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

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

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Chapter 4 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Advice for the Quarterback A quarterback is set up to throw the football to a receiver who is running with a constant velocity directly away from the quarterback and is now a distance away from the quarterback. The quarterback figures that the ball must be thrown at an angle to the horizontal and he estimates that the receiver must catch the ball a time interval after it is thrown to avoid having opposition players prevent the receiver from making the catch. In the following you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is and that the horizontal position of the quaterback is . Use for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem. Part A Find , the vertical component of the velocity of the ball when the quarterback releases it. Express in terms of and . Hint 1. Equation of motion in y direction What is the expression for , the height of the ball as a function of time? Answer in terms of , , and . v r D  tc y = 0 x = 0 g v0y v0y tc g y(t) t g v0y ANSWER: Incorrect; Try Again Hint 2. Height at which the ball is caught, Remember that after time the ball was caught at the same height as it had been released. That is, . ANSWER: Answer Requested Part B Find , the initial horizontal component of velocity of the ball. Express your answer for in terms of , , and . Hint 1. Receiver’s position Find , the receiver’s position before he catches the ball. Answer in terms of , , and . ANSWER: Football’s position y(t) = v0yt− g 1 2 t2 y(tc) tc y(tc) = y0 = 0 v0y = gtc 2 v0x v0x D tc vr xr D vr tc xr = D + vrtc Typesetting math: 100% Find , the horizontal distance that the ball travels before reaching the receiver. Answer in terms of and . ANSWER: ANSWER: Answer Requested Part C Find the speed with which the quarterback must throw the ball. Answer in terms of , , , and . Hint 1. How to approach the problem Remember that velocity is a vector; from solving Parts A and B you have the two components, from which you can find the magnitude of this vector. ANSWER: Answer Requested Part D xc v0x tc xc = v0xtc v0x = + D tc vr v0 D tc vr g v0 = ( + ) + D tc vr 2 ( ) gtc 2 2 −−−−−−−−−−−−−−−−−−−  Typesetting math: 100% Assuming that the quarterback throws the ball with speed , find the angle above the horizontal at which he should throw it. Your solution should contain an inverse trig function (entered as asin, acos, or atan). Give your answer in terms of already known quantities, , , and . Hint 1. Find angle from and Think of velocity as a vector with Cartesian coordinates and . Find the angle that this vector would make with the x axis using the results of Parts A and B. ANSWER: Answer Requested Direction of Velocity at Various Times in Flight for Projectile Motion Conceptual Question For each of the motions described below, determine the algebraic sign (positive, negative, or zero) of the x component and y component of velocity of the object at the time specified. For all of the motions, the positive x axis points to the right and the positive y axis points upward. Alex, a mountaineer, must leap across a wide crevasse. The other side of the crevasse is below the point from which he leaps, as shown in the figure. Alex leaps horizontally and successfully makes the jump. v0  v0x v0y v0  v0x v0y v0xx^ v0yy^   = atan( ) v0y v0x Typesetting math: 100% Part A Determine the algebraic sign of Alex’s x velocity and y velocity at the instant he leaves the ground at the beginning of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Algebraic sign of velocity The algebraic sign of the velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this example, to the right is defined to be the positive x direction and upward the positive y direction. Therefore, any object moving to the right, whether speeding up, slowing down, or even simultaneously moving upward or downward, has a positive x velocity. Similarly, if the object is moving downward, regardless of any other aspect of its motion, its y velocity is negative. Hint 2. Sketch Alex’s initial velocity On the diagram below, sketch the vector representing Alex’s velocity the instant after he leaves the ground at the beginning of the jump. ANSWER: ANSWER: Typesetting math: 100% Answer Requested Part B Determine the algebraic signs of Alex’s x velocity and y velocity the instant before he lands at the end of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Sketch Alex’s final velocity On the diagram below, sketch the vector representing Alex’s velocity the instant before he safely lands on the other side of the crevasse. ANSWER: Answer Requested ANSWER: Answer Requested Typesetting math: 100% At the buzzer, a basketball player shoots a desperation shot. The ball goes in! Part C Determine the algebraic signs of the ball’s x velocity and y velocity the instant after it leaves the player’s hands. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s initial velocity On the diagram below, sketch the vector representing the velocity of the basketball the instant after it leaves the player’s hands. ANSWER: Typesetting math: 100% ANSWER: Correct Part D Determine the algebraic signs of the ball’s x velocity and y velocity at the ball’s maximum height. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s velocity at maximum height Typesetting math: 100% On the diagram below, sketch the vector representing the velocity of the basketball the instant it reaches its maximum height. ANSWER: ANSWER: Answer Requested PSS 4.1 Projectile Motion Problems Learning Goal: Typesetting math: 100% To practice Problem-Solving Strategy 4.1 for projectile motion problems. A rock thrown with speed 9.00 and launch angle 30.0 (above the horizontal) travels a horizontal distance of = 17.0 before hitting the ground. From what height was the rock thrown? Use the value = 9.810 for the free-fall acceleration. PROBLEM-SOLVING STRATEGY 4.1 Projectile motion problems MODEL: Make simplifying assumptions, such as treating the object as a particle. Is it reasonable to ignore air resistance? VISUALIZE: Use a pictorial representation. Establish a coordinate system with the x axis horizontal and the y axis vertical. Show important points in the motion on a sketch. Define symbols, and identify what you are trying to find. SOLVE: The acceleration is known: and . Thus, the problem becomes one of two-dimensional kinematics. The kinematic equations are , . is the same for the horizontal and vertical components of the motion. Find from one component, and then use that value for the other component. ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model Start by making simplifying assumptions: Model the rock as a particle in free fall. You can ignore air resistance because the rock is a relatively heavy object moving relatively slowly. Visualize Part A Which diagram represents an accurate sketch of the rock’s trajectory? Hint 1. The launch angle In a projectile’s motion, the angle of the initial velocity above the horizontal is called the launch angle. ANSWER: m/s  d m g m/s2 ax = 0 ay = −g xf = xi +vixt, yf = yi +viyt− g(t 1 2 )2 vfx = vix = constant, and vfy = viy − gt t t v i Typesetting math: 100% Typesetting math: 100% Correct Part B As stated in the strategy, choose a coordinate system where the x axis is horizontal and the y axis is vertical. Note that in the strategy, the y component of the projectile’s acceleration, , is taken to be negative. This implies that the positive y axis is upward. Use the same convention for your y axis, and take the positive x axis to be to the right. Where you choose your origin doesn’t change the answer to the question, but choosing an origin can make a problem easier to solve (even if only a bit). Usually it is nice if the majority of the quantities you are given and the quantity you are trying to solve for take positive values relative to your chosen origin. Given this goal, what location for the origin of the coordinate system would make this problem easiest? ANSWER: ay At ground level below the point where the rock is launched At the point where the rock strikes the ground At the peak of the trajectory At the point where the rock is released At ground level below the peak of the trajectory Typesetting math: 100% Correct It’s best to place the origin of the coordinate system at ground level below the launching point because in this way all the points of interest (the launching point and the landing point) will have positive coordinates. (Based on your experience, you know that it’s generally easier to work with positive coordinates.) Keep in mind, however, that this is an arbitrary choice. The correct solution of the problem will not depend on the location of the origin of your coordinate system. Now, define symbols representing initial and final position, velocity, and time. Your target variable is , the initial y coordinate of the rock. Your pictorial representation should be complete now, and similar to the picture below: Solve Part C Find the height from which the rock was launched. Express your answer in meters to three significant figures. yi yi Typesetting math: 100% Hint 1. How to approach the problem The time needed to move horizontally to the final position = 17.0 is the same time needed for the rock to rise from the initial position to the peak of its trajectory and then fall to the ground. Use the information you have about motion in the horizontal direction to solve for . Knowing this time will allow you to use the equations of motion for the vertical direction to solve for . Hint 2. Find the time spent in the air How long ( ) is the rock in the air? Express your answer in seconds to three significant figures. Hint 1. Determine which equation to use Which of the equations given in the strategy and shown below is the most appropriate to calculate the time the rock spent in the air? ANSWER: Hint 2. Find the x component of the initial velocity What is the x component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: t xf = d m yi t yi t t xf = xi + vixt yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt vix = 7.79 m/s Typesetting math: 100% Hint 3. Find the y component of the initial velocity What is the y component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: Answer Requested Assess Part D A second rock is thrown straight upward with a speed 4.500 . If this rock takes 2.181 to fall to the ground, from what height was it released? Express your answer in meters to three significant figures. Hint 1. Identify the known variables What are the values of , , , and for the second rock? Take the positive y axis to be upward and the origin to be located on the ground where the rock lands. Express your answers to four significant figures in the units shown to the right, separated by commas. ANSWER: t = 2.18 s viy = 4.50 m/s yi = 13.5 m m/s s H yf viy t a Typesetting math: 100% Answer Requested Hint 2. Determine which equation to use to find the height Which equation should you use to find ? Keep in mind that if the positive y axis is upward and the origin is located on the ground, . ANSWER: ANSWER: Answer Requested Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. Because both rocks were thrown with the same initial vertical velocity, 4.500 , and fell the same vertical distance of 13.5 , they were in the air for the same amount of time. This result was expected and helps to confirm that you did the calculation in Part C correctly. ± Arrow Hits Apple An arrow is shot at an angle of above the horizontal. The arrow hits a tree a horizontal distance away, at the same height above the ground as it was shot. Use for the magnitude of the acceleration due to gravity. Part A , , , = 0,4.500,2.181,-yf viy t a 9.810 m, m/s, s, m/s2 H yi = H yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt = − 2g( − ) v2f y v2i y yf yi H = 13.5 m viy = m/s m  = 45 D = 220 m g = 9.8 m/s2 Typesetting math: 100% Find , the time that the arrow spends in the air. Answer numerically in seconds, to two significant figures. Hint 1. Find the initial upward component of velocity in terms of D. Introduce the (unknown) variables and for the initial components of velocity. Then use kinematics to relate them and solve for . What is the vertical component of the initial velocity? Express your answer symbolically in terms of and . Hint 1. Find Find the horizontal component of the initial velocity. Express your answer symbolically in terms of and given symbolic quantities. ANSWER: Hint 2. Find What is the vertical component of the initial velocity? Express your answer symbolically in terms of . ANSWER: ANSWER: ta vy0 vx0 ta vy0 ta D vx0 vx0 ta vx0 = D ta vy0 vy0 vx0 vy0 = vx0 vy0 = D ta Typesetting math: 100% Hint 2. Find the time of flight in terms of the initial vertical component of velocity. From the change in the vertical component of velocity, you should be able to find in terms of and . Give your answer in terms of and . Hint 1. Find When applied to the y-component of velocity, in this problem the formula for with constant acceleration is What is , the vertical component of velocity when the arrow hits the tree? Answer symbolically in terms of only. ANSWER: ANSWER: Hint 3. Put the algebra together to find symbolically. If you have an expression for the initial vertical velocity component in terms in terms of and , and another in terms of and , you should be able to eliminate this initial component to find an expression for Express your answer symbolically in terms of given variables. ANSWER: ta vy0 g vy0 g vy(ta) v(t) −g vy(t) = vy0 − g t vy(ta ) vy0 vy(ta) = −vy0 ta = 2vy0 g ta D ta g ta ta2 t2 = a 2D g Typesetting math: 100% ANSWER: Answer Requested Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. Part B How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Express your answer numerically in seconds, to two significant figures. Hint 1. When should the apple be dropped The apple should be dropped at the time equal to the total time it takes the arrow to reach the tree minus the time it takes the apple to fall 6.0 meters. Hint 2. Find the time it takes for the apple to fall 6.0 meters How long does it take an apple to fall 6.0 meters? Express your answer numerically in seconds, to two significant figures. ANSWER: Answer Requested ANSWER: ta = 6.7 s tf = 1.1 s td = 5.6 s Typesetting math: 100% Answer Requested Video Tutor: Ball Fired Upward from Accelerating Cart 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 questions at right. You can watch the video again at any point. Part A Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How far behind the cart will the ball land, compared to the distance in the original experiment? Hint 1. Determine how long the ball is in the air How will doubling the initial upward speed of the ball change the time the ball spends in the air? A kinematic equation may be helpful here. The time in the air will ANSWER: be cut in half. stay the same. double. quadruple. Typesetting math: 100% Hint 2. Determine the appropriate kinematic expression Which of the following kinematic equations correctly describes the horizontal distance between the ball and the cart at the moment the ball lands? The cart’s initial horizontal velocity is , its horizontal acceleration is , and is the time elapsed between launch and impact. ANSWER: ANSWER: Correct The ball will spend twice as much time in the air ( , where is the ball’s initial upward velocity), so it will land four times farther behind the cart: (where is the cart’s horizontal acceleration). Video Tutor: Ball Fired Upward from Moving Cart 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 questions at right. You can watch the video again at any point. d v0x ax t d = v0x t d = 1 2 axv0x t2 d = v0x t+ 1 2 axt2 d = 1 2 axt2 the same distance twice as far half as far four times as far by a factor not listed above t = 2v0y/g v0y d = 1 2 axt2 ax Typesetting math: 100% Part A The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the crate? Ignore air resistance. Hint 1. How to approach the problem While the crate is on the plane, it shares the plane’s velocity. What is the crate’s velocity immediately after it is released? Hint 2. What affects the motion of the crate? Gravity will accelerate the crate downward. What, if anything, affects the crate’s horizontal motion? (Keep in mind that we are told to ignore air resistance, even though that’s not very realistic in this situation.) ANSWER: Correct At the moment it is released, the crate shares the plane’s horizontal velocity. In the absence of air resistance, the crate would remain directly below the plane as it fell. Score Summary: Your score on this assignment is 0%. Before the plane is directly over the target After the plane has flown over the target When the plane is directly over the target Typesetting math: 100% You received 0 out of a possible total of 0 points. Typesetting math: 100%

