In case the body have to stay in lower temperature for extended time period (more than 1 hour), how does the body regulate its response?

In case the body have to stay in lower temperature for extended time period (more than 1 hour), how does the body regulate its response?

Arterioles transporting blood to external capillaries beneath the surface of … Read More...
5) _____ The walls of a food storage facility refrigerator are made of a 2-cm-thick layer of wood (k=0.1 W/m-K) in contact with a 5-cm-thick of polyurethane foam (k= 0.03 W/m-K). If the temperature of the inner surface of the wood is -10 °C and the temperature of the outer surface of the polyurethane foam is 20 °C, the temperature where the two layers are in contact is: a) 11°C b) 8 °C c) 3°C d) -2°C e) -7°C

5) _____ The walls of a food storage facility refrigerator are made of a 2-cm-thick layer of wood (k=0.1 W/m-K) in contact with a 5-cm-thick of polyurethane foam (k= 0.03 W/m-K). If the temperature of the inner surface of the wood is -10 °C and the temperature of the outer surface of the polyurethane foam is 20 °C, the temperature where the two layers are in contact is: a) 11°C b) 8 °C c) 3°C d) -2°C e) -7°C

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Chapter 07 Homework Due: 11:59pm on Friday, May 23, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy BioFlix Quiz: The Carbon Cycle Watch the animation at left before answering the questions below. Part A An organism gets carbon by using carbon dioxide in the atmosphere to make sugar molecules. This organism is a Hint 1. Review the animation or your Study Sheet for The Carbon Cycle. ANSWER: Correct During photosynthesis, producers use carbon dioxide to make sugar molecules. Part B Which organisms play a role in returning carbon to the atmosphere? Hint 1. Review the animation or your Study Sheet for The Carbon Cycle. ANSWER: higher-level consumer. producer. primary consumer. decomposer. None of the above Consumers and decomposers, but not producers. Producers only. Decomposers only. Consumers only. Producers, consumers, and decomposers. Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 1 of 7 5/21/2014 8:02 PM Correct Producers, consumers, and decomposers all return carbon dioxide to the atmosphere during cellular respiration. Part C Every carbon atom in the organic molecules that make up your body MUST recently have been part of Hint 1. Review the animation or your Study Sheet for The Carbon Cycle. ANSWER: Correct You are a consumer, and all your carbon comes ultimately from plants and other producers. Part D Imagine following a single carbon atom through the carbon cycle. Which of the following is a possible path for the carbon atom to take? Hint 1. Review the animation or your Study Sheet for The Carbon Cycle. ANSWER: Correct Carbon moves from the atmosphere into a producer (such as a plant), up the food chain, and then back to the atmosphere during cellular respiration. Part E Which process or processes return carbon to the atmosphere? Hint 1. Review the animation. ANSWER: Correct Cellular respiration results in the release of carbon dioxide to the atmosphere. a higher-level consumer. a primary consumer. a decomposer. a producer. a sugar molecule made in one of your chloroplasts. The atmosphere; a plant; a higher-level consumer; then back to the atmosphere. The atmosphere; a plant; an herbivore; another plant; then back to the atmosphere. The atmosphere, a plant, a herbivore, a decomposer, then back to the atmosphere The atmosphere; a decomposer; a higher-level consumer; then back to the atmosphere. The atmosphere; a decomposer; then back to the atmosphere. Cellular respiration only Photosynthesis only Cellular respiration and photosynthesis Breakdown of large organic molecules into smaller organic molecules Cellular respiration and the breakdown of large organic molecules into smaller organic molecules Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 2 of 7 5/21/2014 8:02 PM Activity: The Nitrogen Cycle Click here to complete this activity. Then answer the questions. Part A Nitrifying bacteria convert _____ to _____. ANSWER: Correct Nitrifying bacteria convert ammonium to nitrites. Part B _____ removes nitrogen from the atmosphere. ANSWER: Correct Nitrogen fixation is the conversion of nitrogen gas to a form that can be used by plants (and other organisms). Part C Assimilation is indicated by the letter(s) _____. nitrogen gas … ammonium nitrogen gas … nitrates ammonium … nitrites nitrates … nitrogen gas ammonium … nitrogen gas Denitrification Nitrification Mineralization Nitrogen fixation Assimilation Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 3 of 7 5/21/2014 8:02 PM ANSWER: Correct Assimilation is the uptake of nutrients into an organism. Part D Nitrogen-fixing bacteria is(are) indicated by the letter(s) _____. ANSWER: Correct Both of these pointers are indicating nitrogen-fixing bacteria. Nitrogen fixation is the conversion of nitrogen to a form that plants can use. Part E Nitrification is indicated by the letter(s) _____. ANSWER: C B A D and E C and D B and C A and B D and E C and D A Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 4 of 7 5/21/2014 8:02 PM Correct Nitrification is the conversion of organic nitrogen-containing compounds to nitrites and nitrates. Part F Denitrifying bacteria convert _____ to _____. ANSWER: Correct Denitrifying bacteria convert nitrates to nitrogen gas. Part G Which one of these is a nitrate? ANSWER: Correct NO3 – is a nitrate. Part H Which one of these is a nitrite? ANSWER: Correct This is a nitrite. GeoScience: Earth’s Water and the Hydrologic Cycle A B B and C D and E B and E nitrogen gas … nitrites nitrogen gas … ammonium nitrates … nitrogen gas ammonium … nitrogen gas nitrogen gas … nitrates NO2 – NH4 – NH2 SH NO3 – PO4 – NH2 NH4 – NO2 – NO3 – Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 5 of 7 5/21/2014 8:02 PM When you have finished, answer the questions. Part A The largest percentage of fresh water today is located in: ANSWER: Correct Ice sheets and glaciers are the greatest single repository of fresh water: they contain 77.3% of all Earth’s fresh water and 99.357% of all Earth’s surface fresh water. Part B Earth’s oceans hold: ANSWER: Correct The oceans contain 97.22% of all water, comprising about 1.321 billion cubic kilometers of salt water. This leaves only 2.78% of all of Earth’s water as fresh water (non-oceanic). Part C Which of the following is true of the hydrologic cycle? ANSWER: Correct About 20% of the moisture evaporated from the ocean combines with 2% of land-derived moisture to produce 22% of all precipitation that falls over land. Clearly, the bulk of continental precipitation comes from the oceanic portion of the cycle. Concept Review: Eutrophication Can you sequence the steps in the eutrophication process that occurs in a body of water? Part A Drag each statement to the appropriate location in the flowchart of the eutrophication process. ANSWER: soil. ice sheets and glaciers. the rivers and lakes of the world. groundwater resources. about the same amount of water as all groundwater sources combined. most of the fresh water on Earth. the bulk of all of the water found on Earth. about the same amount of water as all Earth’s rivers and lakes combined. Atmospheric water and surface water do not mix. Most evaporation on Earth occurs over the continents. The bulk of the precipitation occurs over land. Most of the water that falls on the continents is derived from the oceans. Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 6 of 7 5/21/2014 8:02 PM Concept Review: Biogeochemical Cycles Can you sort the items by which biogeochemical cycle they apply to? Part A Drag each description to the appropriate bin. ANSWER: Score Summary: Your score on this assignment is 62.3%. You received 12.45 out of a possible total of 20 points. Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 7 of 7 5/21/2014 8:02 PM

