Lab Report : The purpose of this experiment is to learn about uncertainty in measurement, and how to perform calculations with those uncertainties. The calculation of the density of a wooden block was used as an example.

Lab Report : The purpose of this experiment is to learn about uncertainty in measurement, and how to perform calculations with those uncertainties. The calculation of the density of a wooden block was used as an example.

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The surfaces of the lens have radii of curvature 37.6 cm (located on the object side of the lens) and 55.4 cm (located on the side opposite the object). What is the index of refraction?

The surfaces of the lens have radii of curvature 37.6 cm (located on the object side of the lens) and 55.4 cm (located on the side opposite the object). What is the index of refraction?

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

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

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Comp755 HW 2 – Fall 2015 1. Problem 4.2 (10pts) 2. Problem 4.4 (15pts) 3. Problem 5.25 (10pts) 4. Problem 10.2 (15pts) . Hint: Create three Gantt charts where each square represents 10 time units. a. The first chart should use earliest deadline using the processes that are currently available. Since there is no preemption, some processes will not be scheduled if there start deadline is missed. b. The second chart should schedule strictly by the earliest deadline. The processor will be idle if the process with the earliest deadline has not arrived. c. The third chart should just use FCFS. 5. (Synchronize threads) Write a program that launches 1,000 threads. Each thread adds 1 to a variable sum that initially is 0. You need to pass sum by reference to each thread. In order to pass it by reference, define an Integer wrapper object to hold sum. Run the program with and without synchronization to see its effect (Create a command line argument where passing a 0 means run unsynchronized and passing a 1 means to run synchronized). Submit your entire Netbeans project (50pts).

Comp755 HW 2 – Fall 2015 1. Problem 4.2 (10pts) 2. Problem 4.4 (15pts) 3. Problem 5.25 (10pts) 4. Problem 10.2 (15pts) . Hint: Create three Gantt charts where each square represents 10 time units. a. The first chart should use earliest deadline using the processes that are currently available. Since there is no preemption, some processes will not be scheduled if there start deadline is missed. b. The second chart should schedule strictly by the earliest deadline. The processor will be idle if the process with the earliest deadline has not arrived. c. The third chart should just use FCFS. 5. (Synchronize threads) Write a program that launches 1,000 threads. Each thread adds 1 to a variable sum that initially is 0. You need to pass sum by reference to each thread. In order to pass it by reference, define an Integer wrapper object to hold sum. Run the program with and without synchronization to see its effect (Create a command line argument where passing a 0 means run unsynchronized and passing a 1 means to run synchronized). Submit your entire Netbeans project (50pts).