Chapter 4 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Advice for the Quarterback A quarterback is set up to throw the football to a receiver who is running with a constant velocity directly away from the quarterback and is now a distance away from the quarterback. The quarterback figures that the ball must be thrown at an angle to the horizontal and he estimates that the receiver must catch the ball a time interval after it is thrown to avoid having opposition players prevent the receiver from making the catch. In the following you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is and that the horizontal position of the quaterback is . Use for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem. Part A Find , the vertical component of the velocity of the ball when the quarterback releases it. Express in terms of and . Hint 1. Equation of motion in y direction What is the expression for , the height of the ball as a function of time? Answer in terms of , , and . v r D  tc y = 0 x = 0 g v0y v0y tc g y(t) t g v0y ANSWER: Incorrect; Try Again Hint 2. Height at which the ball is caught, Remember that after time the ball was caught at the same height as it had been released. That is, . ANSWER: Answer Requested Part B Find , the initial horizontal component of velocity of the ball. Express your answer for in terms of , , and . Hint 1. Receiver’s position Find , the receiver’s position before he catches the ball. Answer in terms of , , and . ANSWER: Football’s position y(t) = v0yt− g 1 2 t2 y(tc) tc y(tc) = y0 = 0 v0y = gtc 2 v0x v0x D tc vr xr D vr tc xr = D + vrtc Typesetting math: 100% Find , the horizontal distance that the ball travels before reaching the receiver. Answer in terms of and . ANSWER: ANSWER: Answer Requested Part C Find the speed with which the quarterback must throw the ball. Answer in terms of , , , and . Hint 1. How to approach the problem Remember that velocity is a vector; from solving Parts A and B you have the two components, from which you can find the magnitude of this vector. ANSWER: Answer Requested Part D xc v0x tc xc = v0xtc v0x = + D tc vr v0 D tc vr g v0 = ( + ) + D tc vr 2 ( ) gtc 2 2 −−−−−−−−−−−−−−−−−−−  Typesetting math: 100% Assuming that the quarterback throws the ball with speed , find the angle above the horizontal at which he should throw it. Your solution should contain an inverse trig function (entered as asin, acos, or atan). Give your answer in terms of already known quantities, , , and . Hint 1. Find angle from and Think of velocity as a vector with Cartesian coordinates and . Find the angle that this vector would make with the x axis using the results of Parts A and B. ANSWER: Answer Requested Direction of Velocity at Various Times in Flight for Projectile Motion Conceptual Question For each of the motions described below, determine the algebraic sign (positive, negative, or zero) of the x component and y component of velocity of the object at the time specified. For all of the motions, the positive x axis points to the right and the positive y axis points upward. Alex, a mountaineer, must leap across a wide crevasse. The other side of the crevasse is below the point from which he leaps, as shown in the figure. Alex leaps horizontally and successfully makes the jump. v0  v0x v0y v0  v0x v0y v0xx^ v0yy^   = atan( ) v0y v0x Typesetting math: 100% Part A Determine the algebraic sign of Alex’s x velocity and y velocity at the instant he leaves the ground at the beginning of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Algebraic sign of velocity The algebraic sign of the velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this example, to the right is defined to be the positive x direction and upward the positive y direction. Therefore, any object moving to the right, whether speeding up, slowing down, or even simultaneously moving upward or downward, has a positive x velocity. Similarly, if the object is moving downward, regardless of any other aspect of its motion, its y velocity is negative. Hint 2. Sketch Alex’s initial velocity On the diagram below, sketch the vector representing Alex’s velocity the instant after he leaves the ground at the beginning of the jump. ANSWER: ANSWER: Typesetting math: 100% Answer Requested Part B Determine the algebraic signs of Alex’s x velocity and y velocity the instant before he lands at the end of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Sketch Alex’s final velocity On the diagram below, sketch the vector representing Alex’s velocity the instant before he safely lands on the other side of the crevasse. ANSWER: Answer Requested ANSWER: Answer Requested Typesetting math: 100% At the buzzer, a basketball player shoots a desperation shot. The ball goes in! Part C Determine the algebraic signs of the ball’s x velocity and y velocity the instant after it leaves the player’s hands. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s initial velocity On the diagram below, sketch the vector representing the velocity of the basketball the instant after it leaves the player’s hands. ANSWER: Typesetting math: 100% ANSWER: Correct Part D Determine the algebraic signs of the ball’s x velocity and y velocity at the ball’s maximum height. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s velocity at maximum height Typesetting math: 100% On the diagram below, sketch the vector representing the velocity of the basketball the instant it reaches its maximum height. ANSWER: ANSWER: Answer Requested PSS 4.1 Projectile Motion Problems Learning Goal: Typesetting math: 100% To practice Problem-Solving Strategy 4.1 for projectile motion problems. A rock thrown with speed 9.00 and launch angle 30.0 (above the horizontal) travels a horizontal distance of = 17.0 before hitting the ground. From what height was the rock thrown? Use the value = 9.810 for the free-fall acceleration. PROBLEM-SOLVING STRATEGY 4.1 Projectile motion problems MODEL: Make simplifying assumptions, such as treating the object as a particle. Is it reasonable to ignore air resistance? VISUALIZE: Use a pictorial representation. Establish a coordinate system with the x axis horizontal and the y axis vertical. Show important points in the motion on a sketch. Define symbols, and identify what you are trying to find. SOLVE: The acceleration is known: and . Thus, the problem becomes one of two-dimensional kinematics. The kinematic equations are , . is the same for the horizontal and vertical components of the motion. Find from one component, and then use that value for the other component. ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model Start by making simplifying assumptions: Model the rock as a particle in free fall. You can ignore air resistance because the rock is a relatively heavy object moving relatively slowly. Visualize Part A Which diagram represents an accurate sketch of the rock’s trajectory? Hint 1. The launch angle In a projectile’s motion, the angle of the initial velocity above the horizontal is called the launch angle. ANSWER: m/s  d m g m/s2 ax = 0 ay = −g xf = xi +vixt, yf = yi +viyt− g(t 1 2 )2 vfx = vix = constant, and vfy = viy − gt t t v i Typesetting math: 100% Typesetting math: 100% Correct Part B As stated in the strategy, choose a coordinate system where the x axis is horizontal and the y axis is vertical. Note that in the strategy, the y component of the projectile’s acceleration, , is taken to be negative. This implies that the positive y axis is upward. Use the same convention for your y axis, and take the positive x axis to be to the right. Where you choose your origin doesn’t change the answer to the question, but choosing an origin can make a problem easier to solve (even if only a bit). Usually it is nice if the majority of the quantities you are given and the quantity you are trying to solve for take positive values relative to your chosen origin. Given this goal, what location for the origin of the coordinate system would make this problem easiest? ANSWER: ay At ground level below the point where the rock is launched At the point where the rock strikes the ground At the peak of the trajectory At the point where the rock is released At ground level below the peak of the trajectory Typesetting math: 100% Correct It’s best to place the origin of the coordinate system at ground level below the launching point because in this way all the points of interest (the launching point and the landing point) will have positive coordinates. (Based on your experience, you know that it’s generally easier to work with positive coordinates.) Keep in mind, however, that this is an arbitrary choice. The correct solution of the problem will not depend on the location of the origin of your coordinate system. Now, define symbols representing initial and final position, velocity, and time. Your target variable is , the initial y coordinate of the rock. Your pictorial representation should be complete now, and similar to the picture below: Solve Part C Find the height from which the rock was launched. Express your answer in meters to three significant figures. yi yi Typesetting math: 100% Hint 1. How to approach the problem The time needed to move horizontally to the final position = 17.0 is the same time needed for the rock to rise from the initial position to the peak of its trajectory and then fall to the ground. Use the information you have about motion in the horizontal direction to solve for . Knowing this time will allow you to use the equations of motion for the vertical direction to solve for . Hint 2. Find the time spent in the air How long ( ) is the rock in the air? Express your answer in seconds to three significant figures. Hint 1. Determine which equation to use Which of the equations given in the strategy and shown below is the most appropriate to calculate the time the rock spent in the air? ANSWER: Hint 2. Find the x component of the initial velocity What is the x component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: t xf = d m yi t yi t t xf = xi + vixt yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt vix = 7.79 m/s Typesetting math: 100% Hint 3. Find the y component of the initial velocity What is the y component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: Answer Requested Assess Part D A second rock is thrown straight upward with a speed 4.500 . If this rock takes 2.181 to fall to the ground, from what height was it released? Express your answer in meters to three significant figures. Hint 1. Identify the known variables What are the values of , , , and for the second rock? Take the positive y axis to be upward and the origin to be located on the ground where the rock lands. Express your answers to four significant figures in the units shown to the right, separated by commas. ANSWER: t = 2.18 s viy = 4.50 m/s yi = 13.5 m m/s s H yf viy t a Typesetting math: 100% Answer Requested Hint 2. Determine which equation to use to find the height Which equation should you use to find ? Keep in mind that if the positive y axis is upward and the origin is located on the ground, . ANSWER: ANSWER: Answer Requested Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. Because both rocks were thrown with the same initial vertical velocity, 4.500 , and fell the same vertical distance of 13.5 , they were in the air for the same amount of time. This result was expected and helps to confirm that you did the calculation in Part C correctly. ± Arrow Hits Apple An arrow is shot at an angle of above the horizontal. The arrow hits a tree a horizontal distance away, at the same height above the ground as it was shot. Use for the magnitude of the acceleration due to gravity. Part A , , , = 0,4.500,2.181,-yf viy t a 9.810 m, m/s, s, m/s2 H yi = H yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt = − 2g( − ) v2f y v2i y yf yi H = 13.5 m viy = m/s m  = 45 D = 220 m g = 9.8 m/s2 Typesetting math: 100% Find , the time that the arrow spends in the air. Answer numerically in seconds, to two significant figures. Hint 1. Find the initial upward component of velocity in terms of D. Introduce the (unknown) variables and for the initial components of velocity. Then use kinematics to relate them and solve for . What is the vertical component of the initial velocity? Express your answer symbolically in terms of and . Hint 1. Find Find the horizontal component of the initial velocity. Express your answer symbolically in terms of and given symbolic quantities. ANSWER: Hint 2. Find What is the vertical component of the initial velocity? Express your answer symbolically in terms of . ANSWER: ANSWER: ta vy0 vx0 ta vy0 ta D vx0 vx0 ta vx0 = D ta vy0 vy0 vx0 vy0 = vx0 vy0 = D ta Typesetting math: 100% Hint 2. Find the time of flight in terms of the initial vertical component of velocity. From the change in the vertical component of velocity, you should be able to find in terms of and . Give your answer in terms of and . Hint 1. Find When applied to the y-component of velocity, in this problem the formula for with constant acceleration is What is , the vertical component of velocity when the arrow hits the tree? Answer symbolically in terms of only. ANSWER: ANSWER: Hint 3. Put the algebra together to find symbolically. If you have an expression for the initial vertical velocity component in terms in terms of and , and another in terms of and , you should be able to eliminate this initial component to find an expression for Express your answer symbolically in terms of given variables. ANSWER: ta vy0 g vy0 g vy(ta) v(t) −g vy(t) = vy0 − g t vy(ta ) vy0 vy(ta) = −vy0 ta = 2vy0 g ta D ta g ta ta2 t2 = a 2D g Typesetting math: 100% ANSWER: Answer Requested Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. Part B How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Express your answer numerically in seconds, to two significant figures. Hint 1. When should the apple be dropped The apple should be dropped at the time equal to the total time it takes the arrow to reach the tree minus the time it takes the apple to fall 6.0 meters. Hint 2. Find the time it takes for the apple to fall 6.0 meters How long does it take an apple to fall 6.0 meters? Express your answer numerically in seconds, to two significant figures. ANSWER: Answer Requested ANSWER: ta = 6.7 s tf = 1.1 s td = 5.6 s Typesetting math: 100% Answer Requested Video Tutor: Ball Fired Upward from Accelerating Cart 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 questions at right. You can watch the video again at any point. Part A Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How far behind the cart will the ball land, compared to the distance in the original experiment? Hint 1. Determine how long the ball is in the air How will doubling the initial upward speed of the ball change the time the ball spends in the air? A kinematic equation may be helpful here. The time in the air will ANSWER: be cut in half. stay the same. double. quadruple. Typesetting math: 100% Hint 2. Determine the appropriate kinematic expression Which of the following kinematic equations correctly describes the horizontal distance between the ball and the cart at the moment the ball lands? The cart’s initial horizontal velocity is , its horizontal acceleration is , and is the time elapsed between launch and impact. ANSWER: ANSWER: Correct The ball will spend twice as much time in the air ( , where is the ball’s initial upward velocity), so it will land four times farther behind the cart: (where is the cart’s horizontal acceleration). Video Tutor: Ball Fired Upward from Moving Cart 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 questions at right. You can watch the video again at any point. d v0x ax t d = v0x t d = 1 2 axv0x t2 d = v0x t+ 1 2 axt2 d = 1 2 axt2 the same distance twice as far half as far four times as far by a factor not listed above t = 2v0y/g v0y d = 1 2 axt2 ax Typesetting math: 100% Part A The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the crate? Ignore air resistance. Hint 1. How to approach the problem While the crate is on the plane, it shares the plane’s velocity. What is the crate’s velocity immediately after it is released? Hint 2. What affects the motion of the crate? Gravity will accelerate the crate downward. What, if anything, affects the crate’s horizontal motion? (Keep in mind that we are told to ignore air resistance, even though that’s not very realistic in this situation.) ANSWER: Correct At the moment it is released, the crate shares the plane’s horizontal velocity. In the absence of air resistance, the crate would remain directly below the plane as it fell. Score Summary: Your score on this assignment is 0%. Before the plane is directly over the target After the plane has flown over the target When the plane is directly over the target Typesetting math: 100% You received 0 out of a possible total of 0 points. Typesetting math: 100%