Chapter 07 Homework Due: 11:59pm on Friday, May 23, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy BioFlix Quiz: The Carbon Cycle Watch the animation at left before answering the questions below. Part A An organism gets carbon by using carbon dioxide in the atmosphere to make sugar molecules. This organism is a Hint 1. Review the animation or your Study Sheet for The Carbon Cycle. ANSWER: Correct During photosynthesis, producers use carbon dioxide to make sugar molecules. Part B Which organisms play a role in returning carbon to the atmosphere? Hint 1. Review the animation or your Study Sheet for The Carbon Cycle. ANSWER: higher-level consumer. producer. primary consumer. decomposer. None of the above Consumers and decomposers, but not producers. Producers only. Decomposers only. Consumers only. Producers, consumers, and decomposers. Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 1 of 7 5/21/2014 8:02 PM Correct Producers, consumers, and decomposers all return carbon dioxide to the atmosphere during cellular respiration. Part C Every carbon atom in the organic molecules that make up your body MUST recently have been part of Hint 1. Review the animation or your Study Sheet for The Carbon Cycle. ANSWER: Correct You are a consumer, and all your carbon comes ultimately from plants and other producers. Part D Imagine following a single carbon atom through the carbon cycle. Which of the following is a possible path for the carbon atom to take? Hint 1. Review the animation or your Study Sheet for The Carbon Cycle. ANSWER: Correct Carbon moves from the atmosphere into a producer (such as a plant), up the food chain, and then back to the atmosphere during cellular respiration. Part E Which process or processes return carbon to the atmosphere? Hint 1. Review the animation. ANSWER: Correct Cellular respiration results in the release of carbon dioxide to the atmosphere. a higher-level consumer. a primary consumer. a decomposer. a producer. a sugar molecule made in one of your chloroplasts. The atmosphere; a plant; a higher-level consumer; then back to the atmosphere. The atmosphere; a plant; an herbivore; another plant; then back to the atmosphere. The atmosphere, a plant, a herbivore, a decomposer, then back to the atmosphere The atmosphere; a decomposer; a higher-level consumer; then back to the atmosphere. The atmosphere; a decomposer; then back to the atmosphere. Cellular respiration only Photosynthesis only Cellular respiration and photosynthesis Breakdown of large organic molecules into smaller organic molecules Cellular respiration and the breakdown of large organic molecules into smaller organic molecules Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 2 of 7 5/21/2014 8:02 PM Activity: The Nitrogen Cycle Click here to complete this activity. Then answer the questions. Part A Nitrifying bacteria convert _____ to _____. ANSWER: Correct Nitrifying bacteria convert ammonium to nitrites. Part B _____ removes nitrogen from the atmosphere. ANSWER: Correct Nitrogen fixation is the conversion of nitrogen gas to a form that can be used by plants (and other organisms). Part C Assimilation is indicated by the letter(s) _____. nitrogen gas … ammonium nitrogen gas … nitrates ammonium … nitrites nitrates … nitrogen gas ammonium … nitrogen gas Denitrification Nitrification Mineralization Nitrogen fixation Assimilation Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 3 of 7 5/21/2014 8:02 PM ANSWER: Correct Assimilation is the uptake of nutrients into an organism. Part D Nitrogen-fixing bacteria is(are) indicated by the letter(s) _____. ANSWER: Correct Both of these pointers are indicating nitrogen-fixing bacteria. Nitrogen fixation is the conversion of nitrogen to a form that plants can use. Part E Nitrification is indicated by the letter(s) _____. ANSWER: C B A D and E C and D B and C A and B D and E C and D A Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 4 of 7 5/21/2014 8:02 PM Correct Nitrification is the conversion of organic nitrogen-containing compounds to nitrites and nitrates. Part F Denitrifying bacteria convert _____ to _____. ANSWER: Correct Denitrifying bacteria convert nitrates to nitrogen gas. Part G Which one of these is a nitrate? ANSWER: Correct NO3 – is a nitrate. Part H Which one of these is a nitrite? ANSWER: Correct This is a nitrite. GeoScience: Earth’s Water and the Hydrologic Cycle A B B and C D and E B and E nitrogen gas … nitrites nitrogen gas … ammonium nitrates … nitrogen gas ammonium … nitrogen gas nitrogen gas … nitrates NO2 – NH4 – NH2 SH NO3 – PO4 – NH2 NH4 – NO2 – NO3 – Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 5 of 7 5/21/2014 8:02 PM When you have finished, answer the questions. Part A The largest percentage of fresh water today is located in: ANSWER: Correct Ice sheets and glaciers are the greatest single repository of fresh water: they contain 77.3% of all Earth’s fresh water and 99.357% of all Earth’s surface fresh water. Part B Earth’s oceans hold: ANSWER: Correct The oceans contain 97.22% of all water, comprising about 1.321 billion cubic kilometers of salt water. This leaves only 2.78% of all of Earth’s water as fresh water (non-oceanic). Part C Which of the following is true of the hydrologic cycle? ANSWER: Correct About 20% of the moisture evaporated from the ocean combines with 2% of land-derived moisture to produce 22% of all precipitation that falls over land. Clearly, the bulk of continental precipitation comes from the oceanic portion of the cycle. Concept Review: Eutrophication Can you sequence the steps in the eutrophication process that occurs in a body of water? Part A Drag each statement to the appropriate location in the flowchart of the eutrophication process. ANSWER: soil. ice sheets and glaciers. the rivers and lakes of the world. groundwater resources. about the same amount of water as all groundwater sources combined. most of the fresh water on Earth. the bulk of all of the water found on Earth. about the same amount of water as all Earth’s rivers and lakes combined. Atmospheric water and surface water do not mix. Most evaporation on Earth occurs over the continents. The bulk of the precipitation occurs over land. Most of the water that falls on the continents is derived from the oceans. Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 6 of 7 5/21/2014 8:02 PM Concept Review: Biogeochemical Cycles Can you sort the items by which biogeochemical cycle they apply to? Part A Drag each description to the appropriate bin. ANSWER: Score Summary: Your score on this assignment is 62.3%. You received 12.45 out of a possible total of 20 points. Chapter 07 Homework http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 7 of 7 5/21/2014 8:02 PM

