 T2 ANSWER: Tension in a Massless Rope Learning Goal: To understand the concept of tension and the relationship between tension and force. This problem introduces the concept of tension. The example is a rope, oriented vertically, that is being pulled from both ends. Let and (with u for up and d for down) represent the magnitude of the forces acting on the top and bottom of the rope, respectively. Assume that the rope is massless, so that its weight is negligible compared with the tension. (This is not a ridiculous approximation–modern rope materials such as Kevlar can carry tensions thousands of times greater than the weight of tens of meters of such rope.) Consider the three sections of rope labeled a, b, and c in the figure. At point 1, a downward force of magnitude acts on section a. At point 1, an upward force of magnitude acts on section b. At point 1, the tension in the rope is . At point 2, a downward force of magnitude acts on section b. At point 2, an upward force of magnitude acts on section c. At point 2, the tension in the rope is . Assume, too, that the rope is at equilibrium. Part A What is the magnitude of the downward force on section a? Express your answer in terms of the tension . ANSWER: More information is needed to determine the relationship between and . T1 > T2 T1 = T2 T1 < T2 T1 T2 Fu Fd Fad Fbu T1 Fbd Fcu T2 Fad T1 Part B What is the magnitude of the upward force on section b? Express your answer in terms of the tension . ANSWER: Part C The magnitude of the upward force on c, , and the magnitude of the downward force on b, , are equal because of which of Newton's laws? ANSWER: Part D The magnitude of the force is ____ . ANSWER: Fad = Fbu T1 Fbu = Fcu Fbd 1st 2nd 3rd Fbu Fbd Part E Now consider the forces on the ends of the rope. What is the relationship between the magnitudes of these two forces? You did not open hints for this part. ANSWER: Part F The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straight line. For this situation, which of the following statements are true? Check all that apply. ANSWER: less than greater than equal to Fu > Fd Fu = Fd Fu < Fd The tension in the rope is everywhere the same. The magnitudes of the forces exerted on the two objects by the rope are the same. The forces exerted on the two objects by the rope must be in opposite directions. The forces exerted on the two objects by the rope must be in the direction of the rope. Two Hanging Masses Two blocks with masses and hang one under the other. For this problem, take the positive direction to be upward, and use for the magnitude of the acceleration due to gravity. Case 1: Blocks at rest For Parts A and B assume the blocks are at rest. Part A Find , the tension in the lower rope. Express your answer in terms of some or all of the variables , , and . You did not open hints for this part. ANSWER: M1 M2 g T2 M1 M2 g Part B Find , the tension in the upper rope. Express your answer in terms of some or all of the variables , , and . You did not open hints for this part. ANSWER: Case 2: Accelerating blocks For Parts C and D the blocks are now accelerating upward (due to the tension in the strings) with acceleration of magnitude . Part C Find , the tension in the lower rope. Express your answer in terms of some or all of the variables , , , and . You did not open hints for this part. ANSWER: T2 = T1 M1 M2 g T1 = a T2 M1 M2 a g Part D Find , the tension in the upper rope. Express your answer in terms of some or all of the variables , , , and . You did not open hints for this part. ANSWER: Video Tutor: Suspended Balls: Which String Breaks? First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point. T2 = T1 M1 M2 a g T1 = Part A A heavy crate is attached to the wall by a light rope, as shown in the figure. Another rope hangs off the opposite edge of the box. If you slowly increase the force on the free rope by pulling on it in a horizontal direction, which rope will break? Ignore friction and the mass of the ropes. 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. The rope attached to the wall will break. The rope that you are pulling on will break. Both ropes are equally likely to break.

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1 IN2009: Language Processors Coursework Part 3: The Interpreter Introduction This is the 3rd and final part of the coursework. In the second part of the coursework you created a parser for the Moopl grammar which, given a syntactically correct Moopl program as input, builds an AST representation of the program. In Part 3 you will develop an interpreter which executes Moopl programs by visiting their AST representations. For this part of the coursework we provide functional code (with limitations, see below) for parsing, building a symbol table, type checking and variable allocation. Marks This part of the coursework is worth 12 of the 30 coursework marks for the Language Processors module. This part of the coursework is marked out of 12. Submission deadline This part of the coursework should be handed in before 5pm on Sunday 9th April 2017. In line with school policy, late submissions will be awarded no marks. Return & Feedback Marks and feedback will be available as soon as possible, certainly on or before Wed 3rd May 2017. Plagiarism If you copy the work of others (either that of fellow students or of a third party), with or without their permission, you will score no marks and further disciplinary action will be taken against you. Group working You will be working in the same groups as for the previous parts of the coursework except where group changes have already been approved. Submission: Submit a zip archive (not a rar file) of all your source code (the src folder of your project). We do not want the other parts of your NetBeans project, only the source code. Note 1: Submissions which do not compile will get zero marks. Note 2: You must not change the names or types of any of the existing packages, classes or public methods. 2 Getting started Download either moopl-interp.zip or moopl-interp.tgz from Moodle and extract all files. Key contents to be aware of: • A fully implemented Moopl parser (also implements a parser for the interpreter command language; see below). • A partially implemented Moopl type checker. • Test harnesses for the type checker and interpreter. • A directory of a few example Moopl programs (see Testing below). • Folder interp containing prototype interpreter code. The type-checker is only partially implemented but a more complete implementation will be provided following Session 6. That version is still not fully complete because it doesn’t support inheritance. Part d) below asks you to remove this restriction. The VarAllocator visitor in the interp package uses a simple implementation which only works for methods in which all parameter and local variable names are different. Part e) below asks you to remove this restriction. The three parts below should be attempted in sequence. When you have completed one part you should make a back-up copy of the work and keep it safe, in case you break it in your attempt at the next part. Be sure to test old functionality as well as new (regression testing). We will not assess multiple versions so, if your attempt at part d) or e) breaks previously working code, you may gain a better mark by submitting the earlier version for assessment. c) [8 marks] The Basic Interpreter: complete the implementation of the Interpreter visitor in the interp package. d) [2 marks] Inheritance: extend the type-checker, variable allocator and interpreter to support inheritance. e) [2 marks] Variable Allocation: extend the variable allocator to fully support blockstructure and lexical scoping, removing the requirement that all parameter and local variable names are different. Aim to minimise the number of local variable slots allocated in a stack frame. Note: variable and parameter names declared at the same scope level are still required to be different from each other (a method cannot have two different parameters called x, for example) and this is enforced by the existing typechecking code. But variables declared in different blocks (even when nested) can have the same name. Exceptions Your interpreter will only ever be run on Moopl code which is type-correct (and free from uninitialised local variables). But it is still possible that the Moopl code contains logical errors which may cause runtime errors (such as null-reference or array-bound errors). Your interpreter should throw a MooplRunTimeException with an appropriate error message in these cases. The only kind of exception your interpreter should ever throw is a MooplRunTimeException. 3 Testing The examples folder does not contain a comprehensive test-suite. You need to invent and run your own tests. The document Moopl compared with Java gives a concise summary of how Moopl programs are supposed to behave. You can (and should) also compare the behaviour of your interpreter with that of the online tool: https://smcse.city.ac.uk/student/sj353/langproc/Moopl.html (Note: the online tool checks for uninitialised local variables. Your implementation is not expected to do this.) To test your work, run the top-level Interpret harness, providing the name of a Moopl source file as a command-line argument. When run on a type-correct Moopl source file, Interpret will pretty-print the Moopl program then display a command prompt (>) at which you can enter one of the following commands: :quit This will quit the interpreter. :call main() This will call the top-level proc main, interpreted in the context defined by the Moopl program. (Any top-level proc can be called this way). :eval Exp ; This will evaluate expression Exp, interpreted in the context defined by the Moopl program, and print the result. Note the required terminating semi-colon. Testing your Expression visitors To unit-test your Exp visit methods, run the top-level Interpret harness on a complete Moopl program (though it can be trivial) and use the :eval command. For example, to test your visit methods for the Boolean-literals (ExpTrue and ExpFalse), you would enter the commands > :eval true ; > :eval false ; which should output 1 and 0, respectively. For the most basic cases, the Moopl program is essentially irrelevant (a single top-level proc with empty body may be sufficient). For other cases you will need to write programs containing class definitions (in order, for example, to test object creation and method call). Testing your Statement visitors To unit-test your Stm visit methods, write very simple Moopl programs, each with a top-level proc main() containing just a few lines of code. Run the top-level Interpret harness on these simple programs and enter the command > :call main() You will find a few examples to get you started in the folder examples/unittests. As for the Exp tests, simple cases can be tested using Moopl programs with just a main proc but for the more complex tests you will need to write Moopl programs containing class definitions. 4 Grading criteria Solutions will be graded according to their functional correctness, and the elegance of their implementation. Below are criteria that guide the award of marks. 70 – 100 [1st class] Work that meets all the requirements in full, constructed and presented to a professional standard. Showing evidence of independent reading, thinking and analysis. 60 – 69 [2:1] Work that makes a good attempt to address the requirements, realising all to some extent and most well. Well-structured and well presented. 50 – 59 [2:2] Work that attempts to address requirements realising all to some extent and some well but perhaps also including irrelevant or underdeveloped material. Structure and presentation may not always be clear. 40 – 49 [3rd class] Work that attempts to address the requirements but only realises them to some extent and may not include important elements or be completely accurate. Structure and presentation may lack clarity. 0 – 39 [fail] Unsatisfactory work that does not adequately address the requirements. Structure and presentation may be confused or incoherent.