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Chapter 3 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and 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 , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . 2. The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. A A x A y A Ax Ay |Ax| Ax A x Ax A x A x Ay A x ANSWER: Answer Requested Part B What is the sign of the y component of vector 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, and , of vector shown in the figure. |Ax| = 5 m Ay A positive negative Bx By B Express your answers, separated by a comma, in meters to one significant figure. ANSWER: Correct Vector Components–Review Learning Goal: To introduce you to vectors and the use of sine and cosine for a triangle when resolving components. Vectors are an important part of the language of science, mathematics, and engineering. They are used to discuss multivariable calculus, electrical circuits with oscillating currents, stress and strain in structures and materials, and flows of atmospheres and fluids, and they have many other applications. Resolving a vector into components is a precursor to computing things with or about a vector quantity. Because position, velocity, acceleration, force, momentum, and angular momentum are all vector quantities, resolving vectors into components is the most important skill required in a mechanics course. The figure shows the components of , and , along the x and y axes of the coordinate system, respectively. The components of a vector depend on the coordinate system’s orientation, the key being the angle between the vector and the coordinate axes, often designated . Bx, By = -2,-5 m, m F  Fx Fy  Part A The figure shows the standard way of measuring the angle. is measured to the vector from the x axis, and counterclockwise is positive. Express and in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER:  Fx Fy F  Fx, Fy = Fcos, Fsin Correct In principle, you can determine the components of any vector with these expressions. If lies in one of the other quadrants of the plane, will be an angle larger than 90 degrees (or in radians) and and will have the appropriate signs and values. Unfortunately this way of representing , though mathematically correct, leads to equations that must be simplified using trig identities such as and . These must be used to reduce all trig functions present in your equations to either or . Unless you perform this followup step flawlessly, you will fail to recoginze that , and your equations will not simplify so that you can progress further toward a solution. Therefore, it is best to express all components in terms of either or , with between 0 and 90 degrees (or 0 and in radians), and determine the signs of the trig functions by knowing in which quadrant the vector lies. Part B When you resolve a vector into components, the components must have the form or . The signs depend on which quadrant the vector lies in, and there will be one component with and the other with . In real problems the optimal coordinate system is often rotated so that the x axis is not horizontal. Furthermore, most vectors will not lie in the first quadrant. To assign the sine and cosine correctly for vectors at arbitrary angles, you must figure out which angle is and then properly reorient the definitional triangle. As an example, consider the vector shown in the diagram labeled “tilted axes,” where you know the angle between and the y axis. Which of the various ways of orienting the definitional triangle must be used to resolve into components in the tilted coordinate system shown? (In the figures, the hypotenuse is orange, the side adjacent to is red, and the side opposite is yellow.) F  /2 cos() sin() F  sin(180 + ) = −sin() cos(90 + ) = −sin() sin() cos() sin(180 + ) + cos(270 − ) = 0 sin() cos()  /2 F  |F| cos() |F| sin() sin() cos()  N  N N  Indicate the number of the figure with the correct orientation. Hint 1. Recommended procedure for resolving a vector into components First figure out the sines and cosines of , then figure out the signs from the quadrant the vector is in and write in the signs. Hint 2. Finding the trigonometric functions Sine and cosine are defined according to the following convention, with the key lengths shown in green: The hypotenuse has unit length, the side adjacent to has length , and the   cos() side opposite has length . The colors are chosen to remind you that the vector sum of the two orthogonal sides is the vector whose magnitude is the hypotenuse; red + yellow = orange. ANSWER: Correct Part C Choose the correct procedure for determining the components of a vector in a given coordinate system from this list: ANSWER: sin() 1 2 3 4 Correct Part D The space around a coordinate system is conventionally divided into four numbered quadrants depending on the signs of the x and y coordinates . Consider the following conditions: A. , B. , C. , D. , Which of these lettered conditions are true in which the numbered quadrants shown in ? Write the answer in the following way: If A were true in the third quadrant, B in the second, C in the first, and D in the fourth, enter “3, 2, 1, 4” as your response. ANSWER: Align the adjacent side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and an adjacent side along a coordinate direction with as the included angle. Align the opposite side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and the opposite side along a coordinate direction with as the included angle.     x > 0 y > 0 x > 0 y < 0 x < 0 y > 0 x < 0 y < 0 Correct Part E Now find the components and of in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER: Answer Requested ± Resolving Vector Components with Trigonometry Often a vector is specified by a magnitude and a direction; for example, a rope with tension exerts a force of magnitude in a direction 35 north of east. This is a good way to think of vectors; however, to calculate results with vectors, it is best to select a coordinate system and manipulate the components of the vectors in that coordinate system. Nx Ny N N  Nx, Ny = −Nsin(),Ncos() T  T  Part A Find the components of the vector with length = 1.00 and angle =10.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. What is the x component? Look at the figure shown. points in the positive x direction, so is positive. Also, the magnitude is just the length . ANSWER: Correct Part B Find the components of the vector with length = 1.00 and angle =15.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. A a  A x Ax |Ax| OL = OMcos( ) A  = 0.985,0.174 B b   Hint 1. What is the x component? The x component is still of the same form, that is, . ANSWER: Correct The components of still have the same form, that is, , despite 's placement with respect to the y axis on the drawing. Part C Find the components of the vector with length = 1.00 and angle 35.0 as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. Method 1: Find the angle that makes with the positive x axis Angle = 0.611 differs from the other two angles because it is the angle between the vector and the y axis, unlike the others, which are with respect to the x axis. What is the angle that makes with the positive x axis? Express your answer numerically in degrees. ANSWER: Hint 2. Method 2: Use vector addition Look at the figure shown. Lcos() B = 0.966,0.259 B (Lcos(), Lsin()) B C c  =  C  C 125 1. . 2. . 3. , the x component of is negative, since points in the negative x direction. Use this information to find . Similarly, find . ANSWER: Answer Requested ± Vector Addition and Subtraction In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors , , and . Calculate the following, and express your answers as ordered triplets of values separated by commas. Part A ANSWER: Correct C = C + x C y |C | = length(QR) = c sin() x Cx C C x Cx Cy C  = -0.574,0.819 A = (1, 0,−3) B = (−2, 5, 1) C = (3, 1, 1) A − B  = 3,-5,-4 Part B ANSWER: Correct Part C ANSWER: Correct Part D ANSWER: Correct B − C  = -5,4,0 −A + B − C  = -6,4,3 3A − 2C  = -3,-2,-11 Part E ANSWER: Correct Part F ANSWER: Correct Video Tutor: Balls Take High and Low Tracks 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 questions at right. You can watch the video again at any point. Part A −2A + 3B − C  = -11,14,8 2A − 3(B − C) = 17,-12,-6 Consider the video demonstration that you just watched. Which of the following changes could potentially allow the ball on the straight inclined (yellow) track to win? Ignore air resistance. Select all that apply. Hint 1. How to approach the problem Answers A and B involve changing the steepness of part or all of the track. Answers C and D involve changing the mass of the balls. So, first you should decide which of those factors, if either, can change how fast the ball gets to the end of the track. ANSWER: Correct If the yellow track were tilted steeply enough, its ball could win. How might you go about calculating the necessary change in tilt? Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. A. Increase the tilt of the yellow track. B. Make the downhill and uphill inclines on the red track less steep, while keeping the total distance traveled by the ball the same. C. Increase the mass of the ball on the yellow track. D. Decrease the mass of the ball on the red track.