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Chapter 15 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, May 16, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Fluid Pressure in a U-Tube A U-tube is filled with water, and the two arms are capped. The tube is cylindrical, and the right arm has twice the radius of the left arm. The caps have negligible mass, are watertight, and can freely slide up and down the tube. Part A A one-inch depth of sand is poured onto the cap on each arm. After the caps have moved (if necessary) to reestablish equilibrium, is the right cap higher, lower, or the same height as the left cap? You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Pressure in the Ocean The pressure at 10 below the surface of the ocean is about 2.00×105 . Part A higher lower the same height m Pa Which of the following statements is true? You did not open hints for this part. ANSWER: Part B Now consider the pressure 20 below the surface of the ocean. Which of the following statements is true? You did not open hints for this part. ANSWER: Relating Pressure and Height in a Container Learning Goal: To understand the derivation of the law relating height and pressure in a container. The weight of a column of seawater 1 in cross section and 10 high is about 2.00×105 . The weight of a column of seawater 1 in cross section and 10 high plus the weight of a column of air with the same cross section extending up to the top of the atmosphere is about 2.00×105 . The weight of 1 of seawater at 10 below the surface of the ocean is about 2.00×105 . The density of seawater is about 2.00×105 times the density of air at sea level. m2 m N m2 m N m3 m N m The pressure is twice that at a depth of 10 . The pressure is the same as that at a depth of 10 . The pressure is equal to that at a depth of 10 plus the weight per 1 cross sectional area of a column of seawater 10 high. The pressure is equal to the weight per 1 cross sectional area of a column of seawater 20 high. m m m m2 m m2 m In this problem, you will derive the law relating pressure to height in a container by analyzing a particular system. A container of uniform cross-sectional area is filled with liquid of uniform density . Consider a thin horizontal layer of liquid (thickness ) at a height as measured from the bottom of the container. Let the pressure exerted upward on the bottom of the layer be and the pressure exerted downward on the top be . Assume throughout the problem that the system is in equilibrium (the container has not been recently shaken or moved, etc.). Part A What is , the magnitude of the force exerted upward on the bottom of the liquid? You did not open hints for this part. ANSWER: Part B What is , the magnitude of the force exerted downward on the top of the liquid? A  dy y p p + dp Fup Fup = Fdown You did not open hints for this part. ANSWER: Part C What is the weight of the thin layer of liquid? Express your answer in terms of quantities given in the problem introduction and , the magnitude of the acceleration due to gravity. You did not open hints for this part. ANSWER: Part D Since the liquid is in equilibrium, the net force on the thin layer of liquid is zero. Complete the force equation for the sum of the vertical forces acting on the liquid layer described in the problem introduction. Express your answer in terms of quantities given in the problem introduction and taking upward forces to be positive. You did not open hints for this part. ANSWER: Fdown = wlayer g wlayer = Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). A Submerged Ball A ball of mass and volume is lowered on a string into a fluid of density . Assume that the object would sink to the bottom if it were not supported by the string. Part A  = = i Fy,i mb V f What is the tension in the string when the ball is fully submerged but not touching the bottom, as shown in the figure? Express your answer in terms of any or all of the given quantities and , the magnitude of the acceleration due to gravity. You did not open hints for this part. ANSWER: Archimedes’ Principle Learning Goal: To understand the applications of Archimedes’ principle. Archimedes’ principle is a powerful tool for solving many problems involving equilibrium in fluids. It states the following: When a body is partially or completely submerged in a fluid (either a liquid or a gas), the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body. As a result of the upward Archimedes force (often called the buoyant force), some objects may float in a fluid, and all of them appear to weigh less. This is the familiar phenomenon of buoyancy. Quantitatively, the buoyant force can be found as , where is the force, is the density of the fluid, is the magnitude of the acceleration due to gravity, and is the volume of the displaced fluid. In this problem, you will be asked several qualitative questions that should help you develop a feel for Archimedes’ principle. An object is placed in a fluid and then released. Assume that the object either floats to the surface (settling so that the object is partly above and partly below the fluid surface) or sinks to the bottom. (Note that for Parts A through D, you should assume that the object has settled in equilibrium.) Part A Consider the following statement: The magnitude of the buoyant force is equal to the weight of fluid displaced by the object. Under what circumstances is this statement true? T g T = Fbuoyant = fluidgV Fbuoyant fluid g V You did not open hints for this part. ANSWER: Part B Consider the following statement: The magnitude of the buoyant force is equal to the weight of the amount of fluid that has the same total volume as the object. Under what circumstances is this statement true? You did not open hints for this part. ANSWER: Part C Consider the following statement: The magnitude of the buoyant force equals the weight of the object. Under what circumstances is this statement true? for every object submerged partially or completely in a fluid only for an object that floats only for an object that sinks for no object submerged in a fluid for an object that is partially submerged in a fluid only for an object that floats for an object completely