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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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Assignment 11 Due: 11:59pm on Wednesday, April 30, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 13.2 The gravitational force of a star on orbiting planet 1 is . Planet 2, which is twice as massive as planet 1 and orbits at twice the distance from the star, experiences gravitational force . Part A What is the ratio ? ANSWER: Correct Conceptual Question 13.3 A 1500 satellite and a 2200 satellite follow exactly the same orbit around the earth. Part A What is the ratio of the force on the first satellite to that on the second satellite? ANSWER: Correct F1 F2 F1 F2 = 2 F1 F2 kg kg F1 F2 = 0.682 F1 F2 Part B What is the ratio of the acceleration of the first satellite to that of the second satellite? ANSWER: Correct Problem 13.2 The centers of a 15.0 lead ball and a 90.0 lead ball are separated by 9.00 . Part A What gravitational force does each exert on the other? Express your answer with the appropriate units. ANSWER: Correct Part B What is the ratio of this gravitational force to the weight of the 90.0 ball? ANSWER: a1 a2 = 1 a1 a2 kg g cm 1.11×10−8 N g 1.26×10−8 Typesetting math: 100% Correct Problem 13.6 The space shuttle orbits 310 above the surface of the earth. Part A What is the gravitational force on a 7.5 sphere inside the space shuttle? Express your answer with the appropriate units. ANSWER: Correct ± A Satellite in Orbit A satellite used in a cellular telephone network has a mass of 2310 and is in a circular orbit at a height of 650 above the surface of the earth. Part A What is the gravitational force on the satellite? Take the gravitational constant to be = 6.67×10−11 , the mass of the earth to be = 5.97×1024 , and the radius of the Earth to be = 6.38×106 . Express your answer in newtons. Hint 1. How to approach the problem Use the equation for the law of gravitation to calculate the force on the satellite. Be careful about the units when performing the calculations. km kg Fe on s = 67.0 N kg km Fgrav G N m2/kg2 me kg re m Typesetting math: 100% Hint 2. Law of gravitation According to Newton’s law of gravitation, , where is the gravitational constant, and are the masses of the two objects, and is the distance between the centers of mass of the two objects. Hint 3. Calculate the distance between the centers of mass What is the distance from the center of mass of the satellite to the center of mass of the earth? Express your answer in meters. ANSWER: ANSWER: Correct Part B What fraction is this of the satellite’s weight at the surface of the earth? Take the free-fall acceleration at the surface of the earth to be = 9.80 . Hint 1. How to approach the problem All you need to do is to take the ratio of the gravitational force on the satellite to the weight of the satellite at ground level. There are two ways to do this, depending on how you define the force of gravity at the surface of the earth. ANSWER: F = Gm1m2/r2 G m1 m2 r r = 7.03×10r 6 m = 1.86×10Fgrav 4 N g m/s2 0.824 Typesetting math: 100% Correct Although it is easy to find the weight of the satellite using the constant acceleration due to gravity, it is instructional to consider the weight calculated using the law of gravitation: . Dividing the gravitational force on the satellite by , we find that the ratio of the forces due to the earth’s gravity is simply the square of the ratio of the earth’s radius to the sum of the earth’s radius and the height of the orbit of the satellite above the earth, . This will also be the fraction of the weight of, say, an astronaut in an orbit at the same altitude. Notice that an astronaut’s weight is never zero. When people speak of “weightlessness” in space, what they really mean is “free fall.” Problem 13.8 Part A What is the free-fall acceleration at the surface of the moon? Express your answer with the appropriate units. ANSWER: Correct Part B What is the free-fall acceleration at the surface of the Jupiter? Express your answer with the appropriate units. ANSWER: Correct w = G m/ me r2e Fgrav = Gmem/(re + h)2 w [re/(re + h)]2 gmoon = 1.62 m s2 gJupiter = 25.9 m s2 Typesetting math: 100% Enhanced EOC: Problem 13.14 A rocket is launched straight up from the earth’s surface at a speed of 1.90×104 . You may want to review ( pages 362 – 365) . For help with math skills, you may want to review: Mathematical Expressions Involving Squares Part A What is its speed when it is very far away from the earth? Express your answer with the appropriate units. Hint 1. How to approach the problem What is conserved in this problem? What is the rocket’s initial kinetic energy in terms of its unknown mass, ? What is the rocket’s initial gravitational potential energy in terms of its unknown mass, ? When the rocket is very far away from the Earth, what is its gravitational potential energy? Using conservation of energy, what is the rocket’s kinetic energy when it is very far away from the Earth? Therefore, what is the rocket’s velocity when it is very far away from the Earth? ANSWER: Correct Problem 13.13 Part A m/s m m 1.54×104 ms Typesetting math: 100% What is the escape speed from Venus? Express your answer with the appropriate units. ANSWER: Correct Problem 13.17 The asteroid belt circles the sun between the orbits of Mars and Jupiter. One asteroid has a period of 4.2 earth years. Part A What is the asteroid’s orbital radius? Express your answer with the appropriate units. ANSWER: Correct Part B What is the asteroid’s orbital speed? Express your answer with the appropriate units. ANSWER: vescape = 10.4 km s = 3.89×1011 R m = 1.85×104 v ms Typesetting math: 100% Correct Problem 13.32 Part A At what height above the earth is the acceleration due to gravity 15.0% of its value at the surface? Express your answer with the appropriate units. ANSWER: Correct Part B What is the speed of a satellite orbiting at that height? Express your answer with the appropriate units. ANSWER: Correct Problem 13.36 Two meteoroids are heading for earth. Their speeds as they cross the moon’s orbit are 2 . 1.01×107 m 4920 ms km/s Typesetting math: 100% Part A The first meteoroid is heading straight for earth. What is its speed of impact? Express your answer with the appropriate units. ANSWER: Correct Part B The second misses the earth by 5500 . What is its speed at its closest point? Express your answer with the appropriate units. ANSWER: Incorrect; Try Again Problem 14.2 An air-track glider attached to a spring oscillates between the 11.0 mark and the 67.0 mark on the track. The glider completes 11.0 oscillations in 32.0 . Part A What is the period of the oscillations? Express your answer with the appropriate units. v1 = 11.3 km s km v2 = cm cm s Typesetting math: 100% ANSWER: Correct Part B What is the frequency of the oscillations? Express your answer with the appropriate units. ANSWER: Correct Part C What is the angular frequency of the oscillations? Express your answer with the appropriate units. ANSWER: Correct Part D What is the amplitude? Express your answer with the appropriate units. 2.91 s 0.344 Hz 2.16 rad s Typesetting math: 100% ANSWER: Correct Part E What is the maximum speed of the glider? Express your answer with the appropriate units. ANSWER: Correct Good Vibes: Introduction to Oscillations Learning Goal: To learn the basic terminology and relationships among the main characteristics of simple harmonic motion. Motion that repeats itself over and over is called periodic motion. There are many examples of periodic motion: the earth revolving around the sun, an elastic ball bouncing up and down, or a block attached to a spring oscillating back and forth. The last example differs from the first two, in that it represents a special kind of periodic motion called simple harmonic motion. The conditions that lead to simple harmonic motion are as follows: There must be a position of stable equilibrium. There must be a restoring force acting on the oscillating object. The direction of this force must always point toward the equilibrium, and its magnitude must be directly proportional to the magnitude of the object’s displacement from its equilibrium position. Mathematically, the restoring force is given by , where is the displacement from equilibrium and is a constant that depends on the properties of the oscillating system. The resistive forces in the system must be reasonably small. In this problem, we will introduce some of the basic quantities that describe oscillations and the relationships among them. Consider a block of mass attached to a spring with force constant , as shown in the figure. The spring can be either stretched or compressed. The block slides on a frictionless horizontal surface, as shown. When the spring is relaxed, the block is located at . If the 28.0 cm 60.5 cms F  F = −kx x k m k x = 0 Typesetting math: 100% block is pulled to the right a distance and then released, will be the amplitude of the resulting oscillations. Assume that the mechanical energy of the block-spring system remains unchanged in the subsequent motion of the block. Part A After the block is released from , it will ANSWER: Correct As the block begins its motion to the left, it accelerates. Although the restoring force decreases as the block approaches equilibrium, it still pulls the block to the left, so by the time the equilibrium position is reached, the block has gained some speed. It will, therefore, pass the equilibrium position and keep moving, compressing the spring. The spring will now be pushing the block to the right, and the block will slow down, temporarily coming to rest at . After is reached, the block will begin its motion to the right, pushed by the spring. The block will pass the equilibrium position and continue until it reaches , completing one cycle of motion. The motion will then repeat; if, as we’ve assumed, there is no friction, the motion will repeat indefinitely. The time it takes the block to complete one cycle is called the period. Usually, the period is denoted and is measured in seconds. The frequency, denoted , is the number of cycles that are completed per unit of time: . In SI units, is measured in inverse seconds, or hertz ( ). A A x = A remain at rest. move to the left until it reaches equilibrium and stop there. move to the left until it reaches and stop there. move to the left until it reaches and then begin to move to the right. x = −A x = −A x = −A x = −A x = A T f f = 1/T f Hz Typesetting math: 100% Part B If the period is doubled, the frequency is ANSWER: Correct Part C An oscillating object takes 0.10 to complete one cycle; that is, its period is 0.10 . What is its frequency ? Express your answer in hertz. ANSWER: Correct unchanged. doubled. halved. s s f f = 10 Hz Typesetting math: 100% Part D If the frequency is 40 , what is the period ? Express your answer in seconds. ANSWER: Correct The following questions refer to the figure that graphically depicts the oscillations of the block on the spring. Note that the vertical axis represents the x coordinate of the oscillating object, and the horizontal axis represents time. Part E Which points on the x axis are located a distance from the equilibrium position? ANSWER: Hz T T = 0.025 s A Typesetting math: 100% Correct Part F Suppose that the period is . Which of the following points on the t axis are separated by the time interval ? ANSWER: Correct Now assume for the remaining Parts G – J, that the x coordinate of point R is 0.12 and the t coordinate of point K is 0.0050 . Part G What is the period ? Express your answer in seconds. Hint 1. How to approach the problem In moving from the point to the point K, what fraction of a full wavelength is covered? Call that fraction . Then you can set . Dividing by the fraction will give the R only Q only both R and Q T T K and L K and M K and P L and N M and P m s T t = 0 a aT = 0.005 s a Typesetting math: 100% period . ANSWER: Correct Part H How much time does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement? Express your answer in seconds. ANSWER: Correct Part I What distance does the object cover during one period of oscillation? Express your answer in meters. ANSWER: Correct Part J What distance does the object cover between the moments labeled K and N on the graph? T T = 0.02 s t t = 0.01 s d d = 0.48 m d Typesetting math: 100% Express your answer in meters. ANSWER: Correct Problem 14.4 Part A What is the amplitude of the oscillation shown in the figure? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct d = 0.36 m A = 20.0 cm Typesetting math: 100% Part B What is the frequency of this oscillation? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the phase constant? Express your answer to two significant figures and include the appropriate units. ANSWER: Incorrect; Try Again Problem 14.10 An air-track glider attached to a spring oscillates with a period of 1.50 . At the glider is 4.60 left of the equilibrium position and moving to the right at 33.4 . Part A What is the phase constant? Express your answer to three significant figures and include the appropriate units. ANSWER: f = 0.25 Hz 0 = s t = 0 s cm cm/s Typesetting math: 100% Incorrect; Try Again 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). Part D This question will be shown after you complete previous question(s). Problem 14.12 A 140 air-track glider is attached to a spring. The glider is pushed in 12.2 and released. A student with a stopwatch finds that 14.0 oscillations take 19.0 . Part A What is the spring constant? Express your answer with the appropriate units. ANSWER: 0 = g cm s Typesetting math: 100% Correct Problem 14.14 The position of a 50 g oscillating mass is given by , where is in s. If necessary, round your answers to three significant figures. Determine: Part A The amplitude. Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part B The period. Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part C 3.00 Nm x(t) = (2.0 cm)cos(10t − /4) t 2.00 cm 0.628 s Typesetting math: 100% The spring constant. Express your answer to three significant figures and include the appropriate units. ANSWER: Part D The phase constant. Express your answer to three significant figures and include the appropriate units. ANSWER: Incorrect; Try Again Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G Typesetting math: 100% This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). Part I This question will be shown after you complete previous question(s). Enhanced EOC: Problem 14.17 A spring with spring constant 16 hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 4.0 and released. The ball makes 35 oscillations in 18 seconds. You may want to review ( pages 389 – 391) . For help with math skills, you may want to review: Differentiation of Trigonometric Functions Part A What is its the mass of the ball? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the period of oscillation? What is the angular frequency of the oscillations? How is the angular frequency related to the mass and spring constant? What is the mass? N/m cm s Typesetting math: 100% ANSWER: Correct Part B What is its maximum speed? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the amplitude of the oscillations? How is the maximum speed related to the amplitude of the oscillations and the angular frequency? ANSWER: Correct Changing the Period of a Pendulum A simple pendulum consisting of a bob of mass attached to a string of length swings with a period . Part A If the bob’s mass is doubled, approximately what will the pendulum’s new period be? Hint 1. Period of a simple pendulum The period of a simple pendulum of length is given by m = 110 g vmax = 49 cms m L T Typesetting math: 10T0% L , where is the acceleration due to gravity. ANSWER: Correct Part B If the pendulum is brought on the moon where the gravitational acceleration is about , approximately what will its period now be? Hint 1. How to approach the problem Recall the formula of the period of a simple pendulum. Since the gravitational acceleration appears in the denominator, the period must increase when the gravitational acceleration decreases. ANSWER: T = 2 Lg −−  g T/2 T &2T 2T g/6 T/6 T/&6 &6T 6T Typesetting math: 100% Correct Part C If the pendulum is taken into the orbiting space station what will happen to the bob? Hint 1. How to approach the problem Recall that the oscillations of a simple pendulum occur when a pendulum bob is raised above its equilibrium position and let go, causing the pendulum bob to fall. The gravitational force acts to bring the bob back to its equilibrium position. In the space station, the earth’s gravity acts on both the station and everything inside it, giving them the same acceleration. These objects are said to be in free fall. ANSWER: Correct In the space station, where all objects undergo the same acceleration due to the earth’s gravity, the tension in the string is zero and the bob does not fall relative to the point to which the string is attached. Problem 14.20 A 175 ball is tied to a string. It is pulled to an angle of 8.0 and released to swing as a pendulum. A student with a stopwatch finds that 15 oscillations take 13 . Part A How long is the string? Express your answer to two significant figures and include the appropriate units. It will continue to oscillate in a vertical plane with the same period. It will no longer oscillate because there is no gravity in space. It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall. It will oscillate much faster with a period that approaches zero. g ( s Typesetting math: 100% ANSWER: Correct Problem 14.22 Part A What is the length of a pendulum whose period on the moon matches the period of a 2.1- -long pendulum on the earth? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 14.42 An ultrasonic transducer, of the type used in medical ultrasound imaging, is a very thin disk ( = 0.17 ) driven back and forth in SHM at by an electromagnetic coil. Part A The maximum restoring force that can be applied to the disk without breaking it is 4.4×104 . What is the maximum oscillation amplitude that won’t rupture the disk? Express your answer to two significant figures and include the appropriate units. ANSWER: L = 19 cm m lmoon = 0.35 m m g 1.0 MHz N amax = 6.6 μm Typesetting math: 100% Correct Part B What is the disk’s maximum speed at this amplitude? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 81.4%. You received 117.25 out of a possible total of 144 points. vmax = 41 ms