Chapter 3 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and 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 , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . 2. The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. A A x A y A Ax Ay |Ax| Ax A x Ax A x A x Ay A x ANSWER: Answer Requested Part B What is the sign of the y component of vector 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, and , of vector shown in the figure. |Ax| = 5 m Ay A positive negative Bx By B Express your answers, separated by a comma, in meters to one significant figure. ANSWER: Correct Vector Components–Review Learning Goal: To introduce you to vectors and the use of sine and cosine for a triangle when resolving components. Vectors are an important part of the language of science, mathematics, and engineering. They are used to discuss multivariable calculus, electrical circuits with oscillating currents, stress and strain in structures and materials, and flows of atmospheres and fluids, and they have many other applications. Resolving a vector into components is a precursor to computing things with or about a vector quantity. Because position, velocity, acceleration, force, momentum, and angular momentum are all vector quantities, resolving vectors into components is the most important skill required in a mechanics course. The figure shows the components of , and , along the x and y axes of the coordinate system, respectively. The components of a vector depend on the coordinate system’s orientation, the key being the angle between the vector and the coordinate axes, often designated . Bx, By = -2,-5 m, m F  Fx Fy  Part A The figure shows the standard way of measuring the angle. is measured to the vector from the x axis, and counterclockwise is positive. Express and in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER:  Fx Fy F  Fx, Fy = Fcos, Fsin Correct In principle, you can determine the components of any vector with these expressions. If lies in one of the other quadrants of the plane, will be an angle larger than 90 degrees (or in radians) and and will have the appropriate signs and values. Unfortunately this way of representing , though mathematically correct, leads to equations that must be simplified using trig identities such as and . These must be used to reduce all trig functions present in your equations to either or . Unless you perform this followup step flawlessly, you will fail to recoginze that , and your equations will not simplify so that you can progress further toward a solution. Therefore, it is best to express all components in terms of either or , with between 0 and 90 degrees (or 0 and in radians), and determine the signs of the trig functions by knowing in which quadrant the vector lies. Part B When you resolve a vector into components, the components must have the form or . The signs depend on which quadrant the vector lies in, and there will be one component with and the other with . In real problems the optimal coordinate system is often rotated so that the x axis is not horizontal. Furthermore, most vectors will not lie in the first quadrant. To assign the sine and cosine correctly for vectors at arbitrary angles, you must figure out which angle is and then properly reorient the definitional triangle. As an example, consider the vector shown in the diagram labeled “tilted axes,” where you know the angle between and the y axis. Which of the various ways of orienting the definitional triangle must be used to resolve into components in the tilted coordinate system shown? (In the figures, the hypotenuse is orange, the side adjacent to is red, and the side opposite is yellow.) F  /2 cos() sin() F  sin(180 + ) = −sin() cos(90 + ) = −sin() sin() cos() sin(180 + ) + cos(270 − ) = 0 sin() cos()  /2 F  |F| cos() |F| sin() sin() cos()  N  N N  Indicate the number of the figure with the correct orientation. Hint 1. Recommended procedure for resolving a vector into components First figure out the sines and cosines of , then figure out the signs from the quadrant the vector is in and write in the signs. Hint 2. Finding the trigonometric functions Sine and cosine are defined according to the following convention, with the key lengths shown in green: The hypotenuse has unit length, the side adjacent to has length , and the   cos() side opposite has length . The colors are chosen to remind you that the vector sum of the two orthogonal sides is the vector whose magnitude is the hypotenuse; red + yellow = orange. ANSWER: Correct Part C Choose the correct procedure for determining the components of a vector in a given coordinate system from this list: ANSWER: sin() 1 2 3 4 Correct Part D The space around a coordinate system is conventionally divided into four numbered quadrants depending on the signs of the x and y coordinates . Consider the following conditions: A. , B. , C. , D. , Which of these lettered conditions are true in which the numbered quadrants shown in ? Write the answer in the following way: If A were true in the third quadrant, B in the second, C in the first, and D in the fourth, enter “3, 2, 1, 4” as your response. ANSWER: Align the adjacent side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and an adjacent side along a coordinate direction with as the included angle. Align the opposite side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and the opposite side along a coordinate direction with as the included angle.     x > 0 y > 0 x > 0 y < 0 x < 0 y > 0 x < 0 y < 0 Correct Part E Now find the components and of in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER: Answer Requested ± Resolving Vector Components with Trigonometry Often a vector is specified by a magnitude and a direction; for example, a rope with tension exerts a force of magnitude in a direction 35 north of east. This is a good way to think of vectors; however, to calculate results with vectors, it is best to select a coordinate system and manipulate the components of the vectors in that coordinate system. Nx Ny N N  Nx, Ny = −Nsin(),Ncos() T  T  Part A Find the components of the vector with length = 1.00 and angle =10.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. What is the x component? Look at the figure shown. points in the positive x direction, so is positive. Also, the magnitude is just the length . ANSWER: Correct Part B Find the components of the vector with length = 1.00 and angle =15.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. A a  A x Ax |Ax| OL = OMcos( ) A  = 0.985,0.174 B b   Hint 1. What is the x component? The x component is still of the same form, that is, . ANSWER: Correct The components of still have the same form, that is, , despite 's placement with respect to the y axis on the drawing. Part C Find the components of the vector with length = 1.00 and angle 35.0 as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. Method 1: Find the angle that makes with the positive x axis Angle = 0.611 differs from the other two angles because it is the angle between the vector and the y axis, unlike the others, which are with respect to the x axis. What is the angle that makes with the positive x axis? Express your answer numerically in degrees. ANSWER: Hint 2. Method 2: Use vector addition Look at the figure shown. Lcos() B = 0.966,0.259 B (Lcos(), Lsin()) B C c  =  C  C 125 1. . 2. . 3. , the x component of is negative, since points in the negative x direction. Use this information to find . Similarly, find . ANSWER: Answer Requested ± Vector Addition and Subtraction In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors , , and . Calculate the following, and express your answers as ordered triplets of values separated by commas. Part A ANSWER: Correct C = C + x C y |C | = length(QR) = c sin() x Cx C C x Cx Cy C  = -0.574,0.819 A = (1, 0,−3) B = (−2, 5, 1) C = (3, 1, 1) A − B  = 3,-5,-4 Part B ANSWER: Correct Part C ANSWER: Correct Part D ANSWER: Correct B − C  = -5,4,0 −A + B − C  = -6,4,3 3A − 2C  = -3,-2,-11 Part E ANSWER: Correct Part F ANSWER: Correct Video Tutor: Balls Take High and Low Tracks 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 questions at right. You can watch the video again at any point. Part A −2A + 3B − C  = -11,14,8 2A − 3(B − C) = 17,-12,-6 Consider the video demonstration that you just watched. Which of the following changes could potentially allow the ball on the straight inclined (yellow) track to win? Ignore air resistance. Select all that apply. Hint 1. How to approach the problem Answers A and B involve changing the steepness of part or all of the track. Answers C and D involve changing the mass of the balls. So, first you should decide which of those factors, if either, can change how fast the ball gets to the end of the track. ANSWER: Correct If the yellow track were tilted steeply enough, its ball could win. How might you go about calculating the necessary change in tilt? Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. A. Increase the tilt of the yellow track. B. Make the downhill and uphill inclines on the red track less steep, while keeping the total distance traveled by the ball the same. C. Increase the mass of the ball on the yellow track. D. Decrease the mass of the ball on the red track.