1 IN2009: Language Processors Coursework Part 3: The Interpreter Introduction This is the 3rd and final part of the coursework. In the second part of the coursework you created a parser for the Moopl grammar which, given a syntactically correct Moopl program as input, builds an AST representation of the program. In Part 3 you will develop an interpreter which executes Moopl programs by visiting their AST representations. For this part of the coursework we provide functional code (with limitations, see below) for parsing, building a symbol table, type checking and variable allocation. Marks This part of the coursework is worth 12 of the 30 coursework marks for the Language Processors module. This part of the coursework is marked out of 12. Submission deadline This part of the coursework should be handed in before 5pm on Sunday 9th April 2017. In line with school policy, late submissions will be awarded no marks. Return & Feedback Marks and feedback will be available as soon as possible, certainly on or before Wed 3rd May 2017. Plagiarism If you copy the work of others (either that of fellow students or of a third party), with or without their permission, you will score no marks and further disciplinary action will be taken against you. Group working You will be working in the same groups as for the previous parts of the coursework except where group changes have already been approved. Submission: Submit a zip archive (not a rar file) of all your source code (the src folder of your project). We do not want the other parts of your NetBeans project, only the source code. Note 1: Submissions which do not compile will get zero marks. Note 2: You must not change the names or types of any of the existing packages, classes or public methods. 2 Getting started Download either moopl-interp.zip or moopl-interp.tgz from Moodle and extract all files. Key contents to be aware of: • A fully implemented Moopl parser (also implements a parser for the interpreter command language; see below). • A partially implemented Moopl type checker. • Test harnesses for the type checker and interpreter. • A directory of a few example Moopl programs (see Testing below). • Folder interp containing prototype interpreter code. The type-checker is only partially implemented but a more complete implementation will be provided following Session 6. That version is still not fully complete because it doesn’t support inheritance. Part d) below asks you to remove this restriction. The VarAllocator visitor in the interp package uses a simple implementation which only works for methods in which all parameter and local variable names are different. Part e) below asks you to remove this restriction. The three parts below should be attempted in sequence. When you have completed one part you should make a back-up copy of the work and keep it safe, in case you break it in your attempt at the next part. Be sure to test old functionality as well as new (regression testing). We will not assess multiple versions so, if your attempt at part d) or e) breaks previously working code, you may gain a better mark by submitting the earlier version for assessment. c) [8 marks] The Basic Interpreter: complete the implementation of the Interpreter visitor in the interp package. d) [2 marks] Inheritance: extend the type-checker, variable allocator and interpreter to support inheritance. e) [2 marks] Variable Allocation: extend the variable allocator to fully support blockstructure and lexical scoping, removing the requirement that all parameter and local variable names are different. Aim to minimise the number of local variable slots allocated in a stack frame. Note: variable and parameter names declared at the same scope level are still required to be different from each other (a method cannot have two different parameters called x, for example) and this is enforced by the existing typechecking code. But variables declared in different blocks (even when nested) can have the same name. Exceptions Your interpreter will only ever be run on Moopl code which is type-correct (and free from uninitialised local variables). But it is still possible that the Moopl code contains logical errors which may cause runtime errors (such as null-reference or array-bound errors). Your interpreter should throw a MooplRunTimeException with an appropriate error message in these cases. The only kind of exception your interpreter should ever throw is a MooplRunTimeException. 3 Testing The examples folder does not contain a comprehensive test-suite. You need to invent and run your own tests. The document Moopl compared with Java gives a concise summary of how Moopl programs are supposed to behave. You can (and should) also compare the behaviour of your interpreter with that of the online tool: https://smcse.city.ac.uk/student/sj353/langproc/Moopl.html (Note: the online tool checks for uninitialised local variables. Your implementation is not expected to do this.) To test your work, run the top-level Interpret harness, providing the name of a Moopl source file as a command-line argument. When run on a type-correct Moopl source file, Interpret will pretty-print the Moopl program then display a command prompt (>) at which you can enter one of the following commands: :quit This will quit the interpreter. :call main() This will call the top-level proc main, interpreted in the context defined by the Moopl program. (Any top-level proc can be called this way). :eval Exp ; This will evaluate expression Exp, interpreted in the context defined by the Moopl program, and print the result. Note the required terminating semi-colon. Testing your Expression visitors To unit-test your Exp visit methods, run the top-level Interpret harness on a complete Moopl program (though it can be trivial) and use the :eval command. For example, to test your visit methods for the Boolean-literals (ExpTrue and ExpFalse), you would enter the commands > :eval true ; > :eval false ; which should output 1 and 0, respectively. For the most basic cases, the Moopl program is essentially irrelevant (a single top-level proc with empty body may be sufficient). For other cases you will need to write programs containing class definitions (in order, for example, to test object creation and method call). Testing your Statement visitors To unit-test your Stm visit methods, write very simple Moopl programs, each with a top-level proc main() containing just a few lines of code. Run the top-level Interpret harness on these simple programs and enter the command > :call main() You will find a few examples to get you started in the folder examples/unittests. As for the Exp tests, simple cases can be tested using Moopl programs with just a main proc but for the more complex tests you will need to write Moopl programs containing class definitions. 4 Grading criteria Solutions will be graded according to their functional correctness, and the elegance of their implementation. Below are criteria that guide the award of marks. 70 – 100 [1st class] Work that meets all the requirements in full, constructed and presented to a professional standard. Showing evidence of independent reading, thinking and analysis. 60 – 69 [2:1] Work that makes a good attempt to address the requirements, realising all to some extent and most well. Well-structured and well presented. 50 – 59 [2:2] Work that attempts to address requirements realising all to some extent and some well but perhaps also including irrelevant or underdeveloped material. Structure and presentation may not always be clear. 40 – 49 [3rd class] Work that attempts to address the requirements but only realises them to some extent and may not include important elements or be completely accurate. Structure and presentation may lack clarity. 0 – 39 [fail] Unsatisfactory work that does not adequately address the requirements. Structure and presentation may be confused or incoherent.

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What are two important classes of regular-expression-based applications: lexical analyzers and text search.

What are two important classes of regular-expression-based applications: lexical analyzers and text search.