Assignment 11 Due: 11:59pm on Wednesday, April 30, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 13.2 The gravitational force of a star on orbiting planet 1 is . Planet 2, which is twice as massive as planet 1 and orbits at twice the distance from the star, experiences gravitational force . Part A What is the ratio ? ANSWER: Correct Conceptual Question 13.3 A 1500 satellite and a 2200 satellite follow exactly the same orbit around the earth. Part A What is the ratio of the force on the first satellite to that on the second satellite? ANSWER: Correct F1 F2 F1 F2 = 2 F1 F2 kg kg F1 F2 = 0.682 F1 F2 Part B What is the ratio of the acceleration of the first satellite to that of the second satellite? ANSWER: Correct Problem 13.2 The centers of a 15.0 lead ball and a 90.0 lead ball are separated by 9.00 . Part A What gravitational force does each exert on the other? Express your answer with the appropriate units. ANSWER: Correct Part B What is the ratio of this gravitational force to the weight of the 90.0 ball? ANSWER: a1 a2 = 1 a1 a2 kg g cm 1.11×10−8 N g 1.26×10−8 Typesetting math: 100% Correct Problem 13.6 The space shuttle orbits 310 above the surface of the earth. Part A What is the gravitational force on a 7.5 sphere inside the space shuttle? Express your answer with the appropriate units. ANSWER: Correct ± A Satellite in Orbit A satellite used in a cellular telephone network has a mass of 2310 and is in a circular orbit at a height of 650 above the surface of the earth. Part A What is the gravitational force on the satellite? Take the gravitational constant to be = 6.67×10−11 , the mass of the earth to be = 5.97×1024 , and the radius of the Earth to be = 6.38×106 . Express your answer in newtons. Hint 1. How to approach the problem Use the equation for the law of gravitation to calculate the force on the satellite. Be careful about the units when performing the calculations. km kg Fe on s = 67.0 N kg km Fgrav G N m2/kg2 me kg re m Typesetting math: 100% Hint 2. Law of gravitation According to Newton’s law of gravitation, , where is the gravitational constant, and are the masses of the two objects, and is the distance between the centers of mass of the two objects. Hint 3. Calculate the distance between the centers of mass What is the distance from the center of mass of the satellite to the center of mass of the earth? Express your answer in meters. ANSWER: ANSWER: Correct Part B What fraction is this of the satellite’s weight at the surface of the earth? Take the free-fall acceleration at the surface of the earth to be = 9.80 . Hint 1. How to approach the problem All you need to do is to take the ratio of the gravitational force on the satellite to the weight of the satellite at ground level. There are two ways to do this, depending on how you define the force of gravity at the surface of the earth. ANSWER: F = Gm1m2/r2 G m1 m2 r r = 7.03×10r 6 m = 1.86×10Fgrav 4 N g m/s2 0.824 Typesetting math: 100% Correct Although it is easy to find the weight of the satellite using the constant acceleration due to gravity, it is instructional to consider the weight calculated using the law of gravitation: . Dividing the gravitational force on the satellite by , we find that the ratio of the forces due to the earth’s gravity is simply the square of the ratio of the earth’s radius to the sum of the earth’s radius and the height of the orbit of the satellite above the earth, . This will also be the fraction of the weight of, say, an astronaut in an orbit at the same altitude. Notice that an astronaut’s weight is never zero. When people speak of “weightlessness” in space, what they really mean is “free fall.” Problem 13.8 Part A What is the free-fall acceleration at the surface of the moon? Express your answer with the appropriate units. ANSWER: Correct Part B What is the free-fall acceleration at the surface of the Jupiter? Express your answer with the appropriate units. ANSWER: Correct w = G m/ me r2e Fgrav = Gmem/(re + h)2 w [re/(re + h)]2 gmoon = 1.62 m s2 gJupiter = 25.9 m s2 Typesetting math: 100% Enhanced EOC: Problem 13.14 A rocket is launched straight up from the earth’s surface at a speed of 1.90×104 . You may want to review ( pages 362 – 365) . For help with math skills, you may want to review: Mathematical Expressions Involving Squares Part A What is its speed when it is very far away from the earth? Express your answer with the appropriate units. Hint 1. How to approach the problem What is conserved in this problem? What is the rocket’s initial kinetic energy in terms of its unknown mass, ? What is the rocket’s initial gravitational potential energy in terms of its unknown mass, ? When the rocket is very far away from the Earth, what is its gravitational potential energy? Using conservation of energy, what is the rocket’s kinetic energy when it is very far away from the Earth? Therefore, what is the rocket’s velocity when it is very far away from the Earth? ANSWER: Correct Problem 13.13 Part A m/s m m 1.54×104 ms Typesetting math: 100% What is the escape speed from Venus? Express your answer with the appropriate units. ANSWER: Correct Problem 13.17 The asteroid belt circles the sun between the orbits of Mars and Jupiter. One asteroid has a period of 4.2 earth years. Part A What is the asteroid’s orbital radius? Express your answer with the appropriate units. ANSWER: Correct Part B What is the asteroid’s orbital speed? Express your answer with the appropriate units. ANSWER: vescape = 10.4 km s = 3.89×1011 R m = 1.85×104 v ms Typesetting math: 100% Correct Problem 13.32 Part A At what height above the earth is the acceleration due to gravity 15.0% of its value at the surface? Express your answer with the appropriate units. ANSWER: Correct Part B What is the speed of a satellite orbiting at that height? Express your answer with the appropriate units. ANSWER: Correct Problem 13.36 Two meteoroids are heading for earth. Their speeds as they cross the moon’s orbit are 2 . 1.01×107 m 4920 ms km/s Typesetting math: 100% Part A The first meteoroid is heading straight for earth. What is its speed of impact? Express your answer with the appropriate units. ANSWER: Correct Part B The second misses the earth by 5500 . What is its speed at its closest point? Express your answer with the appropriate units. ANSWER: Incorrect; Try Again Problem 14.2 An air-track glider attached to a spring oscillates between the 11.0 mark and the 67.0 mark on the track. The glider completes 11.0 oscillations in 32.0 . Part A What is the period of the oscillations? Express your answer with the appropriate units. v1 = 11.3 km s km v2 = cm cm s Typesetting math: 100% ANSWER: Correct Part B What is the frequency of the oscillations? Express your answer with the appropriate units. ANSWER: Correct Part C What is the angular frequency of the oscillations? Express your answer with the appropriate units. ANSWER: Correct Part D What is the amplitude? Express your answer with the appropriate units. 2.91 s 0.344 Hz 2.16 rad s Typesetting math: 100% ANSWER: Correct Part E What is the maximum speed of the glider? Express your answer with the appropriate units. ANSWER: Correct Good Vibes: Introduction to Oscillations Learning Goal: To learn the basic terminology and relationships among the main characteristics of simple harmonic motion. Motion that repeats itself over and over is called periodic motion. There are many examples of periodic motion: the earth revolving around the sun, an elastic ball bouncing up and down, or a block attached to a spring oscillating back and forth. The last example differs from the first two, in that it represents a special kind of periodic motion called simple harmonic motion. The conditions that lead to simple harmonic motion are as follows: There must be a position of stable equilibrium. There must be a restoring force acting on the oscillating object. The direction of this force must always point toward the equilibrium, and its magnitude must be directly proportional to the magnitude of the object’s displacement from its equilibrium position. Mathematically, the restoring force is given by , where is the displacement from equilibrium and is a constant that depends on the properties of the oscillating system. The resistive forces in the system must be reasonably small. In this problem, we will introduce some of the basic quantities that describe oscillations and the relationships among them. Consider a block of mass attached to a spring with force constant , as shown in the figure. The spring can be either stretched or compressed. The block slides on a frictionless horizontal surface, as shown. When the spring is relaxed, the block is located at . If the 28.0 cm 60.5 cms F  F = −kx x k m k x = 0 Typesetting math: 100% block is pulled to the right a distance and then released, will be the amplitude of the resulting oscillations. Assume that the mechanical energy of the block-spring system remains unchanged in the subsequent motion of the block. Part A After the block is released from , it will ANSWER: Correct As the block begins its motion to the left, it accelerates. Although the restoring force decreases as the block approaches equilibrium, it still pulls the block to the left, so by the time the equilibrium position is reached, the block has gained some speed. It will, therefore, pass the equilibrium position and keep moving, compressing the spring. The spring will now be pushing the block to the right, and the block will slow down, temporarily coming to rest at . After is reached, the block will begin its motion to the right, pushed by the spring. The block will pass the equilibrium position and continue until it reaches , completing one cycle of motion. The motion will then repeat; if, as we’ve assumed, there is no friction, the motion will repeat indefinitely. The time it takes the block to complete one cycle is called the period. Usually, the period is denoted and is measured in seconds. The frequency, denoted , is the number of cycles that are completed per unit of time: . In SI units, is measured in inverse seconds, or hertz ( ). A A x = A remain at rest. move to the left until it reaches equilibrium and stop there. move to the left until it reaches and stop there. move to the left until it reaches and then begin to move to the right. x = −A x = −A x = −A x = −A x = A T f f = 1/T f Hz Typesetting math: 100% Part B If the period is doubled, the frequency is ANSWER: Correct Part C An oscillating object takes 0.10 to complete one cycle; that is, its period is 0.10 . What is its frequency ? Express your answer in hertz. ANSWER: Correct unchanged. doubled. halved. s s f f = 10 Hz Typesetting math: 100% Part D If the frequency is 40 , what is the period ? Express your answer in seconds. ANSWER: Correct The following questions refer to the figure that graphically depicts the oscillations of the block on the spring. Note that the vertical axis represents the x coordinate of the oscillating object, and the horizontal axis represents time. Part E Which points on the x axis are located a distance from the equilibrium position? ANSWER: Hz T T = 0.025 s A Typesetting math: 100% Correct Part F Suppose that the period is . Which of the following points on the t axis are separated by the time interval ? ANSWER: Correct Now assume for the remaining Parts G – J, that the x coordinate of point R is 0.12 and the t coordinate of point K is 0.0050 . Part G What is the period ? Express your answer in seconds. Hint 1. How to approach the problem In moving from the point to the point K, what fraction of a full wavelength is covered? Call that fraction . Then you can set . Dividing by the fraction will give the R only Q only both R and Q T T K and L K and M K and P L and N M and P m s T t = 0 a aT = 0.005 s a Typesetting math: 100% period . ANSWER: Correct Part H How much time does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement? Express your answer in seconds. ANSWER: Correct Part I What distance does the object cover during one period of oscillation? Express your answer in meters. ANSWER: Correct Part J What distance does the object cover between the moments labeled K and N on the graph? T T = 0.02 s t t = 0.01 s d d = 0.48 m d Typesetting math: 100% Express your answer in meters. ANSWER: Correct Problem 14.4 Part A What is the amplitude of the oscillation shown in the figure? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct d = 0.36 m A = 20.0 cm Typesetting math: 100% Part B What is the frequency of this oscillation? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the phase constant? Express your answer to two significant figures and include the appropriate units. ANSWER: Incorrect; Try Again Problem 14.10 An air-track glider attached to a spring oscillates with a period of 1.50 . At the glider is 4.60 left of the equilibrium position and moving to the right at 33.4 . Part A What is the phase constant? Express your answer to three significant figures and include the appropriate units. ANSWER: f = 0.25 Hz 0 = s t = 0 s cm cm/s Typesetting math: 100% Incorrect; Try Again 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). Part D This question will be shown after you complete previous question(s). Problem 14.12 A 140 air-track glider is attached to a spring. The glider is pushed in 12.2 and released. A student with a stopwatch finds that 14.0 oscillations take 19.0 . Part A What is the spring constant? Express your answer with the appropriate units. ANSWER: 0 = g cm s Typesetting math: 100% Correct Problem 14.14 The position of a 50 g oscillating mass is given by , where is in s. If necessary, round your answers to three significant figures. Determine: Part A The amplitude. Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part B The period. Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part C 3.00 Nm x(t) = (2.0 cm)cos(10t − /4) t 2.00 cm 0.628 s Typesetting math: 100% The spring constant. Express your answer to three significant figures and include the appropriate units. ANSWER: Part D The phase constant. Express your answer to three significant figures and include the appropriate units. ANSWER: Incorrect; Try Again Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G Typesetting math: 100% This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). Part I This question will be shown after you complete previous question(s). Enhanced EOC: Problem 14.17 A spring with spring constant 16 hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 4.0 and released. The ball makes 35 oscillations in 18 seconds. You may want to review ( pages 389 – 391) . For help with math skills, you may want to review: Differentiation of Trigonometric Functions Part A What is its the mass of the ball? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the period of oscillation? What is the angular frequency of the oscillations? How is the angular frequency related to the mass and spring constant? What is the mass? N/m cm s Typesetting math: 100% ANSWER: Correct Part B What is its maximum speed? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the amplitude of the oscillations? How is the maximum speed related to the amplitude of the oscillations and the angular frequency? ANSWER: Correct Changing the Period of a Pendulum A simple pendulum consisting of a bob of mass attached to a string of length swings with a period . Part A If the bob’s mass is doubled, approximately what will the pendulum’s new period be? Hint 1. Period of a simple pendulum The period of a simple pendulum of length is given by m = 110 g vmax = 49 cms m L T Typesetting math: 10T0% L , where is the acceleration due to gravity. ANSWER: Correct Part B If the pendulum is brought on the moon where the gravitational acceleration is about , approximately what will its period now be? Hint 1. How to approach the problem Recall the formula of the period of a simple pendulum. Since the gravitational acceleration appears in the denominator, the period must increase when the gravitational acceleration decreases. ANSWER: T = 2 Lg −−  g T/2 T &2T 2T g/6 T/6 T/&6 &6T 6T Typesetting math: 100% Correct Part C If the pendulum is taken into the orbiting space station what will happen to the bob? Hint 1. How to approach the problem Recall that the oscillations of a simple pendulum occur when a pendulum bob is raised above its equilibrium position and let go, causing the pendulum bob to fall. The gravitational force acts to bring the bob back to its equilibrium position. In the space station, the earth’s gravity acts on both the station and everything inside it, giving them the same acceleration. These objects are said to be in free fall. ANSWER: Correct In the space station, where all objects undergo the same acceleration due to the earth’s gravity, the tension in the string is zero and the bob does not fall relative to the point to which the string is attached. Problem 14.20 A 175 ball is tied to a string. It is pulled to an angle of 8.0 and released to swing as a pendulum. A student with a stopwatch finds that 15 oscillations take 13 . Part A How long is the string? Express your answer to two significant figures and include the appropriate units. It will continue to oscillate in a vertical plane with the same period. It will no longer oscillate because there is no gravity in space. It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall. It will oscillate much faster with a period that approaches zero. g ( s Typesetting math: 100% ANSWER: Correct Problem 14.22 Part A What is the length of a pendulum whose period on the moon matches the period of a 2.1- -long pendulum on the earth? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 14.42 An ultrasonic transducer, of the type used in medical ultrasound imaging, is a very thin disk ( = 0.17 ) driven back and forth in SHM at by an electromagnetic coil. Part A The maximum restoring force that can be applied to the disk without breaking it is 4.4×104 . What is the maximum oscillation amplitude that won’t rupture the disk? Express your answer to two significant figures and include the appropriate units. ANSWER: L = 19 cm m lmoon = 0.35 m m g 1.0 MHz N amax = 6.6 μm Typesetting math: 100% Correct Part B What is the disk’s maximum speed at this amplitude? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 81.4%. You received 117.25 out of a possible total of 144 points. vmax = 41 ms