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Chapter 1 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 5, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Curved Motion Diagram The motion diagram shown in the figure represents a pendulum released from rest at an angle of 45 from the vertical. The dots in the motion diagram represent the positions of the pendulum bob at eleven moments separated by equal time intervals. The green arrows represent the average velocity between adjacent dots. Also given is a “compass rose” in which directions are labeled with the letters of the alphabet.  Part A What is the direction of the acceleration of the object at moment 5? Enter the letter of the arrow with this direction from the compass rose in the figure. Type Z if the acceleration vector has zero length. You did not open hints for this part. ANSWER: Incorrect; Try Again Part B What is the direction of the acceleration of the object at moments 0 and 10? Enter the letters corresponding to the arrows with these directions from the compass rose in the figure, separated by commas. Type Z if the acceleration vector has zero length. You did not open hints for this part. ANSWER: Incorrect; Try Again PSS 1.1 Motion Diagrams Learning Goal: To practice Problem-Solving Strategy 1.1 for motion diagram problems. A car is traveling with constant velocity along a highway. The driver notices he is late for work, so he stomps down on the gas pedal and the car begins to speed up. The car has just achieved double its directions at time step 0, time step 10 = initial velocity when the driver spots a police officer behind him and applies the brakes. The car then slows down, coming to rest at a stoplight ahead. Draw a complete motion diagram for this situation. PROBLEM-SOLVING STRATEGY 1.1 Motion diagrams MODEL: Represent the moving object as a particle. Make simplifying assumptions when interpreting the problem statement. VISUALIZE: A complete motion diagram consists of: The position of the object in each frame of the film, shown as a dot. Use five or six dots to make the motion clear but without overcrowding the picture. More complex motions may need more dots. The average velocity vectors, found by connecting each dot in the motion diagram to the next with a vector arrow. There is one velocity vector linking each set of two position dots. Label the row of velocity vectors . The average acceleration vectors, found using Tactics Box 1.3. There is one acceleration vector linking each set of two velocity vectors. Each acceleration vector is drawn at the dot between the two velocity vectors it links. Use to indicate a point at which the acceleration is zero. Label the row of acceleration vectors . Model It is appropriate to use the particle model for the car. You should also make some simplifying assumptions. v 0 a Part A The car’s motion can be divided into three different stages: its motion before the driver realizes he’s late, its motion after the driver hits the gas (but before he sees the police car), and its motion after the driver sees the police car. Which of the following simplifying assumptions is it reasonable to make in this problem? During each of the three different stages of its motion, the car is moving with constant A. acceleration. B. During each of the three different stages of its motion, the car is moving with constant velocity. C. The highway is straight (i.e., there are no curves). D. The highway is level (i.e., there are no hills or valleys). Enter all the correct answers in alphabetical order without commas. For example, if statements C and D are correct, enter CD. ANSWER: Correct In addition to the assumptions listed above, in the rest of this problem assume that the car is moving in a straight line to the right. Visualize Part B In the three diagrams shown to the left, the position of the car at five subsequent instants of time is represented by black dots, and the car’s average velocity is represented by green arrows. Which of these diagrams best describes the position and the velocity of the car before the driver notices he is late? ANSWER: Correct Part C Which of the diagrams shown to the left best describes the position and the velocity of the car after the driver hits the gas, but before he notices the police officer? ANSWER: Correct A B C A B C Part D Which of the diagrams shown to the left best describes the position and the velocity of the car after the driver notices the police officer? ANSWER: Correct Part E Which of the diagrams shown below most accurately depicts the average acceleration vectors of the car during the events described in the problem introduction? ANSWER: A B C Correct You can now draw a complete motion diagram for the situation described in this problem. Your diagram should look like this: Measurements in SI Units Familiarity with SI units will aid your study of physics and all other sciences. Part A What is the approximate height of the average adult in centimeters? Hint 1. Converting between feet and centimeters The distance from your elbow to your fingertips is typically about 50 . A B C cm ANSWER: Correct If you’re not familiar with metric units of length, you can use your body to develop intuition for them. The average height of an adult is 5 6.4 . The distance from elbow to fingertips on the average adult is about 50 . Ten (1 ) is about the width of this adult’s little finger and 10 is about the width of the average hand. Part B Approximately what is the mass of the average adult in kilograms? Hint 1. Converting between pounds and kilograms Something that weighs 1 has a mass of about . ANSWER: Correct Something that weighs 1 has a mass of about . This is a useful conversion to keep in mind! ± A Trip to Europe 100 200 300 cm cm cm feet inches cm mm cm cm pound 1 kg 2 80 500 1200 kg kg kg pound (1/2) kg Learning Goal: To understand how to use dimensional analysis to solve problems. Dimensional analysis is a useful tool for solving problems that involve unit conversions. Since unit conversion is not limited to physics problems but is part of our everyday life, correct use of conversion factors is essential to working through problems of practical importance. For example, dimensional analysis could be used in problems involving currency exchange. Say you want to calculate how many euros you get if you exchange 3600 ( ), given the exchange rate , that is, 1 to 1.20 . Begin by writing down the starting value, 3600 . This can also be written as a fraction: . Next, convert dollars to euros. This conversion involves multiplying by a simple conversion factor derived from the exchange rate: . Note that the “dollar” unit, , should appear on the bottom of this conversion factor, since appears on the top of the starting value. Finally, since dollars are divided by dollars, the units can be canceled and the final result is . Currency exchange is only one example of many practical situations where dimensional analysis may help you to work through problems. Remember that dimensional analysis involves multiplying a given value by a conversion factor, resulting in a value in the new units. The conversion factor can be the ratio of any two quantities, as long as the ratio is equal to one. You and your friends are organizing a trip to Europe. Your plan is to rent a car and drive through the major European capitals. By consulting a map you estimate that you will cover a total distance of 5000 . Consider the euro-dollar exchange rate given in the introduction and use dimensional analysis to work through these simple problems. Part A You select a rental package that includes a car with an average consumption of 6.00 of fuel per 100 . Considering that in Europe the average fuel cost is 1.063 , how much (in US dollars) will you spend in fuel on your trip? Express your answer numerically in US dollars to three significant figures. You did not open hints for this part. ANSWER: US dollars USD 1 EUR = 1.20 USD euro US dollars USD 3600 USD 1 1.00 EUR 1.20 USD USD USD ( )( ) = 3000 EUR 3600 USD 1 1.00 EUR 1.20 USD km liters km euros/liter Part B How many gallons of fuel would the rental car consume per mile? Express your answer numerically in gallons per mile to three significant figures. You did not open hints for this part. ANSWER: Part C This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. Cost of fuel = USD gallons/mile