submerged in a fluid for no object partially or completely submerged in a fluid You did not open hints for this part. ANSWER: Part D Consider the following statement: The magnitude of the buoyant force is less than the weight of the object. Under what circumstances is this statement true? ANSWER: Now apply what you know to some more complicated situations. Part E An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, filled with a denser liquid. What would you observe? You did not open hints for this part. ANSWER: for every object submerged partially or completely in a fluid for an object that floats only for an object that sinks for no object submerged in a fluid for every object submerged partially or completely in a fluid for an object that floats for an object that sinks for no object submerged in a fluid Part F An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, filled with a less dense liquid. What would you observe? You did not open hints for this part. ANSWER: Part G Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled with a liquid. Both objects float in equilibrium. Less of object T is submerged than of object B, which floats, fully submerged, closer to the bottom of the container. Which of the following statements is true? ANSWER: The object would sink all the way to the bottom. The object would float submerged more deeply than in the first container. The object would float submerged less deeply than in the first container. More than one of these outcomes is possible. The object would sink all the way to the bottom. The object would float submerged more deeply than in the first container. The object would float submerged less deeply than in the first container. More than one of these outcomes is possible. Object T has a greater density than object B. Object B has a greater density than object T. Both objects have the same density. ± Buoyant Force Conceptual Question A rectangular wooden block of weight floats with exactly one-half of its volume below the waterline. Part A What is the buoyant force acting on the block? You did not open hints for this part. ANSWER: Part B W The buoyant force cannot be determined. 2W W 1 W 2 The density of water is 1.00 . What is the density of the block? You did not open hints for this part. ANSWER: Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). g/cm3 2.00 between 1.00 and 2.00 1.00 between 0.50 and 1.00 0.50 The density cannot be determined. g/cm3 g/cm3 g/cm3 g/cm3 g/cm3 Flow Velocity of Blood Conceptual Question Arteriosclerotic plaques forming on the inner walls of arteries can decrease the effective cross-sectional area of an artery. Even small changes in the effective area of an artery can lead to very large changes in the blood pressure in the artery and possibly to the collapse of the blood vessel. Imagine a healthy artery, with blood flow velocity of and mass per unit volume of . The kinetic energy per unit volume of blood is given by Imagine that plaque has narrowed an artery to one-fifth of its normal cross-sectional area (an 80% blockage). Part A Compared to normal blood flow velocity, , what is the velocity of blood as it passes through this blockage? You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C v0 = 0.14 m/s  = 1050 kg/m3 K0 =  . 1 2 v20 v0 80v0 20v0 5v0 v0/5 This question will be shown after you complete previous question(s). For parts D – F imagine that plaque has grown to a 90% blockage. Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). ± Playing with a Water Hose Two children, Ferdinand and Isabella, are playing with a water hose on a sunny summer day. Isabella is holding the hose in her hand 1.0 meters above the ground and is trying to spray Ferdinand, who is standing 10.0 meters away. Part A Will Isabella be able to spray Ferdinand if the water is flowing out of the hose at a constant speed of 3.5 meters per second? Assume that the hose is pointed parallel to the ground and take the magnitude of the acceleration due to gravity to be 9.81 meters per second, per second. You did not open hints for this part. v0 g ANSWER: Part B This question will be shown after you complete previous question(s). Tactics Box 15.2 Finding Whether an Object Floats or Sinks Learning Goal: To practice Tactics Box 15.2 Finding whether an object floats or sinks. If you hold an object underwater and then release it, it can float to the surface, sink, or remain “hanging” in the water, depending on whether the fluid density is larger than, smaller than, or equal to the object’s average density . These conditions are summarized in this Tactics Box. Yes No f avg TACTICS BOX 15.2 Finding whether an object floats or sinks Object sinks Object floats Object has neutral buoyancy An object sinks if it weighs more than the fluid it displaces, that is, if its average density is greater than the density of the fluid: . An object floats on the surface if it weighs less than the fluid it displaces, that is, if its average density is less than the density of the fluid: . An object hangs motionless in the fluid if it weighs exactly the same as the fluid it displaces. It has neutral buoyancy if its average density equals the density of the fluid: . Part A Ice at 0.0 has a density of 917 . A 3.00 ice cube is gently released inside a small container filled with oil and is observed to be neutrally buoyant. What is the density of the oil, ? Express your answer in kilograms per meter cubed to three significant figures. ANSWER: Part B Once the ice cube melts, what happens to the liquid water that it produces? You did not open hints for this part. ANSWER: avg > f avg < f avg = f 'C kg/m3 cm3 oil oil = kg/m3 Part C What happens if some ethyl alcohol of density 790 is poured into the container after the ice cube has melted? ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. The liquid water sinks to the bottom of the container. The liquid water rises to the surface and floats on top of the oil. The liquid water is in static equilibrium at the location where the ice cube was originally placed. kg/m3 A layer of ethyl alcohol forms between the oil and the water. The layer of ethyl alcohol forms at the bottom of the container. The layer of ethyl alcohol forms on the surface.