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The Geographic Grid The Geographic grid is based on angular measurements from the center of the earth. Latitude measures the angular distance north and south of the plane of the earth’s equator. longitude measures the angular distance east and west of the prime meridian. The angular distance between any two locations on the earth’s surface is simply the latitudinal or longitudinal difference the two locations. Both longitude and latitude are measured in degrees ( o ), minutes ( ‘ ) and seconds ( ” ) of arc. Figure 1 shows the latitude and longitude coordinates of the earth in 10o degree intervals. There are 60′ minutes of arc in 1o . There are 60″ seconds of arc in 1′ minute of arc. One could therefore express the latitude and longitude of a place as 39o 50′ 10″ N, 77o 35′ 15″ W. 1. Using latitude and longitude coordinates, determine the location of the following places. a) Toronto, Ontario ( Canada ) ____________________________________________________________ b) Billings, Montana _____________________________________________________________________ c) Chicago, Illinois ______________________________________________________________________ d) Westminster, England _________________________________________________________________ e) Venice, Italy _________________________________________________________________________ f) Baghdad, Iraq _______________________________________________________________________ g) Tokyo, Japan ________________________________________________________________________ h) Rio de Janeiro, Brazil _________________________________________________________________ 2. Provide the name of the following places located at the given coordinates below. a) 13o 09′ 50.19″ S, 72o 32′ 45.58″ W ______________________________________________________ b) 33o 51′ 35.90″ S, 151o 12′ 40″ E ______________________________________________________ c) 71o 17′ 07.62″ N, 156o 45′ 57.98″ W _____________________________________________________ d) 41o 53′ 29.84″ N, 87o 36′ 01.78″ W ______________________________________________________ The geographic grid uses circles of two different types, great circles and small circles. All great circles pass through the geographic center of the earth. All meridians, the equator, and the circle of illumination are great circles. All parallels other than the equator are small circles. The direction of any grid lines ( meridians and parallels ) can be determined as either an azimuth or a bearing. Azimuths are read in a clockwise direction as degrees ranging from 0o at the North Pole, to 90o at East, to 180o at the South Pole, to 270o at West, and back to 360o at the North Pole. Azimuths only give a direction such as 45o . Bearings are read as quadrants from either the North or South Poles. Hence east is 90o from either North or South. West is also 90o from either North or South. A bearing shows the direction one is traveling as well as the magnitude of the angle from either North or South. Hence a azimuth of 45o is read as a bearing of N 45o E. An azimuth of 150o would as a bearing read S 30o E. Figure 1.2 shows the relationship between azimuths and bearings. 3. Convert the values below. Bearing Azimuth 20o S 500 20′ E 265o 30’ N 20o 20 W Longitudinal and latitudinal distances vary as a result of trying to fit a flat grid onto a spherical surface such as the earth’s curved surface. The grid is constant in a north-south direction, but varies in the east-west direction. The data below shows how the grid values vary ( rounded off to the nearest mile of distance ). 1o of longitude at the equator = 69 miles 1o of longitude at the 30th parallel = 60 miles 1o of longitude at the 60th parallel = 35 miles 1o of longitude at the 90th parallel = 0 miles 1o of latitude along any meridian = 69 miles 4. Determine by calculation the linear ( miles ) and angular ( degree ) distances between the following places. From To Angular Linear Quito, Ecuador Macapa, Brazil 0o S, 78o W 0o N, 51o W Cairo, Egypt Shiraz, Iran 30o N, 31o E 30o N, 52o E Seward, Alaska St Petersburg, Russia 60o N, 149o W 60o N, 30o E Hamhung, North Korea Ankara, Turkey 39o N, 127o E 39o N, 32o E Detroit, Michigan Morristown, Tennessee 42o N, 83o W 36o N, 83o W Sample problem We see that between Quito and Macapa that there is no latitude difference as both places are at 0 degrees latitude. Therefore the angular difference between the two places can only be calculated based upon the longitudinal difference. Seeing that both places are west longitude we simply subtract 78-51 = 27 degrees. Hence the angular distance between Quito and Macapa is 27 degrees. The linear distance is the number of miles ( feet,yards, kilometers, etc.) Quito and Macapa. Bothe places are on the equator so consulting the table above we find that in 1 degree of longitude ate the equator is equal to 69 miles. So multiplying the 69 miles/degree X 27 degrees, the degrees cancel and the remaining units are miles, and the numerical value is 1,863 linear miles.