Chapter 1 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 5, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Curved Motion Diagram The motion diagram shown in the figure represents a pendulum released from rest at an angle of 45 from the vertical. The dots in the motion diagram represent the positions of the pendulum bob at eleven moments separated by equal time intervals. The green arrows represent the average velocity between adjacent dots. Also given is a “compass rose” in which directions are labeled with the letters of the alphabet.  Part A What is the direction of the acceleration of the object at moment 5? Enter the letter of the arrow with this direction from the compass rose in the figure. Type Z if the acceleration vector has zero length. You did not open hints for this part. ANSWER: Incorrect; Try Again Part B What is the direction of the acceleration of the object at moments 0 and 10? Enter the letters corresponding to the arrows with these directions from the compass rose in the figure, separated by commas. Type Z if the acceleration vector has zero length. You did not open hints for this part. ANSWER: Incorrect; Try Again PSS 1.1 Motion Diagrams Learning Goal: To practice Problem-Solving Strategy 1.1 for motion diagram problems. A car is traveling with constant velocity along a highway. The driver notices he is late for work, so he stomps down on the gas pedal and the car begins to speed up. The car has just achieved double its directions at time step 0, time step 10 = initial velocity when the driver spots a police officer behind him and applies the brakes. The car then slows down, coming to rest at a stoplight ahead. Draw a complete motion diagram for this situation. PROBLEM-SOLVING STRATEGY 1.1 Motion diagrams MODEL: Represent the moving object as a particle. Make simplifying assumptions when interpreting the problem statement. VISUALIZE: A complete motion diagram consists of: The position of the object in each frame of the film, shown as a dot. Use five or six dots to make the motion clear but without overcrowding the picture. More complex motions may need more dots. The average velocity vectors, found by connecting each dot in the motion diagram to the next with a vector arrow. There is one velocity vector linking each set of two position dots. Label the row of velocity vectors . The average acceleration vectors, found using Tactics Box 1.3. There is one acceleration vector linking each set of two velocity vectors. Each acceleration vector is drawn at the dot between the two velocity vectors it links. Use to indicate a point at which the acceleration is zero. Label the row of acceleration vectors . Model It is appropriate to use the particle model for the car. You should also make some simplifying assumptions. v 0 a Part A The car’s motion can be divided into three different stages: its motion before the driver realizes he’s late, its motion after the driver hits the gas (but before he sees the police car), and its motion after the driver sees the police car. Which of the following simplifying assumptions is it reasonable to make in this problem? During each of the three different stages of its motion, the car is moving with constant A. acceleration. B. During each of the three different stages of its motion, the car is moving with constant velocity. C. The highway is straight (i.e., there are no curves). D. The highway is level (i.e., there are no hills or valleys). Enter all the correct answers in alphabetical order without commas. For example, if statements C and D are correct, enter CD. ANSWER: Correct In addition to the assumptions listed above, in the rest of this problem assume that the car is moving in a straight line to the right. Visualize Part B In the three diagrams shown to the left, the position of the car at five subsequent instants of time is represented by black dots, and the car’s average velocity is represented by green arrows. Which of these diagrams best describes the position and the velocity of the car before the driver notices he is late? ANSWER: Correct Part C Which of the diagrams shown to the left best describes the position and the velocity of the car after the driver hits the gas, but before he notices the police officer? ANSWER: Correct A B C A B C Part D Which of the diagrams shown to the left best describes the position and the velocity of the car after the driver notices the police officer? ANSWER: Correct Part E Which of the diagrams shown below most accurately depicts the average acceleration vectors of the car during the events described in the problem introduction? ANSWER: A B C Correct You can now draw a complete motion diagram for the situation described in this problem. Your diagram should look like this: Measurements in SI Units Familiarity with SI units will aid your study of physics and all other sciences. Part A What is the approximate height of the average adult in centimeters? Hint 1. Converting between feet and centimeters The distance from your elbow to your fingertips is typically about 50 . A B C cm ANSWER: Correct If you’re not familiar with metric units of length, you can use your body to develop intuition for them. The average height of an adult is 5 6.4 . The distance from elbow to fingertips on the average adult is about 50 . Ten (1 ) is about the width of this adult’s little finger and 10 is about the width of the average hand. Part B Approximately what is the mass of the average adult in kilograms? Hint 1. Converting between pounds and kilograms Something that weighs 1 has a mass of about . ANSWER: Correct Something that weighs 1 has a mass of about . This is a useful conversion to keep in mind! ± A Trip to Europe 100 200 300 cm cm cm feet inches cm mm cm cm pound 1 kg 2 80 500 1200 kg kg kg pound (1/2) kg Learning Goal: To understand how to use dimensional analysis to solve problems. Dimensional analysis is a useful tool for solving problems that involve unit conversions. Since unit conversion is not limited to physics problems but is part of our everyday life, correct use of conversion factors is essential to working through problems of practical importance. For example, dimensional analysis could be used in problems involving currency exchange. Say you want to calculate how many euros you get if you exchange 3600 ( ), given the exchange rate , that is, 1 to 1.20 . Begin by writing down the starting value, 3600 . This can also be written as a fraction: . Next, convert dollars to euros. This conversion involves multiplying by a simple conversion factor derived from the exchange rate: . Note that the “dollar” unit, , should appear on the bottom of this conversion factor, since appears on the top of the starting value. Finally, since dollars are divided by dollars, the units can be canceled and the final result is . Currency exchange is only one example of many practical situations where dimensional analysis may help you to work through problems. Remember that dimensional analysis involves multiplying a given value by a conversion factor, resulting in a value in the new units. The conversion factor can be the ratio of any two quantities, as long as the ratio is equal to one. You and your friends are organizing a trip to Europe. Your plan is to rent a car and drive through the major European capitals. By consulting a map you estimate that you will cover a total distance of 5000 . Consider the euro-dollar exchange rate given in the introduction and use dimensional analysis to work through these simple problems. Part A You select a rental package that includes a car with an average consumption of 6.00 of fuel per 100 . Considering that in Europe the average fuel cost is 1.063 , how much (in US dollars) will you spend in fuel on your trip? Express your answer numerically in US dollars to three significant figures. You did not open hints for this part. ANSWER: US dollars USD 1 EUR = 1.20 USD euro US dollars USD 3600 USD 1 1.00 EUR 1.20 USD USD USD ( )( ) = 3000 EUR 3600 USD 1 1.00 EUR 1.20 USD km liters km euros/liter Part B How many gallons of fuel would the rental car consume per mile? Express your answer numerically in gallons per mile to three significant figures. You did not open hints for this part. ANSWER: Part C This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. Cost of fuel = USD gallons/mile

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

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

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Part I. (20%) Multiple Choice Questions (only one answer) 1. What is the SS7 A-link? a. Signaling Link between SSP and STP b. Signaling Link between STP and STP c. Signaling Link between SCP and SCP d. Signaling Link between SSP and SSP 2. How is a TDM trunk identified on the SS7 network? It is identified by _____. a. IP Address b. ISDN B channel c. UDP port number d. Circuit Identificaiton Code (CIC) 3. What is the layer-3 protocol for ISUP? a. IP b. Q.931 c. MTP3 d. TCAP 4. Which of the following signaling is required to support Local Number Portability (LNP)? a. SIP b. Q.931 c. SS7 d. MGCP 5. Which of the following signlaing is NOT supported on the local telephone switch? a. ISDN/Q.931 b. SIP c. SS7 d. Station Signaling 6. Which of the following protocol is used by Advanced Intelligent Network (AIN) for remote database query? One of the AIN features is 1-800/1-900 calls. a. TCAP b. ISUP c. MGCP d. SIP 7. Which of the following is an NNI signaling? a. ISDN/Q.931 b. SIP c. MGCP d. SS7 8. Which of the following message is used by Access Media Gateway (MG) to inform Media Gateway Controller (MGC) that a phone is off-hook? a. NTFY b. RQNT c. CRCX d. MDCX 9. Which of the following device supports codec (such as G.711)? a. Media Gateway Contoller b. Media Gateway (Access Gateway) c. Analog Phone d. Signaling Transfer Point (STP) 10. In MGCP, a connection is _______________ ? a. The same as a TCP connection b. The same as a phone call. c. An end-point and its associated RTP session. d. A TDM channel identified by CIC Part II Questions (10%). 1. Provide two reasons for separating Media Gateway Controller (MGC) and Media Gateway. Media gateways are often controlled by a separate Media Gateway Controller which provides the call control and signaling functionality. 2. Identify the signaling between the network devices. (enter N/A if not applicable) 1 Media Gateway Controller Media Gateway Controller 2 Media Gateway Controller Media Gateway 3 Media Gateway Controller Class-5 (local) Switch 4 Media Gateway Controller SIP Proxy 5 Media Gateway Media Gateway Part III (20%) Call Flow Diagram and protocol stacks Media Gateway Controller (MGC) also has the function of SS7 Signaling Gateway. 1. Show the call flow diagram from Phone-22 (SIP) to Phone-33 (SIP). 2. Show the call flow diagram from Phone-41 (analog) to Phone-42(analog) over the VoIP Carrier Network. Note: each signaling message is an arrow with its own label. Do not use one arrow and one label to represent multiple messages.