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Assignment 2 Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 2.6 Part A The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Is the object moving the slowest? Is the object moving the fastest? Is the object at rest? Drag the appropriate items to their respective bins. ANSWER: Correct Part B At which lettered point or points is the object moving to the negative direction? ANSWER: Correct Conceptual Question 2.7 The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Part A Is the object moving the fastest? ANSWER: A B C D E Correct Part B Is the object speeding up? ANSWER: Correct Part C Is the object moving to the left and turning around? ANSWER: A B C D E F A B C D E F Correct Kinematic Vocabulary One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings, among them velocity, acceleration, position, speed, and displacement. The series of questions in this problem is designed to get you to try to think of these quantities like a physicist. Answer the questions in this problem using words from the following list: A. position B. direction C. displacement D. coordinates E. velocity F. acceleration G. distance H. magnitude I. vector J. scalar K. components Part A Velocity differs from speed in that velocity indicates a particle’s __________ of motion. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part B Unlike speed, velocity is a __________ quantity. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part C A vector has, by definition, both __________ and direction. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part D Once you have selected a coordinate system, you can express a two-dimensional vector using a pair of quantities known collectively as __________. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part E Speed differs from velocity in the same way that __________ differs from displacement. Enter the letter from the list given in the problem introduction that best completes the sentence. Hint 1. Definition of displacement Displacement is the vector that indicates the difference of two positions (e.g., the final position from the initial position). Being a vector, it is independent of the coordinate system used to describe it (although its vector components depend on the coordinate system). ANSWER: Correct Part F Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate systems that are identical except that they have different origins. The __________ of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the __________ from A to B is/are the same as expressed in both coordinate systems. Type the letters from the list given in the problem introduction that best complete the sentence. Separate the letters with commas. There is more than one correct answer, but you should only enter one pair of comma-separated letters. For example, if the words “vector” and “scalar” fit best in the blanks, enter I,J. ANSWER: Correct The coordinates of a point will depend on the coordinate system that is chosen, but there are several other quantities that are independent of the choice of origin for a coordinate system: in particular, distance, displacement, direction, and velocity. In working physics problems, unless you are interested in the position of an object or event relative to a specific origin, you can usually choose the coordinate system origin to be wherever is most convenient or intuitive. Note that the vector indicating a displacement from A to B is usually represented as . Part G Identify the following physical quantities as scalars or vectors. ANSWER: rB A = rB − rA Correct Problem 2.4 The figure is the position-versus-time graph of a jogger. Part A What is the jogger’s velocity at = 10 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Part B What is the jogger’s velocity at = 25 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the jogger’s velocity at = 35 ? Express your answer to two significant figures and include the appropriate units. ANSWER: t s v = 1.3 ms t s v = 0 ms t s v = -5.0 ms Correct Analyzing Position versus Time Graphs: Conceptual Question Two cars travel on the parallel lanes of a two-lane road. The cars’ motions are represented by the position versus time graph shown in the figure. Answer the questions using the times from the graph indicated by letters. Part A At which of the times do the two cars pass each other? Hint 1. Two cars passing Two objects can pass each other only if they have the same position at the same time. ANSWER: Correct Part B Are the two cars traveling in the same direction when they pass each other? ANSWER: Correct Part C At which of the lettered times, if any, does car #1 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E None Cannot be determined yes no Correct Part D At which of the lettered times, if any, does car #2 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E none cannot be determined A B C D E none cannot be determined Correct Part E At which of the lettered times are the cars moving with nearly identical velocity? Hint 1. Determining Velocity from a Position versus Time Graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: Correct Problem 2.6 A particle starts from 10 at = 0 and moves with the velocity graph shown in the figure. A B C D E None Cannot be determined m t0 Part A Does this particle have a turning point? ANSWER: Correct Part B If so, at what time? Express your answer using two significant figures and include the appropriate units. ANSWER: Correct Part C What is the object’s position at = 2, 3, 4 ? Yes No t = 1.0 s t s Express your answers using two significant figures separated by commas. ANSWER: Correct Overcoming a Head Start Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance beyond the starting line at . The starting line is at . Car A travels at a constant speed . Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed , which is greater than . Part A How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities. Hint 1. Consider the kinematics relation Write an expression for the displacement of Car A from the starting line at a time after Car B starts. (Note that we are taking this time to be .) Answer in terms of , , , and for time, and take at the starting line. Hint 1. What is the acceleration of Car A? The acceleration of Car A is zero, so the general formula has at least one term equal to zero. ANSWER: Hint 2. What is the relation between the positions of the two cars? x2 , x3 , x4 = 10,16,26 m DA t = 0 x = 0 vA vB vA t t = 0 vA vB DA t x = 0 x(t) = x0 + v0t + (1/2)at2 xA(t) = DA + vAt The positions of the two cars are equal at time . Hint 3. Consider Car B’s position as a function of time Write down an expression for the position of Car B at time after starting. Give your answer in terms of any variables needed (use for time). ANSWER: ANSWER: Correct Part B How far from Car B’s starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities. (You may use as well.) Hint 1. Which expression should you use? Just use your expression for the position of either car after time , and substitute in the correct value for (found in the previous part). ANSWER: Correct tcatch t t xB(t) = vBt tcatch = DA vB−vA tcatch t = 0 tcatch dpass = vBDA vB−vA Problem 2.11 The figure shows the velocity graph of a particle moving along the x-axis. Its initial position is at . At = 2 , what are the particle’s (a) position, (b) velocity, and (c) acceleration? Part A Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Express your answer to two significant figures and include the appropriate units. ANSWER: x0 = 2 m t0 = 0 t s x = 6.0 m vx = 4.0 ms Correct Part C Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 2.13 A jet plane is cruising at 300 when suddenly the pilot turns the engines up to full throttle. After traveling 3.9 , the jet is moving with a speed of 400 . Part A What is the jet’s acceleration, assuming it to be a constant acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 2.20 A rock is tossed straight up with a velocity of 22 When it returns, it falls into a hole deep. You may want to review ( pages 51 – 54) . ax = 2.0 m s2 m/s km m/s a = 9.0 m s2 m/s 10 m For help with math skills, you may want to review: Quadratic Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. Part A What is the rock’s velocity as it hits the bottom of the hole? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the rock, including its launch point, initial direction, and end point in the hole. Choose a coordinate system, and indicate it on your picture. Where is ? What is the positive direction? What is the position of the launch point and the bottom of the hole? In this coordinate system, what is the sign of the initial velocity and the sign of the acceleration? Calling the launch time , what is the equation for as a function of time? What is the position at the bottom of the hole? This will lead to a quadratic equation for the time when the rock hits the bottom of the hole. The quadratic equation has two solutions for the time. Not all mathematical solutions make sense physically. Which solution makes sense physically in terms of the picture that you drew at the beginning? Keeping the same coordinate system, what is the velocity in the direction as a function of time? What is the velocity when the rock hits the bottom of the hole? ANSWER: Correct Part B How long is the rock in the air, from the instant it is released until it hits the bottom of the hole? Express your answer with the appropriate units. y = 0 m y t = 0 y y t y y v = -26.1 ms Hint 1. How to approach the problem How is the time the rock was in the air related to the time at which the rock hit the ground in Part A? ANSWER: Correct Enhanced EOC: Problem 2.23 A particle moving along the x-axis has its position described by the function 2.00 5.00 5.00 , where is in s. At = 4.00, what are the particle’s (a) position, (b) velocity, and (c) acceleration? You may want to review ( pages 38 – 42) . For help with math skills, you may want to review: Differentiation of Polynomial Functions t = 4.90 s x = ( t3 − t + ) m t t Part A Express your answer with the appropriate units. Hint 1. How to approach the problem Evaluate the position at time = 4.00 . ANSWER: Correct Part B Express your answer with the appropriate units. Hint 1. How to approach the problem How do you determine the velocity as a function of time, , from the position, ? What calculus operation do you have to perform? Once you have , how do you determine at a particular time? ANSWER: Correct Part C Express your answer with the appropriate units. t s 113 m v(t) x(t) v(t) v 91.0 ms Hint 1. How to approach the problem How do you determine the acceleration as a function of time, , from the velocity, ? What calculus operation do you have to perform? Once you have , how do you determine the acceleration at a particular time? ANSWER: Correct Problem 2.26 A particle’s position on the x-axis is given by the function 6.00 6.00 , where is in s. Part A Where is the particle when = 4.00 ? Express your answer with the appropriate units. ANSWER: Correct Problem 2.30 A particle’s velocity is described by the function = , where is in . a(t) v(t) a(t) 48.0 m s2 x = (t2 − t + ) m t vx m/s 1.00 m vx t2 − 7t + 7 m/s t s Part A How many turning points does the particle reach. Express your answer as an integer. ANSWER: Correct Part B At what times does the particle reach its turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct Part C What is the particle’s acceleration at each of the turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct 2 t1 , t2 = 5.8,1.2 s a1 , a2 = 4.6,-4.6 m/s2 Problem 2.49 A 200 weather rocket is loaded with 100 of fuel and fired straight up. It accelerates upward at 35 for 30 , then runs out of fuel. Ignore any air resistance effects. Part A What is the rocket’s maximum altitude? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How long is the rocket in the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Problem 2.52 A hotel elevator ascends with maximum speed of . Its acceleration and deceleration both have a magnitude of . Part A How far does the elevator move while accelerating to full speed from rest? kg kg m/s2 s h = 72 km t = 260 s 200 m 5 m/s 1.0 m/s2 Express your answer with the appropriate units. ANSWER: Correct Part B How long does it take to make the complete trip from bottom to top? Express your answer with the appropriate units. ANSWER: Answer Requested Components of Vectors Shown is a 10 by 10 grid, with coordinate axes x and y . The grid runs from -5 to 5 on both axes. Drawn on this grid are four vectors, labeled through . This problem will ask you various questions about these vectors. All answers should be in decimal notation, unless otherwise specified. 12.5 m 45.0 s A D Part A What is the x component of ? Express your answer to two significant figures. Hint 1. How to derive the component A component of a vector is its length (but with appropriate sign) along a particular coordinate axis, the axes being specfied in advance. You are asked for the component of that lies along the x axis, which is horizontal in this problem. Imagine two lines perpendicular to the x axis running from the head (end with the arrow) and tail of down to the x axis. The length of the x axis between the points where these lines intersect is the x component of . In this problem, the x component is the x coordinate at which the perpendicular from the head of the vector hits the origin (because the tail of the vector is at the origin). ANSWER: Correct Part B What is the y component of ? Express your answer to the nearest integer. ANSWER: Correct A A A A Ax = 2.5 A Ay = 3 Part C What is the y component of ? Express your answer to the nearest integer. Hint 1. Consider the direction Don’t forget the sign. ANSWER: Correct Part D What is the component of ? Express your answer to the nearest integer. Hint 1. How to find the start and end points of the vector components A vector is defined only by its magnitude and direction. The starting point of the vector is of no consequence to its definition. Therefore, you need to somehow eliminate the starting point from your answer. You can run two perpendiculars to the x axis, one from the head (end with the arrow) of , and another to the tail, with the x component being the difference between x coordinates of head and tail (negative if the tail is to the right of the head). Another way is to imagine bringing the tail of to the origin, and then using the same procedure you used before to find the components of and . This is equivalent to the previous method, but it might be easier to visualize. ANSWER: B By = -3 x C C C A B Cx = -2 Correct The following questions will ask you to give both components of vectors using the ordered pairs method. In this method, the x component is written first, followed by a comma, and then the y component. For example, the components of would be written 2.5,3 in ordered pair notation. The answers below are all integers, so estimate the components to the nearest whole number. Part E In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part F In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part G What is true about and ? Choose from the pulldown list below. A B Bx, By = 2,-3 D Dx, Dy = 2,-3 B D ANSWER: Correct Problem 3.6 Find x- and y-components of the following vectors. Part A Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Part B Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: They have different components and are not the same vectors. They have the same components but are not the same vectors. They are the same vectors. = (r 430m, 60& below positive x − axis) rx, ry = 210,-370 m v = (610m/s, 23& above positive x − axis) Correct Part C Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Problem 3.10 Part A Draw . Draw the vector with its tail at the origin. ANSWER: vx, vy = 560,240 m/s a = (7.3m/s2 , negative y − direction) ax, ay = 0,-7.3 m/s2 B = −4 + 4 ı ^  ^ Correct Part B Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct B B = 5.7 Part C Find the direction of . Express your answer using two significant figures. ANSWER: Correct Part D Draw . Draw the vector with its tail at the origin. ANSWER: B = 45 above the B negative x-axis & = (−2.0 − 1.0 ) cm r ı ^  ^ Correct Part E Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct r r = 2.2 cm Part F Find the direction of . ANSWER: Correct Part G Draw . Draw the vector with its tail at the origin. ANSWER: r = 26.6 below the r negative x-axis & = (−10 − 100 ) m/s v ı ^  ^ Correct Part H Find the magnitude of . Express your answer using four significant figures. ANSWER: Correct v v = 100.5 m/s Part I Find the direction of . ANSWER: Correct Part J Draw . Draw the vector with it’s tail at the origin. ANSWER: v = 84.3 below the v negative x-axis & = (20 + 10 ) m/ a ı ^  ^ s2 Correct Part K Find the magnitude of . ANSWER: Correct Part L a a = 22.4 m/s2 Find the direction of . ANSWER: Correct Problem 3.14 Let , , and . Part A What is the component form of vector ? ANSWER: Correct Part B What is the magnitude of vector ? ANSWER: a = 26.6 above the a positive x-axis & A = 5 − 2 ı ^  ^ B = −2 + 6 ı ^  ^ D = A − B D D = 7 − 8 ı ^  ^ D = −7 − 5 ı ^  ^ D = 7 + 8 ı ^  ^ D = 4 + 5 ı ^  ^ D Correct Part C What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.15 Let , , and . Part A Write vector in component form. ANSWER: D = 10.6 D  = 49 & below positive x-axis A = 4 − 2 ı ^  ^ B = −3 + 5 ı ^  ^ E = 4A + 2B E E = 10 + 2 ı ^  ^ E = + 10 ı ^  ^ E = −10 ^ E = 10 − 2 ı ^  ^ Correct Part B Draw vectors , , and . Draw the vectors with their tails at the origin. ANSWER: Correct Part C A B E What is the magnitude of vector ? Express your answer using two significant figures. ANSWER: Correct Part D What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.24 Part A What is the angle between vectors and in the figure? Express your answer with the appropriate units. E E = 10.0 E  = 11 & counterclockwise from positive direction of x-axis  E F ANSWER: Correct Part B Use components to determine the magnitude of . ANSWER: Correct Part C Use components to determine the direction of . Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 91.3%.  = 71.6 & G = E + F  G = 3.00 G = E + F   = 90.0 & You received 129.62 out of a possible total of 142 points.