Chapter 15 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, May 16, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Fluid Pressure in a U-Tube A U-tube is filled with water, and the two arms are capped. The tube is cylindrical, and the right arm has twice the radius of the left arm. The caps have negligible mass, are watertight, and can freely slide up and down the tube. Part A A one-inch depth of sand is poured onto the cap on each arm. After the caps have moved (if necessary) to reestablish equilibrium, is the right cap higher, lower, or the same height as the left cap? You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Pressure in the Ocean The pressure at 10 below the surface of the ocean is about 2.00×105 . Part A higher lower the same height m Pa Which of the following statements is true? You did not open hints for this part. ANSWER: Part B Now consider the pressure 20 below the surface of the ocean. Which of the following statements is true? You did not open hints for this part. ANSWER: Relating Pressure and Height in a Container Learning Goal: To understand the derivation of the law relating height and pressure in a container. The weight of a column of seawater 1 in cross section and 10 high is about 2.00×105 . The weight of a column of seawater 1 in cross section and 10 high plus the weight of a column of air with the same cross section extending up to the top of the atmosphere is about 2.00×105 . The weight of 1 of seawater at 10 below the surface of the ocean is about 2.00×105 . The density of seawater is about 2.00×105 times the density of air at sea level. m2 m N m2 m N m3 m N m The pressure is twice that at a depth of 10 . The pressure is the same as that at a depth of 10 . The pressure is equal to that at a depth of 10 plus the weight per 1 cross sectional area of a column of seawater 10 high. The pressure is equal to the weight per 1 cross sectional area of a column of seawater 20 high. m m m m2 m m2 m In this problem, you will derive the law relating pressure to height in a container by analyzing a particular system. A container of uniform cross-sectional area is filled with liquid of uniform density . Consider a thin horizontal layer of liquid (thickness ) at a height as measured from the bottom of the container. Let the pressure exerted upward on the bottom of the layer be and the pressure exerted downward on the top be . Assume throughout the problem that the system is in equilibrium (the container has not been recently shaken or moved, etc.). Part A What is , the magnitude of the force exerted upward on the bottom of the liquid? You did not open hints for this part. ANSWER: Part B What is , the magnitude of the force exerted downward on the top of the liquid? A  dy y p p + dp Fup Fup = Fdown You did not open hints for this part. ANSWER: Part C What is the weight of the thin layer of liquid? Express your answer in terms of quantities given in the problem introduction and , the magnitude of the acceleration due to gravity. You did not open hints for this part. ANSWER: Part D Since the liquid is in equilibrium, the net force on the thin layer of liquid is zero. Complete the force equation for the sum of the vertical forces acting on the liquid layer described in the problem introduction. Express your answer in terms of quantities given in the problem introduction and taking upward forces to be positive. You did not open hints for this part. ANSWER: Fdown = wlayer g wlayer = Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). A Submerged Ball A ball of mass and volume is lowered on a string into a fluid of density . Assume that the object would sink to the bottom if it were not supported by the string. Part A  = = i Fy,i mb V f What is the tension in the string when the ball is fully submerged but not touching the bottom, as shown in the figure? Express your answer in terms of any or all of the given quantities and , the magnitude of the acceleration due to gravity. You did not open hints for this part. ANSWER: Archimedes’ Principle Learning Goal: To understand the applications of Archimedes’ principle. Archimedes’ principle is a powerful tool for solving many problems involving equilibrium in fluids. It states the following: When a body is partially or completely submerged in a fluid (either a liquid or a gas), the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body. As a result of the upward Archimedes force (often called the buoyant force), some objects may float in a fluid, and all of them appear to weigh less. This is the familiar phenomenon of buoyancy. Quantitatively, the buoyant force can be found as , where is the force, is the density of the fluid, is the magnitude of the acceleration due to gravity, and is the volume of the displaced fluid. In this problem, you will be asked several qualitative questions that should help you develop a feel for Archimedes’ principle. An object is placed in a fluid and then released. Assume that the object either floats to the surface (settling so that the object is partly above and partly below the fluid surface) or sinks to the bottom. (Note that for Parts A through D, you should assume that the object has settled in equilibrium.) Part A Consider the following statement: The magnitude of the buoyant force is equal to the weight of fluid displaced by the object. Under what circumstances is this statement true? T g T = Fbuoyant = fluidgV Fbuoyant fluid g V You did not open hints for this part. ANSWER: Part B Consider the following statement: The magnitude of the buoyant force is equal to the weight of the amount of fluid that has the same total volume as the object. Under what circumstances is this statement true? You did not open hints for this part. ANSWER: Part C Consider the following statement: The magnitude of the buoyant force equals the weight of the object. Under what circumstances is this statement true? for every object submerged partially or completely in a fluid only for an object that floats only for an object that sinks for no object submerged in a fluid for an object that is partially submerged in a fluid only for an object that floats for an object completely submerged in a fluid for no