Part I. (20%) Multiple Choice Questions (only one answer) 1. What is the SS7 A-link? a. Signaling Link between SSP and STP b. Signaling Link between STP and STP c. Signaling Link between SCP and SCP d. Signaling Link between SSP and SSP 2. How is a TDM trunk identified on the SS7 network? It is identified by _____. a. IP Address b. ISDN B channel c. UDP port number d. Circuit Identificaiton Code (CIC) 3. What is the layer-3 protocol for ISUP? a. IP b. Q.931 c. MTP3 d. TCAP 4. Which of the following signaling is required to support Local Number Portability (LNP)? a. SIP b. Q.931 c. SS7 d. MGCP 5. Which of the following signlaing is NOT supported on the local telephone switch? a. ISDN/Q.931 b. SIP c. SS7 d. Station Signaling 6. Which of the following protocol is used by Advanced Intelligent Network (AIN) for remote database query? One of the AIN features is 1-800/1-900 calls. a. TCAP b. ISUP c. MGCP d. SIP 7. Which of the following is an NNI signaling? a. ISDN/Q.931 b. SIP c. MGCP d. SS7 8. Which of the following message is used by Access Media Gateway (MG) to inform Media Gateway Controller (MGC) that a phone is off-hook? a. NTFY b. RQNT c. CRCX d. MDCX 9. Which of the following device supports codec (such as G.711)? a. Media Gateway Contoller b. Media Gateway (Access Gateway) c. Analog Phone d. Signaling Transfer Point (STP) 10. In MGCP, a connection is _______________ ? a. The same as a TCP connection b. The same as a phone call. c. An end-point and its associated RTP session. d. A TDM channel identified by CIC Part II Questions (10%). 1. Provide two reasons for separating Media Gateway Controller (MGC) and Media Gateway. Media gateways are often controlled by a separate Media Gateway Controller which provides the call control and signaling functionality. 2. Identify the signaling between the network devices. (enter N/A if not applicable) 1 Media Gateway Controller Media Gateway Controller 2 Media Gateway Controller Media Gateway 3 Media Gateway Controller Class-5 (local) Switch 4 Media Gateway Controller SIP Proxy 5 Media Gateway Media Gateway Part III (20%) Call Flow Diagram and protocol stacks Media Gateway Controller (MGC) also has the function of SS7 Signaling Gateway. 1. Show the call flow diagram from Phone-22 (SIP) to Phone-33 (SIP). 2. Show the call flow diagram from Phone-41 (analog) to Phone-42(analog) over the VoIP Carrier Network. Note: each signaling message is an arrow with its own label. Do not use one arrow and one label to represent multiple messages.

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Doppler Shift 73 Because of the Doppler Effect, light emitted by an object can appear to change wavelength due to its motion toward or away from an observer. When the observer and the source of light are moving toward each other, the light is shifted to shorter wavelengths (blueshifted). When the observer and the source of light are moving away from each other, the light is shifted to longer wavelengths (redshifted). Part I: Motion of Source Star is not . rnovrng r ABCD 1) Consider the situations shown (A—D). a) In which situation will the observer receive light that is shifted to shorter wavelengths? b) Will this light be blueshifted or redshifted for this case? c) What direction is the star moving relative to the observer for this case? 2) Consider the situations shown (A—D). a) In which situation will the observer receive light that is shifted to longer wavelengths? b) Will this light be blueshifted or redshifted for this case? c) What direction is the star moving relative to the observer for this case? . 74 Doppler Shift 3) In which of the srtuations shown (A—D) will theobserver receive light that Is not Doppler Shifted at all? Explain your reasoning. – 4) Imagine our solar system Is moving In the Milky Way toward a group of three stars. Star A is a blue star that is slightly closer to us than the other two. Star B is a red star that is farthest away from us. Star C is a yellow star that is halfway between Stars A end B. a) Which of these three stars, if any, will give off light that appears to be blueshifted? Explain your reasoning. . / b) Which of these three stars, if any, will give off light that appears to be redshifted? Explain your reasoning. c) Which of these three stars, if any, will give off light that appears to have no shift? Explain your reasoning. — 5) You overhear two students discussing the topic of Doppler Shift. Student 1: Since Betelgeuse is a red star, it must be going away from us, and since Rigel is a blue star it must be coming toward us. Student 2: 1 disagree, the color of the star does not tell you if it is moving. You have to look at the shift in wavelength of the lines in the star’s absorption spectrum to determine whether it’s moving toward or away from you. Do you agree or disagree with either or both of the students? Explain your reasoning. 5 Part II: Shift in Absorption Spectra When we study an astronomical object like a star or galaxy, we examine the spectrum of light it gives off. Since the lines of a spectrum occur at specific wavelengths we can determine that an object is moving when we see that the lines have been shifted to either longer or shorter wavelengths. For the absorption line spectra shown on the next page, short-wavelength light (the blue end of the spectrum) is shown on the left-hand side and long-wavelength light (the red end of the spectrum) is shown on the right-hand side. Doppler Shift 75 For the three absorption line spectra shown below (A, B, and C), one of the spectra corresponds to a star that is not moving relative to you, one of the spectra is from a star that is moving toward you, and one of the spectra is from a star that is moving away from you. A B Blue J___ ..‘ C 6) Which of the three spectra above corresponds with the star moving toward you? Explain your reasoning. If two sources of llght are moving relative to an observer, the light from the star that is moving faster will appear to undergo a greater Doppler Consider the four spectra at the right. The spectrum labeled F is an absorption line spectrum from a star that is at rest. Again, note that short-wavelength (blue) light is shown on the left-hand side of each spectrum and long-wavelength (red) light is shown on the right-hand side of each spectrum. 7) Which of the three spectra corresponds with the star moving away from you? Explain your reasoning. Part 111: Size of Shift and Speed Blue Red . – 76 Doppler Shift 8) Which of the four spectra would be from the star that is moving the fastest? Would this star be moving toward or away from the observer? 9) Of the stars that are moving, which spectra would be from the star that is moving the slowest? Describe the motion of this star, – (fJ 1O)An Important line In the absorption spectrum of stars occurs at a wavelength of 656 nm for stars at rest. Irna me that you observe five stars (H—L) from Earth and discover that this Important absorption line Is measured at the wavelength shown in the table below for each of the five stars, Star Wavelength of Absorption Line H 649nm I 660 nm J 656nrn K 658nrn L 647nm a) Which of the stars are gMng off light that appears blueshifted? Explain your reasoning. b) Which of the stars are gMng off light that appears redshifted? Explain your reasoning. d) Which star is moving the fastest? Is it moving toward or away from the observer? Explain your reasoning. , . . c) Which star is giving off light that appears shifted by the greatest amount? Is this light shifted to longer or shorter wavelengths? Explain your reasoning. a) Which planets will receive a radio signal that Is redshifted? Explain your reasoning. b) Which planets wfll receive a radio signal that is shifted to shorter wavelengths? Explain your reasoning. a a . ii) The figure at right shows a spaceprobe and five planets. The motion of the spaceprobe is indicated by the arrow. The spaceprobe is continuously broadcasting a radio signal in all directions. 4 C E not to scale c) Will all the planets receive radio signals from the spaceprobe that are Doppler shifted? Explain your reasoning. d) How will the size of the Doppler Shift in the radio signals detected at Planets A and B compare? Explain your reasoning. Cats r , ‘, e) How Will the slz of 1h Dupler Shift in the radio signals deteed °lane E and B compare? Explain your reasoning. ‘

Doppler Shift 73 Because of the Doppler Effect, light emitted by an object can appear to change wavelength due to its motion toward or away from an observer. When the observer and the source of light are moving toward each other, the light is shifted to shorter wavelengths (blueshifted). When the observer and the source of light are moving away from each other, the light is shifted to longer wavelengths (redshifted). Part I: Motion of Source Star is not . rnovrng r ABCD 1) Consider the situations shown (A—D). a) In which situation will the observer receive light that is shifted to shorter wavelengths? b) Will this light be blueshifted or redshifted for this case? c) What direction is the star moving relative to the observer for this case? 2) Consider the situations shown (A—D). a) In which situation will the observer receive light that is shifted to longer wavelengths? b) Will this light be blueshifted or redshifted for this case? c) What direction is the star moving relative to the observer for this case? . 74 Doppler Shift 3) In which of the srtuations shown (A—D) will theobserver receive light that Is not Doppler Shifted at all? Explain your reasoning. – 4) Imagine our solar system Is moving In the Milky Way toward a group of three stars. Star A is a blue star that is slightly closer to us than the other two. Star B is a red star that is farthest away from us. Star C is a yellow star that is halfway between Stars A end B. a) Which of these three stars, if any, will give off light that appears to be blueshifted? Explain your reasoning. . / b) Which of these three stars, if any, will give off light that appears to be redshifted? Explain your reasoning. c) Which of these three stars, if any, will give off light that appears to have no shift? Explain your reasoning. — 5) You overhear two students discussing the topic of Doppler Shift. Student 1: Since Betelgeuse is a red star, it must be going away from us, and since Rigel is a blue star it must be coming toward us. Student 2: 1 disagree, the color of the star does not tell you if it is moving. You have to look at the shift in wavelength of the lines in the star’s absorption spectrum to determine whether it’s moving toward or away from you. Do you agree or disagree with either or both of the students? Explain your reasoning. 5 Part II: Shift in Absorption Spectra When we study an astronomical object like a star or galaxy, we examine the spectrum of light it gives off. Since the lines of a spectrum occur at specific wavelengths we can determine that an object is moving when we see that the lines have been shifted to either longer or shorter wavelengths. For the absorption line spectra shown on the next page, short-wavelength light (the blue end of the spectrum) is shown on the left-hand side and long-wavelength light (the red end of the spectrum) is shown on the right-hand side. Doppler Shift 75 For the three absorption line spectra shown below (A, B, and C), one of the spectra corresponds to a star that is not moving relative to you, one of the spectra is from a star that is moving toward you, and one of the spectra is from a star that is moving away from you. A B Blue J___ ..‘ C 6) Which of the three spectra above corresponds with the star moving toward you? Explain your reasoning. If two sources of llght are moving relative to an observer, the light from the star that is moving faster will appear to undergo a greater Doppler Consider the four spectra at the right. The spectrum labeled F is an absorption line spectrum from a star that is at rest. Again, note that short-wavelength (blue) light is shown on the left-hand side of each spectrum and long-wavelength (red) light is shown on the right-hand side of each spectrum. 7) Which of the three spectra corresponds with the star moving away from you? Explain your reasoning. Part 111: Size of Shift and Speed Blue Red . – 76 Doppler Shift 8) Which of the four spectra would be from the star that is moving the fastest? Would this star be moving toward or away from the observer? 9) Of the stars that are moving, which spectra would be from the star that is moving the slowest? Describe the motion of this star, – (fJ 1O)An Important line In the absorption spectrum of stars occurs at a wavelength of 656 nm for stars at rest. Irna me that you observe five stars (H—L) from Earth and discover that this Important absorption line Is measured at the wavelength shown in the table below for each of the five stars, Star Wavelength of Absorption Line H 649nm I 660 nm J 656nrn K 658nrn L 647nm a) Which of the stars are gMng off light that appears blueshifted? Explain your reasoning. b) Which of the stars are gMng off light that appears redshifted? Explain your reasoning. d) Which star is moving the fastest? Is it moving toward or away from the observer? Explain your reasoning. , . . c) Which star is giving off light that appears shifted by the greatest amount? Is this light shifted to longer or shorter wavelengths? Explain your reasoning. a) Which planets will receive a radio signal that Is redshifted? Explain your reasoning. b) Which planets wfll receive a radio signal that is shifted to shorter wavelengths? Explain your reasoning. a a . ii) The figure at right shows a spaceprobe and five planets. The motion of the spaceprobe is indicated by the arrow. The spaceprobe is continuously broadcasting a radio signal in all directions. 4 C E not to scale c) Will all the planets receive radio signals from the spaceprobe that are Doppler shifted? Explain your reasoning. d) How will the size of the Doppler Shift in the radio signals detected at Planets A and B compare? Explain your reasoning. Cats r , ‘, e) How Will the slz of 1h Dupler Shift in the radio signals deteed °lane E and B compare? Explain your reasoning. ‘