Assignment 2 Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 2.6 Part A The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Is the object moving the slowest? Is the object moving the fastest? Is the object at rest? Drag the appropriate items to their respective bins. ANSWER: Correct Part B At which lettered point or points is the object moving to the negative direction? ANSWER: Correct Conceptual Question 2.7 The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Part A Is the object moving the fastest? ANSWER: A B C D E Correct Part B Is the object speeding up? ANSWER: Correct Part C Is the object moving to the left and turning around? ANSWER: A B C D E F A B C D E F Correct Kinematic Vocabulary One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings, among them velocity, acceleration, position, speed, and displacement. The series of questions in this problem is designed to get you to try to think of these quantities like a physicist. Answer the questions in this problem using words from the following list: A. position B. direction C. displacement D. coordinates E. velocity F. acceleration G. distance H. magnitude I. vector J. scalar K. components Part A Velocity differs from speed in that velocity indicates a particle’s __________ of motion. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part B Unlike speed, velocity is a __________ quantity. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part C A vector has, by definition, both __________ and direction. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part D Once you have selected a coordinate system, you can express a two-dimensional vector using a pair of quantities known collectively as __________. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part E Speed differs from velocity in the same way that __________ differs from displacement. Enter the letter from the list given in the problem introduction that best completes the sentence. Hint 1. Definition of displacement Displacement is the vector that indicates the difference of two positions (e.g., the final position from the initial position). Being a vector, it is independent of the coordinate system used to describe it (although its vector components depend on the coordinate system). ANSWER: Correct Part F Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate systems that are identical except that they have different origins. The __________ of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the __________ from A to B is/are the same as expressed in both coordinate systems. Type the letters from the list given in the problem introduction that best complete the sentence. Separate the letters with commas. There is more than one correct answer, but you should only enter one pair of comma-separated letters. For example, if the words “vector” and “scalar” fit best in the blanks, enter I,J. ANSWER: Correct The coordinates of a point will depend on the coordinate system that is chosen, but there are several other quantities that are independent of the choice of origin for a coordinate system: in particular, distance, displacement, direction, and velocity. In working physics problems, unless you are interested in the position of an object or event relative to a specific origin, you can usually choose the coordinate system origin to be wherever is most convenient or intuitive. Note that the vector indicating a displacement from A to B is usually represented as . Part G Identify the following physical quantities as scalars or vectors. ANSWER: rB A = rB − rA Correct Problem 2.4 The figure is the position-versus-time graph of a jogger. Part A What is the jogger’s velocity at = 10 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Part B What is the jogger’s velocity at = 25 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the jogger’s velocity at = 35 ? Express your answer to two significant figures and include the appropriate units. ANSWER: t s v = 1.3 ms t s v = 0 ms t s v = -5.0 ms Correct Analyzing Position versus Time Graphs: Conceptual Question Two cars travel on the parallel lanes of a two-lane road. The cars’ motions are represented by the position versus time graph shown in the figure. Answer the questions using the times from the graph indicated by letters. Part A At which of the times do the two cars pass each other? Hint 1. Two cars passing Two objects can pass each other only if they have the same position at the same time. ANSWER: Correct Part B Are the two cars traveling in the same direction when they pass each other? ANSWER: Correct Part C At which of the lettered times, if any, does car #1 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E None Cannot be determined yes no Correct Part D At which of the lettered times, if any, does car #2 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E none cannot be determined A B C D E none cannot be determined Correct Part E At which of the lettered times are the cars moving with nearly identical velocity? Hint 1. Determining Velocity from a Position versus Time Graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: Correct Problem 2.6 A particle starts from 10 at = 0 and moves with the velocity graph shown in the figure. A B C D E None Cannot be determined m t0 Part A Does this particle have a turning point? ANSWER: Correct Part B If so, at what time? Express your answer using two significant figures and include the appropriate units. ANSWER: Correct Part C What is the object’s position at = 2, 3, 4 ? Yes No t = 1.0 s t s Express your answers using two significant figures separated by commas. ANSWER: Correct Overcoming a Head Start Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance beyond the starting line at . The starting line is at . Car A travels at a constant speed . Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed , which is greater than . Part A How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities. Hint 1. Consider the kinematics relation Write an expression for the displacement of Car A from the starting line at a time after Car B starts. (Note that we are taking this time to be .) Answer in terms of , , , and for time, and take at the starting line. Hint 1. What is the acceleration of Car A? The acceleration of Car A is zero, so the general formula has at least one term equal to zero. ANSWER: Hint 2. What is the relation between the positions of the two cars? x2 , x3 , x4 = 10,16,26 m DA t = 0 x = 0 vA vB vA t t = 0 vA vB DA t x = 0 x(t) = x0 + v0t + (1/2)at2 xA(t) = DA + vAt The positions of the two cars are equal at time . Hint 3. Consider Car B’s position as a function of time Write down an expression for the position of Car B at time after starting. Give your answer in terms of any variables needed (use for time). ANSWER: ANSWER: Correct Part B How far from Car B’s starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities. (You may use as well.) Hint 1. Which expression should you use? Just use your expression for the position of either car after time , and substitute in the correct value for (found in the previous part). ANSWER: Correct tcatch t t xB(t) = vBt tcatch = DA vB−vA tcatch t = 0 tcatch dpass = vBDA vB−vA Problem 2.11 The figure shows the velocity graph of a particle moving along the x-axis. Its initial position is at . At = 2 , what are the particle’s (a) position, (b) velocity, and (c) acceleration? Part A Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Express your answer to two significant figures and include the appropriate units. ANSWER: x0 = 2 m t0 = 0 t s x = 6.0 m vx = 4.0 ms Correct Part C Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 2.13 A jet plane is cruising at 300 when suddenly the pilot turns the engines up to full throttle. After traveling 3.9 , the jet is moving with a speed of 400 . Part A What is the jet’s acceleration, assuming it to be a constant acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 2.20 A rock is tossed straight up with a velocity of 22 When it returns, it falls into a hole deep. You may want to review ( pages 51 – 54) . ax = 2.0 m s2 m/s km m/s a = 9.0 m s2 m/s 10 m For help with math skills, you may want to review: Quadratic Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. Part A What is the rock’s velocity as it hits the bottom of the hole? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the rock, including its launch point, initial direction, and end point in the hole. Choose a coordinate system, and indicate it on your picture. Where is ? What is the positive direction? What is the position of the launch point and the bottom of the hole? In this coordinate system, what is the sign of the initial velocity and the sign of the acceleration? Calling the launch time , what is the equation for as a function of time? What is the position at the bottom of the hole? This will lead to a quadratic equation for the time when the rock hits the bottom of the hole. The quadratic equation has two solutions for the time. Not all mathematical solutions make sense physically. Which solution makes sense physically in terms of the picture that you drew at the beginning? Keeping the same coordinate system, what is the velocity in the direction as a function of time? What is the velocity when the rock hits the bottom of the hole? ANSWER: Correct Part B How long is the rock in the air, from the instant it is released until it hits the bottom of the hole? Express your answer with the appropriate units. y = 0 m y t = 0 y y t y y v = -26.1 ms Hint 1. How to approach the problem How is the time the rock was in the air related to the time at which the rock hit the ground in Part A? ANSWER: Correct Enhanced EOC: Problem 2.23 A particle moving along the x-axis has its position described by the function 2.00 5.00 5.00 , where is in s. At = 4.00, what are the particle’s (a) position, (b) velocity, and (c) acceleration? You may want to review ( pages 38 – 42) . For help with math skills, you may want to review: Differentiation of Polynomial Functions t = 4.90 s x = ( t3 − t + ) m t t Part A Express your answer with the appropriate units. Hint 1. How to approach the problem Evaluate the position at time = 4.00 . ANSWER: Correct Part B Express your answer with the appropriate units. Hint 1. How to approach the problem How do you determine the velocity as a function of time, , from the position, ? What calculus operation do you have to perform? Once you have , how do you determine at a particular time? ANSWER: Correct Part C Express your answer with the appropriate units. t s 113 m v(t) x(t) v(t) v 91.0 ms Hint 1. How to approach the problem How do you determine the acceleration as a function of time, , from the velocity, ? What calculus operation do you have to perform? Once you have , how do you determine the acceleration at a particular time? ANSWER: Correct Problem 2.26 A particle’s position on the x-axis is given by the function 6.00 6.00 , where is in s. Part A Where is the particle when = 4.00 ? Express your answer with the appropriate units. ANSWER: Correct Problem 2.30 A particle’s velocity is described by the function = , where is in . a(t) v(t) a(t) 48.0 m s2 x = (t2 − t + ) m t vx m/s 1.00 m vx t2 − 7t + 7 m/s t s Part A How many turning points does the particle reach. Express your answer as an integer. ANSWER: Correct Part B At what times does the particle reach its turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct Part C What is the particle’s acceleration at each of the turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct 2 t1 , t2 = 5.8,1.2 s a1 , a2 = 4.6,-4.6 m/s2 Problem 2.49 A 200 weather rocket is loaded with 100 of fuel and fired straight up. It accelerates upward at 35 for 30 , then runs out of fuel. Ignore any air resistance effects. Part A What is the rocket’s maximum altitude? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How long is the rocket in the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Problem 2.52 A hotel elevator ascends with maximum speed of . Its acceleration and deceleration both have a magnitude of . Part A How far does the elevator move while accelerating to full speed from rest? kg kg m/s2 s h = 72 km t = 260 s 200 m 5 m/s 1.0 m/s2 Express your answer with the appropriate units. ANSWER: Correct Part B How long does it take to make the complete trip from bottom to top? Express your answer with the appropriate units. ANSWER: Answer Requested Components of Vectors Shown is a 10 by 10 grid, with coordinate axes x and y . The grid runs from -5 to 5 on both axes. Drawn on this grid are four vectors, labeled through . This problem will ask you various questions about these vectors. All answers should be in decimal notation, unless otherwise specified. 12.5 m 45.0 s A D Part A What is the x component of ? Express your answer to two significant figures. Hint 1. How to derive the component A component of a vector is its length (but with appropriate sign) along a particular coordinate axis, the axes being specfied in advance. You are asked for the component of that lies along the x axis, which is horizontal in this problem. Imagine two lines perpendicular to the x axis running from the head (end with the arrow) and tail of down to the x axis. The length of the x axis between the points where these lines intersect is the x component of . In this problem, the x component is the x coordinate at which the perpendicular from the head of the vector hits the origin (because the tail of the vector is at the origin). ANSWER: Correct Part B What is the y component of ? Express your answer to the nearest integer. ANSWER: Correct A A A A Ax = 2.5 A Ay = 3 Part C What is the y component of ? Express your answer to the nearest integer. Hint 1. Consider the direction Don’t forget the sign. ANSWER: Correct Part D What is the component of ? Express your answer to the nearest integer. Hint 1. How to find the start and end points of the vector components A vector is defined only by its magnitude and direction. The starting point of the vector is of no consequence to its definition. Therefore, you need to somehow eliminate the starting point from your answer. You can run two perpendiculars to the x axis, one from the head (end with the arrow) of , and another to the tail, with the x component being the difference between x coordinates of head and tail (negative if the tail is to the right of the head). Another way is to imagine bringing the tail of to the origin, and then using the same procedure you used before to find the components of and . This is equivalent to the previous method, but it might be easier to visualize. ANSWER: B By = -3 x C C C A B Cx = -2 Correct The following questions will ask you to give both components of vectors using the ordered pairs method. In this method, the x component is written first, followed by a comma, and then the y component. For example, the components of would be written 2.5,3 in ordered pair notation. The answers below are all integers, so estimate the components to the nearest whole number. Part E In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part F In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part G What is true about and ? Choose from the pulldown list below. A B Bx, By = 2,-3 D Dx, Dy = 2,-3 B D ANSWER: Correct Problem 3.6 Find x- and y-components of the following vectors. Part A Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Part B Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: They have different components and are not the same vectors. They have the same components but are not the same vectors. They are the same vectors. = (r 430m, 60& below positive x − axis) rx, ry = 210,-370 m v = (610m/s, 23& above positive x − axis) Correct Part C Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Problem 3.10 Part A Draw . Draw the vector with its tail at the origin. ANSWER: vx, vy = 560,240 m/s a = (7.3m/s2 , negative y − direction) ax, ay = 0,-7.3 m/s2 B = −4 + 4 ı ^  ^ Correct Part B Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct B B = 5.7 Part C Find the direction of . Express your answer using two significant figures. ANSWER: Correct Part D Draw . Draw the vector with its tail at the origin. ANSWER: B = 45 above the B negative x-axis & = (−2.0 − 1.0 ) cm r ı ^  ^ Correct Part E Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct r r = 2.2 cm Part F Find the direction of . ANSWER: Correct Part G Draw . Draw the vector with its tail at the origin. ANSWER: r = 26.6 below the r negative x-axis & = (−10 − 100 ) m/s v ı ^  ^ Correct Part H Find the magnitude of . Express your answer using four significant figures. ANSWER: Correct v v = 100.5 m/s Part I Find the direction of . ANSWER: Correct Part J Draw . Draw the vector with it’s tail at the origin. ANSWER: v = 84.3 below the v negative x-axis & = (20 + 10 ) m/ a ı ^  ^ s2 Correct Part K Find the magnitude of . ANSWER: Correct Part L a a = 22.4 m/s2 Find the direction of . ANSWER: Correct Problem 3.14 Let , , and . Part A What is the component form of vector ? ANSWER: Correct Part B What is the magnitude of vector ? ANSWER: a = 26.6 above the a positive x-axis & A = 5 − 2 ı ^  ^ B = −2 + 6 ı ^  ^ D = A − B D D = 7 − 8 ı ^  ^ D = −7 − 5 ı ^  ^ D = 7 + 8 ı ^  ^ D = 4 + 5 ı ^  ^ D Correct Part C What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.15 Let , , and . Part A Write vector in component form. ANSWER: D = 10.6 D  = 49 & below positive x-axis A = 4 − 2 ı ^  ^ B = −3 + 5 ı ^  ^ E = 4A + 2B E E = 10 + 2 ı ^  ^ E = + 10 ı ^  ^ E = −10 ^ E = 10 − 2 ı ^  ^ Correct Part B Draw vectors , , and . Draw the vectors with their tails at the origin. ANSWER: Correct Part C A B E What is the magnitude of vector ? Express your answer using two significant figures. ANSWER: Correct Part D What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.24 Part A What is the angle between vectors and in the figure? Express your answer with the appropriate units. E E = 10.0 E  = 11 & counterclockwise from positive direction of x-axis  E F ANSWER: Correct Part B Use components to determine the magnitude of . ANSWER: Correct Part C Use components to determine the direction of . Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 91.3%.  = 71.6 & G = E + F  G = 3.00 G = E + F   = 90.0 & You received 129.62 out of a possible total of 142 points.