object partially or completely submerged in a fluid You did not open hints for this part. ANSWER: Part D Consider the following statement: The magnitude of the buoyant force is less than the weight of the object. Under what circumstances is this statement true? ANSWER: Now apply what you know to some more complicated situations. Part E An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, filled with a denser liquid. What would you observe? You did not open hints for this part. ANSWER: for every object submerged partially or completely in a fluid for an object that floats only for an object that sinks for no object submerged in a fluid for every object submerged partially or completely in a fluid for an object that floats for an object that sinks for no object submerged in a fluid Part F An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, filled with a less dense liquid. What would you observe? You did not open hints for this part. ANSWER: Part G Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled with a liquid. Both objects float in equilibrium. Less of object T is submerged than of object B, which floats, fully submerged, closer to the bottom of the container. Which of the following statements is true? ANSWER: The object would sink all the way to the bottom. The object would float submerged more deeply than in the first container. The object would float submerged less deeply than in the first container. More than one of these outcomes is possible. The object would sink all the way to the bottom. The object would float submerged more deeply than in the first container. The object would float submerged less deeply than in the first container. More than one of these outcomes is possible. Object T has a greater density than object B. Object B has a greater density than object T. Both objects have the same density. ± Buoyant Force Conceptual Question A rectangular wooden block of weight floats with exactly one-half of its volume below the waterline. Part A What is the buoyant force acting on the block? You did not open hints for this part. ANSWER: Part B W The buoyant force cannot be determined. 2W W 1 W 2 The density of water is 1.00 . What is the density of the block? You did not open hints for this part. ANSWER: Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). g/cm3 2.00 between 1.00 and 2.00 1.00 between 0.50 and 1.00 0.50 The density cannot be determined. g/cm3 g/cm3 g/cm3 g/cm3 g/cm3 Flow Velocity of Blood Conceptual Question Arteriosclerotic plaques forming on the inner walls of arteries can decrease the effective cross-sectional area of an artery. Even small changes in the effective area of an artery can lead to very large changes in the blood pressure in the artery and possibly to the collapse of the blood vessel. Imagine a healthy artery, with blood flow velocity of and mass per unit volume of . The kinetic energy per unit volume of blood is given by Imagine that plaque has narrowed an artery to one-fifth of its normal cross-sectional area (an 80% blockage). Part A Compared to normal blood flow velocity, , what is the velocity of blood as it passes through this blockage? You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C v0 = 0.14 m/s  = 1050 kg/m3 K0 =  . 1 2 v20 v0 80v0 20v0 5v0 v0/5 This question will be shown after you complete previous question(s). For parts D – F imagine that plaque has grown to a 90% blockage. Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). ± Playing with a Water Hose Two children, Ferdinand and Isabella, are playing with a water hose on a sunny summer day. Isabella is holding the hose in her hand 1.0 meters above the ground and is trying to spray Ferdinand, who is standing 10.0 meters away. Part A Will Isabella be able to spray Ferdinand if the water is flowing out of the hose at a constant speed of 3.5 meters per second? Assume that the hose is pointed parallel to the ground and take the magnitude of the acceleration due to gravity to be 9.81 meters per second, per second. You did not open hints for this part. v0 g ANSWER: Part B This question will be shown after you complete previous question(s). Tactics Box 15.2 Finding Whether an Object Floats or Sinks Learning Goal: To practice Tactics Box 15.2 Finding whether an object floats or sinks. If you hold an object underwater and then release it, it can float to the surface, sink, or remain “hanging” in the water, depending on whether the fluid density is larger than, smaller than, or equal to the object’s average density . These conditions are summarized in this Tactics Box. Yes No f avg TACTICS BOX 15.2 Finding whether an object floats or sinks Object sinks Object floats Object has neutral buoyancy An object sinks if it weighs more than the fluid it displaces, that is, if its average density is greater than the density of the fluid: . An object floats on the surface if it weighs less than the fluid it displaces, that is, if its average density is less than the density of the fluid: . An object hangs motionless in the fluid if it weighs exactly the same as the fluid it displaces. It has neutral buoyancy if its average density equals the density of the fluid: . Part A Ice at 0.0 has a density of 917 . A 3.00 ice cube is gently released inside a small container filled with oil and is observed to be neutrally buoyant. What is the density of the oil, ? Express your answer in kilograms per meter cubed to three significant figures. ANSWER: Part B Once the ice cube melts, what happens to the liquid water that it produces? You did not open hints for this part. ANSWER: avg > f avg < f avg = f 'C kg/m3 cm3 oil oil = kg/m3 Part C What happens if some ethyl alcohol of density 790 is poured into the container after the ice cube has melted? ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. The liquid water sinks to the bottom of the container. The liquid water rises to the surface and floats on top of the oil. The liquid water is in static equilibrium at the location where the ice cube was originally placed. kg/m3 A layer of ethyl alcohol forms between the oil and the water. The layer of ethyl alcohol forms at the bottom of the container. The layer of ethyl alcohol forms on the surface.