  ANSWERS Part 1 1 C is the answer because … Read More...
Operational amplifiers are often used to amplify a sensor output. This problem will walk you through the design of a simple temperature measuring device based on a platinum wire sensor. Goal: Build a circuit that will provide a calibrated output between .32V and 2.12V for temperatures sensed between 0°C and 100°C. (The final circuit can be seen at the end of this homework but we will work out each stage in turn.) The platinum wire sensor has a resistance of 100Ω at 0°C and 138.5Ω at 100°C, or a change of 0.385Ω/°C. (The arrow through the resistor in the circuit indicates it is a variable resistor.) A 0.5mA source is used to excite the platinum wire resistor to obtain a voltage. The first stage of our circuit will be to buffer the output of the sensor so we do not load the sensor circuit by drawing off any of the .5mA current to the op amp. A. (10 points) What is V1 when the temperature is 0°C, 1°C, 20°C and 100°C? (Use at least four decimal points.) B. (10 points)The output voltage of the resistor changes by I*ΔRT where I =0.5mA and ΔRT = 0.385Ω/°C. It is too small, so we need amplify this so the V2 output in the second stage of this circuit will be 5mV per degree using a non-inverting amplifier. So we want 5mV = (I*ΔRT * gain) per degree centigrade. What is the required gain for this circuit? Choose values of R1 and R2 between 1k and 100kΩ to achieve this. Choose Rin to be 1K-10kΩ. C. (10 points) What is V2 for a temperature of 0°C and for 1°C. What is the difference between the two voltages? (Hint: The difference should be exactly 5mV! The resistance of the platinum wire will be 100.385Ω @ 1°C.) D. (10 extra credit points) We would like for the output voltage, V2, to be 0V when the temperature is 0°C. This can be done by adding a third stage with an offset voltage in the circuit below. Find Voffset so that VC = 0V when the temperature is 0°C. Let R3 =R4 = R5 and pick appropriate values of the resistors between 1k and 10kΩ. (Hint: V2= the voltage when the temperature is 0°C you found in part C. Find Voffset so that Vc =0V. Superposition may a good technique to use here. You can analyze the circuit when V2=0 and the offset is activated and then you can analyze the circuit when V2 = the value from part C and the offset voltage is zero.) What is Voffset? What values did you chose for the resistors? What is VC when the temperature is is 0°C, 1°C, 20°C and 100°C? E. The voltage VC should now be 0V when the temperature is 0°C and increase by 10mV for every degree centigrade. We need to multiply this by 1.8, the factor to convert a degree Centigrade to a degree Fahrenheit. The output of this stage, V4, should range from 0V to 1.80V and then we will add an offset to change the VF output range to 0.32V to 2.12V in the last stage. The following circuit can be used. Note that this circuit uses inverting amplifiers instead of non-inverting amplifiers. (10 extra credit points) Find the correct resistor values for R7, R8 so that V4 will range between 0V to -1.80V when the temperature is sensed between 0°C and 100°C. (10 extra credit points) Find Voffset2 and then determine VF at 0°C, 1°C, 20°C and 100°C so the VF will range from 0.32V to 2.12V. (Hint: When VC = 0V at 0°C, the V4 output of this stage will also be 0V. Determine the offset voltage so VF = 0.32V. Choose R9 = R10 = R11 = 1kΩ, so at 0°C, VF = 0.32V = VR10 with the current going from the output back through R10 to zero volts, then down through R11 and the Voffset2 source. Since VR10=VR11=0.32V, determine Voffset2. When the temperature is 100°C the output should be 2.12V.)

Operational amplifiers are often used to amplify a sensor output. This problem will walk you through the design of a simple temperature measuring device based on a platinum wire sensor. Goal: Build a circuit that will provide a calibrated output between .32V and 2.12V for temperatures sensed between 0°C and 100°C. (The final circuit can be seen at the end of this homework but we will work out each stage in turn.) The platinum wire sensor has a resistance of 100Ω at 0°C and 138.5Ω at 100°C, or a change of 0.385Ω/°C. (The arrow through the resistor in the circuit indicates it is a variable resistor.) A 0.5mA source is used to excite the platinum wire resistor to obtain a voltage. The first stage of our circuit will be to buffer the output of the sensor so we do not load the sensor circuit by drawing off any of the .5mA current to the op amp. A. (10 points) What is V1 when the temperature is 0°C, 1°C, 20°C and 100°C? (Use at least four decimal points.) B. (10 points)The output voltage of the resistor changes by I*ΔRT where I =0.5mA and ΔRT = 0.385Ω/°C. It is too small, so we need amplify this so the V2 output in the second stage of this circuit will be 5mV per degree using a non-inverting amplifier. So we want 5mV = (I*ΔRT * gain) per degree centigrade. What is the required gain for this circuit? Choose values of R1 and R2 between 1k and 100kΩ to achieve this. Choose Rin to be 1K-10kΩ. C. (10 points) What is V2 for a temperature of 0°C and for 1°C. What is the difference between the two voltages? (Hint: The difference should be exactly 5mV! The resistance of the platinum wire will be 100.385Ω @ 1°C.) D. (10 extra credit points) We would like for the output voltage, V2, to be 0V when the temperature is 0°C. This can be done by adding a third stage with an offset voltage in the circuit below. Find Voffset so that VC = 0V when the temperature is 0°C. Let R3 =R4 = R5 and pick appropriate values of the resistors between 1k and 10kΩ. (Hint: V2= the voltage when the temperature is 0°C you found in part C. Find Voffset so that Vc =0V. Superposition may a good technique to use here. You can analyze the circuit when V2=0 and the offset is activated and then you can analyze the circuit when V2 = the value from part C and the offset voltage is zero.) What is Voffset? What values did you chose for the resistors? What is VC when the temperature is is 0°C, 1°C, 20°C and 100°C? E. The voltage VC should now be 0V when the temperature is 0°C and increase by 10mV for every degree centigrade. We need to multiply this by 1.8, the factor to convert a degree Centigrade to a degree Fahrenheit. The output of this stage, V4, should range from 0V to 1.80V and then we will add an offset to change the VF output range to 0.32V to 2.12V in the last stage. The following circuit can be used. Note that this circuit uses inverting amplifiers instead of non-inverting amplifiers. (10 extra credit points) Find the correct resistor values for R7, R8 so that V4 will range between 0V to -1.80V when the temperature is sensed between 0°C and 100°C. (10 extra credit points) Find Voffset2 and then determine VF at 0°C, 1°C, 20°C and 100°C so the VF will range from 0.32V to 2.12V. (Hint: When VC = 0V at 0°C, the V4 output of this stage will also be 0V. Determine the offset voltage so VF = 0.32V. Choose R9 = R10 = R11 = 1kΩ, so at 0°C, VF = 0.32V = VR10 with the current going from the output back through R10 to zero volts, then down through R11 and the Voffset2 source. Since VR10=VR11=0.32V, determine Voffset2. When the temperature is 100°C the output should be 2.12V.)

A) At 0 deg C —> R =100 Ohm, I … Read More...
Count the atoms on both sides of the arrow to demonstrate that these equations are balanced. a. 2 C3H8(g) + 7 O2(g) →6CO(g) + 8HO(l) b. 2 C8H18(g) + 25O2(g) → 16 C02(g) + 18 H2O(l)

Count the atoms on both sides of the arrow to demonstrate that these equations are balanced. a. 2 C3H8(g) + 7 O2(g) →6CO(g) + 8HO(l) b. 2 C8H18(g) + 25O2(g) → 16 C02(g) + 18 H2O(l)

Both sides have 16 carbon , 36 hydrogen and 50 … Read More...