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(b) Based on the lessons learned, best practices and any additional steps you came up with in part (a), what if project manager X then got a job at Bank of America. Would it be possible for him/her to implement lean in the banking industry based on experience from the previous positions held at the automotive plant and the pharmaceutical company? Please state yes or no and explain the logic clearly for the same. Also, explain the steps that project manager X could take to implement lean at Bank of America (in the service industry) [10 points] You can refer to your class notes and will also have to do research online for both parts (a) and (b). Please state all the references used for each question.

(b) Based on the lessons learned, best practices and any additional steps you came up with in part (a), what if project manager X then got a job at Bank of America. Would it be possible for him/her to implement lean in the banking industry based on experience from the previous positions held at the automotive plant and the pharmaceutical company? Please state yes or no and explain the logic clearly for the same. Also, explain the steps that project manager X could take to implement lean at Bank of America (in the service industry) [10 points] You can refer to your class notes and will also have to do research online for both parts (a) and (b). Please state all the references used for each question.

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

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

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Q3b.Explain the possible consequences if the above mentioned principles are not followed

Q3b.Explain the possible consequences if the above mentioned principles are not followed

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Chapter 10 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 A One-Dimensional Inelastic Collision Block 1, of mass = 3.70 , moves along a frictionless air track with speed = 15.0 . It collides with block 2, of mass = 19.0 , 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. You did not open hints for this part. ANSWER: m1 kg v1 m/s m2 kg pi 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: Part C What is the change in the two-block system’s kinetic energy due to the collision? Express your answer numerically in joules. You did not open hints for this part. ANSWER: pi = kg m/s vf vf = m/s K = Kfinal − Kinitial K = J Conservation of Energy Ranking Task Six pendulums of various masses are released from various heights above a tabletop, as shown in the figures below. All the pendulums have the same length and are mounted such that at the vertical position their lowest points are the height of the tabletop and just do not strike the tabletop when released. Assume that the size of each bob is negligible. Part A Rank each pendulum on the basis of its initial gravitational potential energy (before being released) relative to the tabletop. Rank from largest to smallest To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: m h Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Momentum and Kinetic Energy Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed and has mass . Object 2 moves with speed and has mass . Part A Which object has the larger magnitude of its momentum? You did not open hints for this part. ANSWER: Part B Which object has the larger kinetic energy? You did not open hints for this part. ANSWER: v1 = v m1 = 2m v2 = 2v m2 = m Object 1 has the greater magnitude of its momentum. Object 2 has the greater magnitude of its momentum. Both objects have the same magnitude of their momenta. Object 1 has the greater kinetic energy. Object 2 has the greater kinetic energy. The objects have the same kinetic energy. Projectile Motion and Conservation of Energy Ranking Task Part A Six baseball throws are shown below. In each case the baseball is thrown at the same initial speed and from the same height above the ground. Assume that the effects of air resistance are negligible. Rank these throws according to the speed of the baseball the instant before it hits the ground. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: H PSS 10.1 Conservation of Mechanical Energy Learning Goal: To practice Problem-Solving Strategy 10.1 for conservation of mechanical energy problems. Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 that makes an angle of 45 with the vertical, steps off his tree limb, and swings down and then up to Jane’s open arms. When he arrives, his vine makes an angle of 30 with the vertical. Determine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan’s speed just before he reaches Jane. You can ignore air resistance and the mass of the vine. PROBLEM-SOLVING STRATEGY 10.1 Conservation of mechanical energy MODEL: Choose a system without friction or other losses of mechanical energy. m   VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you’re trying to find. SOLVE: The mathematical representation is based on the law of conservation of mechanical energy: . ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model The problem does not involve friction, nor are there losses of mechanical energy, so conservation of mechanical energy applies. Model Tarzan and the vine as a pendulum. Visualize Part A Which of the following sketches can be used in drawing a before-and-after pictorial representation? ANSWER: Kf + Uf = Ki + Ui Solve Part B What is Tarzan’s speed just before he reaches Jane? Express your answer in meters per second to two significant figures. You did not open hints for this part. ANSWER: Assess Part C This question will be shown after you complete previous question(s). Bungee Jumping Diagram A Diagram B Diagram C Diagram D vf vf = m/s Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass , and the surface of the bridge is a height above the water. The bungee cord, which has length when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant . Kate doesn’t actually jump but simply steps off the edge of the bridge and falls straight downward. Kate’s height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use for the magnitude of the acceleration due to gravity. Part A How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn’t touch the water. Express the distance in terms of quantities given in the problem introduction. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Spinning Mass on a Spring An object of mass is attached to a spring with spring constant whose unstretched length is , and whose far end is fixed to a shaft that is rotating with angular speed . Neglect gravity and assume that the mass rotates with angular speed as shown. When solving this problem use an inertial coordinate system, as drawn here. m h L k g d = M k L Part A Given the angular speed , find the radius at which the mass rotates without moving toward or away from the origin. Express the radius in terms of , , , and . You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C R( ) k L M R( ) = This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). ± Baby Bounce with a Hooke One of the pioneers of modern science, Sir Robert Hooke (1635-1703), studied the elastic properties of springs and formulated the law that bears his name. Hooke found the relationship among the force a spring exerts, , the distance from equilibrium the end of the spring is displaced, , and a number called the spring constant (or, sometimes, the force constant of the spring). According to Hooke, the force of the spring is directly proportional to its displacement from equilibrium, or . In its scalar form, this equation is simply . The negative sign indicates that the force that the spring exerts and its displacement have opposite directions. The value of depends on the geometry and the material of the spring; it can be easily determined experimentally using this scalar equation. Toy makers have always been interested in springs for the entertainment value of the motion they produce. One well-known application is a baby bouncer,which consists of a harness seat for a toddler, attached to a spring. The entire contraption hooks onto the top of a doorway. The idea is for the baby to hang in the seat with his or her feet just touching the ground so that a good push up will get the baby bouncing, providing potentially hours of entertainment. F  x k F = −kx F = −kx k Part A The following chart and accompanying graph depict an experiment to determine the spring constant for a baby bouncer. Displacement from equilibrium, ( ) Force exerted on the spring, ( ) 0 0 0.005 2.5 0.010 5.0 0.015 7.5 0.020 10 What is the spring constant of the spring being tested for the baby bouncer? Express your answer to two significant figures in newtons per meter. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Shooting a ball into a box Two children are trying to shoot a marble of mass into a small box using a spring-loaded gun that is fixed on a table and shoots horizontally from the edge of the table. The edge of the table is a height above the top of the box (the height of which is negligibly small), and the center of the box is a distance from the edge of the table. x m F N k k = N/m m H d The spring has a spring constant . The first child compresses the spring a distance and finds that the marble falls short of its target by a horizontal distance . Part A By what distance, , should the second child compress the spring so that the marble lands in the middle of the box? (Assume that height of the box is negligible, so that there is no chance that the marble will hit the side of the box before it lands in the bottom.) Express the distance in terms of , , , , and . You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). k x1 d12 x2 m k g H d x2 = Elastic Collision in One Dimension Block 1, of mass , moves across a frictionless surface with speed . It collides elastically with block 2, of mass , which is at rest ( ). After the collision, block 1 moves with speed , while block 2 moves with speed . Assume that , so that after the collision, the two objects move off in the direction of the first object before the collision. Part A This collision is elastic. What quantities, if any, are conserved in this collision? You did not open hints for this part. ANSWER: Part B What is the final speed of block 1? m1 ui m2 vi = 0 uf vf m1 > m2 kinetic energy only momentum only kinetic energy and momentum uf Express in terms of , , and . You did not open hints for this part. ANSWER: Part C What is the final speed of block 2? Express in terms of , , and . You did not open hints for this part. ANSWER: Ballistic Pendulum In a ballistic pendulum an object of mass is fired with an initial speed at a pendulum bob. The bob has a mass , which is suspended by a rod of length and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement as shown . uf m1 m2 ui uf = vf vf m1 m2 ui vf = m v0 M L  Part A Find an expression for , the initial speed of the fired object. Express your answer in terms of some or all of the variables , , , and and the acceleration due to gravity, . You did not open hints for this part. ANSWER: Part B An experiment is done to compare the initial speed of bullets fired from different handguns: a 9.0 and a .44 caliber. The guns are fired into a 10- pendulum bob of length . Assume that the 9.0- bullet has a mass of 6.0 and the .44-caliber bullet has a mass of 12 . If the 9.0- bullet causes the pendulum to swing to a maximum angular displacement of 4.3 and the .44-caliber bullet causes a displacement of 10.1 , find the ratio of the initial speed of the 9.0- bullet to the speed of the .44-caliber bullet, . Express your answer numerically. You did not open hints for this part. ANSWER: v0 m M L  g v0 = mm kg L mm g g mm   mm (v /( 0 )9.0 v0)44 Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. (v0 )9.0/(v0 )44 =