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

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

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Chapter 6 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 PSS 6.1 Equilibrium Problems Learning Goal: To practice Problem-Solving Strategy 6.1 for equilibrium problems. A pair of students are lifting a heavy trunk on move-in day. Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity . Each rope makes an angle with respect to the vertical. The gravitational force acting on the trunk has magnitude . Find the tension in each rope. PROBLEM-SOLVING STRATEGY 6.1 Equilibrium problems MODEL: Make simplifying assumptions. VISUALIZE: Establish a coordinate system, define symbols, and identify what the problem is asking you to find. This is the process of translating words into symbols. Identify all forces acting on the object, and show them on a free-body diagram. These elements form the pictorial representation of the problem. SOLVE: The mathematical representation is based on Newton’s first law: . The vector sum of the forces is found directly from the free-body diagram. v  FG T F  = = net i F  i 0

Chapter 6 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 PSS 6.1 Equilibrium Problems Learning Goal: To practice Problem-Solving Strategy 6.1 for equilibrium problems. A pair of students are lifting a heavy trunk on move-in day. Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity . Each rope makes an angle with respect to the vertical. The gravitational force acting on the trunk has magnitude . Find the tension in each rope. PROBLEM-SOLVING STRATEGY 6.1 Equilibrium problems MODEL: Make simplifying assumptions. VISUALIZE: Establish a coordinate system, define symbols, and identify what the problem is asking you to find. This is the process of translating words into symbols. Identify all forces acting on the object, and show them on a free-body diagram. These elements form the pictorial representation of the problem. SOLVE: The mathematical representation is based on Newton’s first law: . The vector sum of the forces is found directly from the free-body diagram. v  FG T F  = = net i F  i 0