The Geographic Grid The Geographic grid is based on angular measurements from the center of the earth. Latitude measures the angular distance north and south of the plane of the earth’s equator. longitude measures the angular distance east and west of the prime meridian. The angular distance between any two locations on the earth’s surface is simply the latitudinal or longitudinal difference the two locations. Both longitude and latitude are measured in degrees ( o ), minutes ( ‘ ) and seconds ( ” ) of arc. Figure 1 shows the latitude and longitude coordinates of the earth in 10o degree intervals. There are 60′ minutes of arc in 1o . There are 60″ seconds of arc in 1′ minute of arc. One could therefore express the latitude and longitude of a place as 39o 50′ 10″ N, 77o 35′ 15″ W. 1. Using latitude and longitude coordinates, determine the location of the following places. a) Toronto, Ontario ( Canada ) ____________________________________________________________ b) Billings, Montana _____________________________________________________________________ c) Chicago, Illinois ______________________________________________________________________ d) Westminster, England _________________________________________________________________ e) Venice, Italy _________________________________________________________________________ f) Baghdad, Iraq _______________________________________________________________________ g) Tokyo, Japan ________________________________________________________________________ h) Rio de Janeiro, Brazil _________________________________________________________________ 2. Provide the name of the following places located at the given coordinates below. a) 13o 09′ 50.19″ S, 72o 32′ 45.58″ W ______________________________________________________ b) 33o 51′ 35.90″ S, 151o 12′ 40″ E ______________________________________________________ c) 71o 17′ 07.62″ N, 156o 45′ 57.98″ W _____________________________________________________ d) 41o 53′ 29.84″ N, 87o 36′ 01.78″ W ______________________________________________________ The geographic grid uses circles of two different types, great circles and small circles. All great circles pass through the geographic center of the earth. All meridians, the equator, and the circle of illumination are great circles. All parallels other than the equator are small circles. The direction of any grid lines ( meridians and parallels ) can be determined as either an azimuth or a bearing. Azimuths are read in a clockwise direction as degrees ranging from 0o at the North Pole, to 90o at East, to 180o at the South Pole, to 270o at West, and back to 360o at the North Pole. Azimuths only give a direction such as 45o . Bearings are read as quadrants from either the North or South Poles. Hence east is 90o from either North or South. West is also 90o from either North or South. A bearing shows the direction one is traveling as well as the magnitude of the angle from either North or South. Hence a azimuth of 45o is read as a bearing of N 45o E. An azimuth of 150o would as a bearing read S 30o E. Figure 1.2 shows the relationship between azimuths and bearings. 3. Convert the values below. Bearing Azimuth 20o S 500 20′ E 265o 30’ N 20o 20 W Longitudinal and latitudinal distances vary as a result of trying to fit a flat grid onto a spherical surface such as the earth’s curved surface. The grid is constant in a north-south direction, but varies in the east-west direction. The data below shows how the grid values vary ( rounded off to the nearest mile of distance ). 1o of longitude at the equator = 69 miles 1o of longitude at the 30th parallel = 60 miles 1o of longitude at the 60th parallel = 35 miles 1o of longitude at the 90th parallel = 0 miles 1o of latitude along any meridian = 69 miles 4. Determine by calculation the linear ( miles ) and angular ( degree ) distances between the following places. From To Angular Linear Quito, Ecuador Macapa, Brazil 0o S, 78o W 0o N, 51o W Cairo, Egypt Shiraz, Iran 30o N, 31o E 30o N, 52o E Seward, Alaska St Petersburg, Russia 60o N, 149o W 60o N, 30o E Hamhung, North Korea Ankara, Turkey 39o N, 127o E 39o N, 32o E Detroit, Michigan Morristown, Tennessee 42o N, 83o W 36o N, 83o W Sample problem We see that between Quito and Macapa that there is no latitude difference as both places are at 0 degrees latitude. Therefore the angular difference between the two places can only be calculated based upon the longitudinal difference. Seeing that both places are west longitude we simply subtract 78-51 = 27 degrees. Hence the angular distance between Quito and Macapa is 27 degrees. The linear distance is the number of miles ( feet,yards, kilometers, etc.) Quito and Macapa. Bothe places are on the equator so consulting the table above we find that in 1 degree of longitude ate the equator is equal to 69 miles. So multiplying the 69 miles/degree X 27 degrees, the degrees cancel and the remaining units are miles, and the numerical value is 1,863 linear miles.