Chapter 10 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 A One-Dimensional Inelastic Collision Block 1, of mass = 3.70 , moves along a frictionless air track with speed = 15.0 . It collides with block 2, of mass = 19.0 , 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. You did not open hints for this part. ANSWER: m1 kg v1 m/s m2 kg pi 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: Part C What is the change in the two-block system’s kinetic energy due to the collision? Express your answer numerically in joules. You did not open hints for this part. ANSWER: pi = kg m/s vf vf = m/s K = Kfinal − Kinitial K = J Conservation of Energy Ranking Task Six pendulums of various masses are released from various heights above a tabletop, as shown in the figures below. All the pendulums have the same length and are mounted such that at the vertical position their lowest points are the height of the tabletop and just do not strike the tabletop when released. Assume that the size of each bob is negligible. Part A Rank each pendulum on the basis of its initial gravitational potential energy (before being released) relative to the tabletop. Rank from largest to smallest To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: m h Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Momentum and Kinetic Energy Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed and has mass . Object 2 moves with speed and has mass . Part A Which object has the larger magnitude of its momentum? You did not open hints for this part. ANSWER: Part B Which object has the larger kinetic energy? You did not open hints for this part. ANSWER: v1 = v m1 = 2m v2 = 2v m2 = m Object 1 has the greater magnitude of its momentum. Object 2 has the greater magnitude of its momentum. Both objects have the same magnitude of their momenta. Object 1 has the greater kinetic energy. Object 2 has the greater kinetic energy. The objects have the same kinetic energy. Projectile Motion and Conservation of Energy Ranking Task Part A Six baseball throws are shown below. In each case the baseball is thrown at the same initial speed and from the same height above the ground. Assume that the effects of air resistance are negligible. Rank these throws according to the speed of the baseball the instant before it hits the ground. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: H PSS 10.1 Conservation of Mechanical Energy Learning Goal: To practice Problem-Solving Strategy 10.1 for conservation of mechanical energy problems. Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 that makes an angle of 45 with the vertical, steps off his tree limb, and swings down and then up to Jane’s open arms. When he arrives, his vine makes an angle of 30 with the vertical. Determine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan’s speed just before he reaches Jane. You can ignore air resistance and the mass of the vine. PROBLEM-SOLVING STRATEGY 10.1 Conservation of mechanical energy MODEL: Choose a system without friction or other losses of mechanical energy. m   VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you’re trying to find. SOLVE: The mathematical representation is based on the law of conservation of mechanical energy: . ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model The problem does not involve friction, nor are there losses of mechanical energy, so conservation of mechanical energy applies. Model Tarzan and the vine as a pendulum. Visualize Part A Which of the following sketches can be used in drawing a before-and-after pictorial representation? ANSWER: Kf + Uf = Ki + Ui Solve Part B What is Tarzan’s speed just before he reaches Jane? Express your answer in meters per second to two significant figures. You did not open hints for this part. ANSWER: Assess Part C This question will be shown after you complete previous question(s). Bungee Jumping Diagram A Diagram B Diagram C Diagram D vf vf = m/s Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass , and the surface of the bridge is a height above the water. The bungee cord, which has length when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant . Kate doesn’t actually jump but simply steps off the edge of the bridge and falls straight downward. Kate’s height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use for the magnitude of the acceleration due to gravity. Part A How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn’t touch the water. Express the distance in terms of quantities given in the problem introduction. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Spinning Mass on a Spring An object of mass is attached to a spring with spring constant whose unstretched length is , and whose far end is fixed to a shaft that is rotating with angular speed . Neglect gravity and assume that the mass rotates with angular speed as shown. When solving this problem use an inertial coordinate system, as drawn here. m h L k g d = M k L Part A Given the angular speed , find the radius at which the mass rotates without moving toward or away from the origin. Express the radius in terms of , , , and . You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C R( ) k L M R( ) = This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). ± Baby Bounce with a Hooke One of the pioneers of modern science, Sir Robert Hooke (1635-1703), studied the elastic properties of springs and formulated the law that bears his name. Hooke found the relationship among the force a spring exerts, , the distance from equilibrium the end of the spring is displaced, , and a number called the spring constant (or, sometimes, the force constant of the spring). According to Hooke, the force of the spring is directly proportional to its displacement from equilibrium, or . In its scalar form, this equation is simply . The negative sign indicates that the force that the spring exerts and its displacement have opposite directions. The value of depends on the geometry and the material of the spring; it can be easily determined experimentally using this scalar equation. Toy makers have always been interested in springs for the entertainment value of the motion they produce. One well-known application is a baby bouncer,which consists of a harness seat for a toddler, attached to a spring. The entire contraption hooks onto the top of a doorway. The idea is for the baby to hang in the seat with his or her feet just touching the ground so that a good push up will get the baby bouncing, providing potentially hours of entertainment. F  x k F = −kx F = −kx k Part A The following chart and accompanying graph depict an experiment to determine the spring constant for a baby bouncer. Displacement from equilibrium, ( ) Force exerted on the spring, ( ) 0 0 0.005 2.5 0.010 5.0 0.015 7.5 0.020 10 What is the spring constant of the spring being tested for the baby bouncer? Express your answer to two significant figures in newtons per meter. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Shooting a ball into a box Two children are trying to shoot a marble of mass into a small box using a spring-loaded gun that is fixed on a table and shoots horizontally from the edge of the table. The edge of the table is a height above the top of the box (the height of which is negligibly small), and the center of the box is a distance from the edge of the table. x m F N k k = N/m m H d The spring has a spring constant . The first child compresses the spring a distance and finds that the marble falls short of its target by a horizontal distance . Part A By what distance, , should the second child compress the spring so that the marble lands in the middle of the box? (Assume that height of the box is negligible, so that there is no chance that the marble will hit the side of the box before it lands in the bottom.) Express the distance in terms of , , , , and . You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). k x1 d12 x2 m k g H d x2 = Elastic Collision in One Dimension Block 1, of mass , moves across a frictionless surface with speed . It collides elastically with block 2, of mass , which is at rest ( ). After the collision, block 1 moves with speed , while block 2 moves with speed . Assume that , so that after the collision, the two objects move off in the direction of the first object before the collision. Part A This collision is elastic. What quantities, if any, are conserved in this collision? You did not open hints for this part. ANSWER: Part B What is the final speed of block 1? m1 ui m2 vi = 0 uf vf m1 > m2 kinetic energy only momentum only kinetic energy and momentum uf Express in terms of , , and . You did not open hints for this part. ANSWER: Part C What is the final speed of block 2? Express in terms of , , and . You did not open hints for this part. ANSWER: Ballistic Pendulum In a ballistic pendulum an object of mass is fired with an initial speed at a pendulum bob. The bob has a mass , which is suspended by a rod of length and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement as shown . uf m1 m2 ui uf = vf vf m1 m2 ui vf = m v0 M L  Part A Find an expression for , the initial speed of the fired object. Express your answer in terms of some or all of the variables , , , and and the acceleration due to gravity, . You did not open hints for this part. ANSWER: Part B An experiment is done to compare the initial speed of bullets fired from different handguns: a 9.0 and a .44 caliber. The guns are fired into a 10- pendulum bob of length . Assume that the 9.0- bullet has a mass of 6.0 and the .44-caliber bullet has a mass of 12 . If the 9.0- bullet causes the pendulum to swing to a maximum angular displacement of 4.3 and the .44-caliber bullet causes a displacement of 10.1 , find the ratio of the initial speed of the 9.0- bullet to the speed of the .44-caliber bullet, . Express your answer numerically. You did not open hints for this part. ANSWER: v0 m M L  g v0 = mm kg L mm g g mm   mm (v /( 0 )9.0 v0)44 Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. (v0 )9.0/(v0 )44 =

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Essay – Athlete’s high salaries. Should they be paid that amount or not?

Essay – Athlete’s high salaries. Should they be paid that amount or not?

Athlete’s high salaries: Should they be paid that amount or … Read More...