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Question 10 (1 point) In contrast to Freud’s theory, object relations theorists Question 10 options: focus on internal drives and conflicts. are interested in the intellectual and emotional development of the infant. are interested in an infant’s relationship with his or her parents. do not believe that children develop unconscious representations of significant objects in their environment. ________________________________________ Question 11 (1 point) The psychologists who developed the frustration aggression hypothesis used or adapted each of the following concepts from Freudian theory except one. Which one? Question 11 options: displacement sublimation catharsis reinforcement Question 12 (1 point) Although he changed his mind during his career, which of the following did Freud eventually decide was the cause of human aggression? Question 12 options: a death instinct frustration projection unresolved Oedipal conflicts Question 13 (1 point) Freud wrote about all of the following types of anxiety except one. Which one? Question 13 options: reality anxiety neurotic anxiety moral anxiety performance anxiety Question 14 (1 point) Which of the following is true about neurotic anxiety, as conceived by Freud? Question 14 options: It is experienced when id impulses are close to breaking into consciousness. It prevents the ego from utilizing defense mechanisms. It is created when id impulses violate society’s moral code. People experiencing neurotic anxiety usually are aware of what is making them anxious. Question 15 (1 point) One explanation for why aggression leads to more aggression is that it is reinforced by the cathartic release of tension. Question 15 options: True False ________________________________________ ________________________________________ Question 1 (1 point) A man is said to have one personality trait that dominates his personality. Allport would identify this personality trait as a Question 1 options: 1) common trait. 2) central trait. 3) cardinal trait. 4) secondary trait. Question 2 (1 point) Which of the following is true about the trait approach to personality? Question 2 options: 1) Trait researchers generally are not interested in understanding and predicting the behavior of a single individual. 2) It is not easy to make comparisons across people with the trait approach. 3) The trait approach has been responsible for generating a number of useful approaches to psychotherapy. 4) Trait theorists place a greater emphasis on discovering the mechanisms underlying behavior than do theorists from other approaches to personality. Question 3 (1 point) Many researchers fail to produce strong links between personality traits and behavior. Epstein has argued that the reason for this failure is because Question 3 options: 1) researchers don’t perform the correct statistical analysis. 2) researchers don’t measure personality traits correctly. 3) researchers don’t measure behavior correctly. 4) none of the above Question 4 (1 point) Which theorist had a strong influence on Henry Murray’s theorizing about personality? Question 4 options: 1) Gordon Allport 2) Alfred Adler 3) Sigmund Freud 4) Carl Jung Question 5 (1 point) Sometimes test makers include the same test questions more than once on the test. This is done to detect which potential problem? Question 5 options: 1) faking good 2) faking bad 3) carelessness and sabotage 4) social desirability

Question 10 (1 point) In contrast to Freud’s theory, object relations theorists Question 10 options: focus on internal drives and conflicts. are interested in the intellectual and emotional development of the infant. are interested in an infant’s relationship with his or her parents. do not believe that children develop unconscious representations of significant objects in their environment. ________________________________________ Question 11 (1 point) The psychologists who developed the frustration aggression hypothesis used or adapted each of the following concepts from Freudian theory except one. Which one? Question 11 options: displacement sublimation catharsis reinforcement Question 12 (1 point) Although he changed his mind during his career, which of the following did Freud eventually decide was the cause of human aggression? Question 12 options: a death instinct frustration projection unresolved Oedipal conflicts Question 13 (1 point) Freud wrote about all of the following types of anxiety except one. Which one? Question 13 options: reality anxiety neurotic anxiety moral anxiety performance anxiety Question 14 (1 point) Which of the following is true about neurotic anxiety, as conceived by Freud? Question 14 options: It is experienced when id impulses are close to breaking into consciousness. It prevents the ego from utilizing defense mechanisms. It is created when id impulses violate society’s moral code. People experiencing neurotic anxiety usually are aware of what is making them anxious. Question 15 (1 point) One explanation for why aggression leads to more aggression is that it is reinforced by the cathartic release of tension. Question 15 options: True False ________________________________________ ________________________________________ Question 1 (1 point) A man is said to have one personality trait that dominates his personality. Allport would identify this personality trait as a Question 1 options: 1) common trait. 2) central trait. 3) cardinal trait. 4) secondary trait. Question 2 (1 point) Which of the following is true about the trait approach to personality? Question 2 options: 1) Trait researchers generally are not interested in understanding and predicting the behavior of a single individual. 2) It is not easy to make comparisons across people with the trait approach. 3) The trait approach has been responsible for generating a number of useful approaches to psychotherapy. 4) Trait theorists place a greater emphasis on discovering the mechanisms underlying behavior than do theorists from other approaches to personality. Question 3 (1 point) Many researchers fail to produce strong links between personality traits and behavior. Epstein has argued that the reason for this failure is because Question 3 options: 1) researchers don’t perform the correct statistical analysis. 2) researchers don’t measure personality traits correctly. 3) researchers don’t measure behavior correctly. 4) none of the above Question 4 (1 point) Which theorist had a strong influence on Henry Murray’s theorizing about personality? Question 4 options: 1) Gordon Allport 2) Alfred Adler 3) Sigmund Freud 4) Carl Jung Question 5 (1 point) Sometimes test makers include the same test questions more than once on the test. This is done to detect which potential problem? Question 5 options: 1) faking good 2) faking bad 3) carelessness and sabotage 4) social desirability

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