F7.10 The flame spread rate through porous solids increases with concurrent wind velocity. decreases with concurrent wind velocity. is independent of concurrent wind velocity. F7.11 Surface tension accelerates opposed-flow flame spread over liquid fuels. True False F7.12 Opposed-flow flame spread rates over a solid surface are typically much smaller than 1 mm/s. around 1mm/s. much greater than 1 mm/s. F7.13 Upward flame spread rate over a vertical surface is typically between 10 and 1000 mm/s. True False F7.14 The Steiner tunnel test described in ASTM standard E 84 is used to assess the fire performance of interior finish materials based on lateral flame spread over a vertical sample. True False F8.1 Describe the triad of fire growth. F8.2 Liquid pool fires reach steady burning conditions within seconds after ignition. True False F8.3 The heat of gasification of liquid fuels is typically less than 1 kJ/g. between 1 and 3 kJ/g. greater than 3 kJ/g. F8.4 The heat flux from the flame to the surface of real burning objects can usually be determined with sufficient accuracy so that reasonable burning rate predictions can be made. True False F8.5 The mass burning flux generally associated with extinction is 0.5 g/m2s. 5 g/m2s. 50 g/m2s. F8.6 The mass burning flux of a liquid pool fire is a function of only the pool diameter. only the fuel type. pool diameter and fuel type. F8.7 The energy release rate of real objects can be measured in an oxygen bomb calorimeter. an oxygen consumption calorimeter. a room/corner test. F8.8 The peak energy release rate of typical domestic upholstered furniture can be as high as 3000 kW. True False F8.9 Draw a typical curve of the mass burning flux of a char forming fuel as a function of time. F8.10 A fast fire as defined in NFPA 72B grows proportionally to t2 and reaches an energy release rate of 1 MW in 75 sec. 150 sec. 300 sec. F9.1 Air entrainment into turbulent pool fire flames is due to buoyancy. True False F9.2 The frequency of vortex shedding in turbulent pool fire flames increases with pool diameter. decreases with pool diameter. is independent of pool diameter. F9.3 The height of turbulent jet flames for a given fuel type and orifice size is independent of energy release rate. True False F9.4 The exit velocities of fuel vapors leaving a solid or liquid pool fire surface are responsible for entrainment of air in the plume. True False F9.5 The height of a turbulent pool fire flame is a function of only energy release rate. only pool diameter. energy release rate and pool diameter. F9.6 Turbulent pool fire flame heights fluctuate in time within a factor of 2. True False F9.7 The Q* value for jet fires is 102 or greater. 104 or greater. 106 or greater. F9.8 The temperature in the continuous flame region of moderate size turbulent pool fires is approximately 820°C. True False F9.9 The temperature at the maximum flame height of a turbulent pool fire flame is approximately 1200°C. 800°C. 300°C. F9.10 The adiabatic flame temperature of hydrocarbon fuels is 1700-2000°C. 2000-2300°C. 2300-2600°C. F10.1 The stoichiometric air to fuel mass ratio of hydrocarbon fuels is of the order of 1.5 g/g. 15 g/g. 150 g/g. F10.2 Give two examples of products of incomplete combustion that occur in fires. F10.3 Slight amounts of products of incomplete combustion are generated in overventilated fires. True False F10.4 The CO yield of a fire is a function of only the fuel involved. only the ventilation conditions. the fuel and the ventilation conditions. F10.5 A carboxyhemoglobin level of 40% in the blood is usually lethal. True (doubt) False F10.6 Carbon monoxide is the leading killer of people in fires. True False F10.7 HCN is a narcotic gas. an irritant gas. a fuel vapor. F10.8 The hazard to humans from narcotic gases is a function of only the concentration of the gas. only the duration of exposure. the product of concentration and duration of exposure. F10.9 The effects on lethality of CO, HCN, and reduced O2 are additive. True False F10.10 Irritant gases typically cause post-exposure fatalities. True False F10.11 Visibility through smoke improves with increasing optical density. True False F10.12 Heat stress occurs when the skin is exposed to a heat flux of 1 kW/m2. the skin reaches a temperature of 45°C. the body’s core temperature reaches 41°C.

F7.10 The flame spread rate through porous solids increases with concurrent wind velocity. decreases with concurrent wind velocity. is independent of concurrent wind velocity. F7.11 Surface tension accelerates opposed-flow flame spread over liquid fuels. True False F7.12 Opposed-flow flame spread rates over a solid surface are typically much smaller than 1 mm/s. around 1mm/s. much greater than 1 mm/s. F7.13 Upward flame spread rate over a vertical surface is typically between 10 and 1000 mm/s. True False F7.14 The Steiner tunnel test described in ASTM standard E 84 is used to assess the fire performance of interior finish materials based on lateral flame spread over a vertical sample. True False F8.1 Describe the triad of fire growth. F8.2 Liquid pool fires reach steady burning conditions within seconds after ignition. True False F8.3 The heat of gasification of liquid fuels is typically less than 1 kJ/g. between 1 and 3 kJ/g. greater than 3 kJ/g. F8.4 The heat flux from the flame to the surface of real burning objects can usually be determined with sufficient accuracy so that reasonable burning rate predictions can be made. True False F8.5 The mass burning flux generally associated with extinction is 0.5 g/m2s. 5 g/m2s. 50 g/m2s. F8.6 The mass burning flux of a liquid pool fire is a function of only the pool diameter. only the fuel type. pool diameter and fuel type. F8.7 The energy release rate of real objects can be measured in an oxygen bomb calorimeter. an oxygen consumption calorimeter. a room/corner test. F8.8 The peak energy release rate of typical domestic upholstered furniture can be as high as 3000 kW. True False F8.9 Draw a typical curve of the mass burning flux of a char forming fuel as a function of time. F8.10 A fast fire as defined in NFPA 72B grows proportionally to t2 and reaches an energy release rate of 1 MW in 75 sec. 150 sec. 300 sec. F9.1 Air entrainment into turbulent pool fire flames is due to buoyancy. True False F9.2 The frequency of vortex shedding in turbulent pool fire flames increases with pool diameter. decreases with pool diameter. is independent of pool diameter. F9.3 The height of turbulent jet flames for a given fuel type and orifice size is independent of energy release rate. True False F9.4 The exit velocities of fuel vapors leaving a solid or liquid pool fire surface are responsible for entrainment of air in the plume. True False F9.5 The height of a turbulent pool fire flame is a function of only energy release rate. only pool diameter. energy release rate and pool diameter. F9.6 Turbulent pool fire flame heights fluctuate in time within a factor of 2. True False F9.7 The Q* value for jet fires is 102 or greater. 104 or greater. 106 or greater. F9.8 The temperature in the continuous flame region of moderate size turbulent pool fires is approximately 820°C. True False F9.9 The temperature at the maximum flame height of a turbulent pool fire flame is approximately 1200°C. 800°C. 300°C. F9.10 The adiabatic flame temperature of hydrocarbon fuels is 1700-2000°C. 2000-2300°C. 2300-2600°C. F10.1 The stoichiometric air to fuel mass ratio of hydrocarbon fuels is of the order of 1.5 g/g. 15 g/g. 150 g/g. F10.2 Give two examples of products of incomplete combustion that occur in fires. F10.3 Slight amounts of products of incomplete combustion are generated in overventilated fires. True False F10.4 The CO yield of a fire is a function of only the fuel involved. only the ventilation conditions. the fuel and the ventilation conditions. F10.5 A carboxyhemoglobin level of 40% in the blood is usually lethal. True (doubt) False F10.6 Carbon monoxide is the leading killer of people in fires. True False F10.7 HCN is a narcotic gas. an irritant gas. a fuel vapor. F10.8 The hazard to humans from narcotic gases is a function of only the concentration of the gas. only the duration of exposure. the product of concentration and duration of exposure. F10.9 The effects on lethality of CO, HCN, and reduced O2 are additive. True False F10.10 Irritant gases typically cause post-exposure fatalities. True False F10.11 Visibility through smoke improves with increasing optical density. True False F10.12 Heat stress occurs when the skin is exposed to a heat flux of 1 kW/m2. the skin reaches a temperature of 45°C. the body’s core temperature reaches 41°C.

F7.10 The flame spread rate through porous solids increases with … Read More...
MAE 214 – Fall 2015 Homework 3 Due: October 1, 2015 – Thursday by 1:00 p.m. Total Problems: 4 (including Extra Credit), Total Points: 105 1. Make a solid works part model from the given figure below. All dimensions are in millimeters. All sketches must be fully defined. Also create a drawing sheet and dimension it as shown. You can use a hole call out option under annotation to dimension a counter bore hole. (30 points) Save your part files as follows: My Documents/Homework 3 Folder/Prob1_LastName.SLDPRT My Documents/ Homework 3 Folder/Prob1_LastName.SLDDRW 2. Make a solid works part of the given figure below and also make a drawing sheet – front, top and right side views using 3rd angle projection method. Dimension the views with appropriate dimension technique. All dimensions are in mm. (30 points) Save your part file and drawing sheet as follows: Documents/Homework 3 folder/Problem 2_Last Name.SLDPRT Documents/Homework 3 folder/Problem 2_Last Name.SLDDRW 3. Make a solid works part file for the given figure below. All sketches must be fully defined. Your design tree menu must have advanced features i.e. plane, mirror, and fillet. The spot facing (SF) must be defined in a problem. The inclined cut must be created with an offset sketch and extrude cut or a suitable sketch that uses “up to surface” option. (40 points) Your part model must stick to the isometric view as it is shown here. Save your part file into: My documents/Homework 3 Folder/Problem3_Last Name.SLDPRT Given: A = 76 B = 127 Unit: MMGS ALL ROUNDS (FILLET) EQUAL 6 MM 4. (Extra Credit) Make a solid works part from the given figure below. All sketches must be fully defined. Save your part file to Documents/Homework3 Folder/Prob#4_Last Name.SLDPRT All dimensions are in millimeters. (5 points)

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

1 CE 321 PRINCIPLES ENVIRONMENTAL ENGINEERING LAB WORKSHEET No. 1 Due: One (1) Week After Each Lab Section, respectively MICROBIOLOGY Environmental engineers employ microbiology in a variety of applications. Testing for coliform bacteria is used to assess whether pathogens may be present in a water or wastewater sample. Coliforms are a type of bacteria that live in the intestines of warm blooded mammals, such as humans and cattle. They are not pathogens, but if they are present in a sample, it is taken as an indication that fecal material from humans or cattle has contacted the water. If fecal material is present, pathogens may be present, too. In water treatment, coliform counts must average less than one colony per 100 milliliters of sample tested. In wastewater treatment, typical acceptable levels might be 200-colonies/100 mL. There are two standard ways to test for coliforms, the Most Probable Number test, MPN (also called the multiple tube fermentation technique, MTF) and the membrane filter test, MF. Several companies market testing systems that are somewhat simpler, but these cannot be used by treatment plants until they receive EPA approval. Two recently accepted methods are the Minimal Media Test (Colilert system), and the Presence-Absence coliform test (P-A test). Wastewater treatment plant operators study the microorganism composition of the activated sludge units in order to assess and predict the performance of the biological floc. A sample of mixed liquor from the aeration basin is examined under the microscope, and based on the relative predominance of a variety of organisms that might be present; the operator can tell if the BOD application rates and wasting rates are as they should be. For your worksheet, please submit the items requested below (10 pts. each): 1. Examine a sample of activated sludge under the microscope (To be done together in class). Use the Atlas, Standard Methods, or other references to identify at least 5 different organisms you observed. List them and sketch them neatly on unlined paper. Describe their motility and any other distinctive characteristics as you observed it. 2. Explain what types of organisms you might expect to find in sludge with a high mean cell residence time (MCRT), and explain why these would predominate over the other types. 3. How can the predominance of a certain kind of microorganism in activated sludge affect the settling characteristics of the sludge? Give several examples. 2 4. Explain why coliforms are used as “indicator organisms” for water and wastewater testing. Name two pathogenic bacteria, two pathogenic viruses, and one pathogenic protozoan sometimes found in water supplies. 5. There is also a test for fecal coliforms. Use your class notes and outside references to explain the distinctions between the tests for total and fecal coliforms. Explain why one would use the fecal coliform test instead of the test for total coliforms. 6. Using outside references, indicate typical coliform limits for surface waters used for swimming and fishing; potable water; and wastewater treatment plant effluent. 7). In the recent past, EPA instituted regulations designed to insure that Giardia are removed from the water. Using your text or other references, explain what kind of organism this is, and explain the way in which EPA has set standards to insure they are removed during water treatment. 8. What is meant by “population dynamics”? What two factors usually control the population dynamics of a mixed culture? 9. Use the MPN test data from the samples prepared for class prior to determine the number of coliforms present in the wastewater samples. Please show your work and explain your reasoning. Total Coliforms Raw Intermediate Effluent Sample Volume No. Positive No. Positive No. Positive 10 5 5 4 1 5 5 2 0.1 5 3 1 0.01 5 1 0 0.001 2 1 —- 0.0001 1 —- —- FecalColiforms Raw Intermediate Effluent Sample Volume No. Positive No. Positive No. Positive 10 5 5 2 1 5 4 0 0.1 5 2 1 0.01 1 0 0 0.001 0 0 —– 0.0001 2 —– —– 3 10. Use the membrane filter test data given in class to determine the number of total coliforms and fecal coliforms present in the sample. Please show your work and explain your reasoning. Total Coliforms Fecal Coliforms Dilution Colonies Dilution Colonies Raw Influent 0.1 mL/100 mL 58 1 mL/100 mL 47 Intermediate 1 mL/100 mL 13 10 mL/100 mL 28 Wetland Effluent 10 mL/100 mL 10 100 mL/100 mL 15

1 CE 321 PRINCIPLES ENVIRONMENTAL ENGINEERING LAB WORKSHEET No. 1 Due: One (1) Week After Each Lab Section, respectively MICROBIOLOGY Environmental engineers employ microbiology in a variety of applications. Testing for coliform bacteria is used to assess whether pathogens may be present in a water or wastewater sample. Coliforms are a type of bacteria that live in the intestines of warm blooded mammals, such as humans and cattle. They are not pathogens, but if they are present in a sample, it is taken as an indication that fecal material from humans or cattle has contacted the water. If fecal material is present, pathogens may be present, too. In water treatment, coliform counts must average less than one colony per 100 milliliters of sample tested. In wastewater treatment, typical acceptable levels might be 200-colonies/100 mL. There are two standard ways to test for coliforms, the Most Probable Number test, MPN (also called the multiple tube fermentation technique, MTF) and the membrane filter test, MF. Several companies market testing systems that are somewhat simpler, but these cannot be used by treatment plants until they receive EPA approval. Two recently accepted methods are the Minimal Media Test (Colilert system), and the Presence-Absence coliform test (P-A test). Wastewater treatment plant operators study the microorganism composition of the activated sludge units in order to assess and predict the performance of the biological floc. A sample of mixed liquor from the aeration basin is examined under the microscope, and based on the relative predominance of a variety of organisms that might be present; the operator can tell if the BOD application rates and wasting rates are as they should be. For your worksheet, please submit the items requested below (10 pts. each): 1. Examine a sample of activated sludge under the microscope (To be done together in class). Use the Atlas, Standard Methods, or other references to identify at least 5 different organisms you observed. List them and sketch them neatly on unlined paper. Describe their motility and any other distinctive characteristics as you observed it. 2. Explain what types of organisms you might expect to find in sludge with a high mean cell residence time (MCRT), and explain why these would predominate over the other types. 3. How can the predominance of a certain kind of microorganism in activated sludge affect the settling characteristics of the sludge? Give several examples. 2 4. Explain why coliforms are used as “indicator organisms” for water and wastewater testing. Name two pathogenic bacteria, two pathogenic viruses, and one pathogenic protozoan sometimes found in water supplies. 5. There is also a test for fecal coliforms. Use your class notes and outside references to explain the distinctions between the tests for total and fecal coliforms. Explain why one would use the fecal coliform test instead of the test for total coliforms. 6. Using outside references, indicate typical coliform limits for surface waters used for swimming and fishing; potable water; and wastewater treatment plant effluent. 7). In the recent past, EPA instituted regulations designed to insure that Giardia are removed from the water. Using your text or other references, explain what kind of organism this is, and explain the way in which EPA has set standards to insure they are removed during water treatment. 8. What is meant by “population dynamics”? What two factors usually control the population dynamics of a mixed culture? 9. Use the MPN test data from the samples prepared for class prior to determine the number of coliforms present in the wastewater samples. Please show your work and explain your reasoning. Total Coliforms Raw Intermediate Effluent Sample Volume No. Positive No. Positive No. Positive 10 5 5 4 1 5 5 2 0.1 5 3 1 0.01 5 1 0 0.001 2 1 —- 0.0001 1 —- —- FecalColiforms Raw Intermediate Effluent Sample Volume No. Positive No. Positive No. Positive 10 5 5 2 1 5 4 0 0.1 5 2 1 0.01 1 0 0 0.001 0 0 —– 0.0001 2 —– —– 3 10. Use the membrane filter test data given in class to determine the number of total coliforms and fecal coliforms present in the sample. Please show your work and explain your reasoning. Total Coliforms Fecal Coliforms Dilution Colonies Dilution Colonies Raw Influent 0.1 mL/100 mL 58 1 mL/100 mL 47 Intermediate 1 mL/100 mL 13 10 mL/100 mL 28 Wetland Effluent 10 mL/100 mL 10 100 mL/100 mL 15

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Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

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