## Will carbon dioxide and or carbon monoxide bond to activated rock carbon in an environment of 350 to 1200 degrees F. with a psi of up to 50 or so.

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## You will receive no credit for items you complete after the assignment is due. Grading Policy Exercise 2.5 Starting from the front door of your ranch house, you walk 60.0 due east to your windmill, and then you turn around and slowly walk 35.0 west to a bench where you sit and watch the sunrise. It takes you 27.0 to walk from your house to the windmill and then 49.0 to walk from the windmill to the bench. Part A For the entire trip from your front door to the bench, what is your average velocity? Express your answer with the appropriate units. ANSWER: Correct Part B For the entire trip from your front door to the bench, what is your average speed? Express your answer with the appropriate units. ANSWER: Correct Exercise 2.7 A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by , where = 2.40 and = 0.110 . = -0.329 average speed = 1.25 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 1 of 16 3/23/2015 11:12 AM Part A Calculate the average velocity of the car for the time interval = 0 to = 10.0 . ANSWER: Correct Part B Calculate the instantaneous velocity of the car at =0. ANSWER: Correct Part C Calculate the instantaneous velocity of the car at =5.00 . ANSWER: Correct Part D Calculate the instantaneous velocity of the car at =10.0 . ANSWER: Correct Part E How long after starting from rest is the car again at rest? ANSWER: = 13.0 = 0 = 15.8 = 15.0 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 2 of 16 3/23/2015 11:12 AM Correct Exercise 2.9 A ball moves in a straight line (the x-axis). The graph in the figure shows this ball’s velocity as a function of time. Part A What are the ball’s average velocity during the first 2.8 ? Express your answer using two significant figures. ANSWER: Answer Requested Part B What are the ball’s average speed during the first 2.8 ? Express your answer using two significant figures. ANSWER: Correct = 14.5 = 2.3 = 2.3 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 3 of 16 3/23/2015 11:12 AM Part C Suppose that the ball moved in such a way that the graph segment after 2.0 was -3.0 instead of +3.0 . Find the ball’s and average velocity during the first 2.8 in this case. Express your answer using two significant figures. ANSWER: All attempts used; correct answer displayed Part D Suppose that the ball moved in such a way that the graph segment after 2.0 was -3.0 instead of +3.0 . Find the ball’s average speed during the first 2.8 in this case. Express your answer using two significant figures. ANSWER: Correct Exercise 2.13 Part A The table shows test data for the Bugatti Veyron, the fastest car made. The car is moving in a straight line (the x-axis). Time 0 2.10 20.0 53.0 Speed 0 60.0 205 259 Calculate the car’s average acceleration (in ) between 0 and 2.1 . ANSWER: Correct Part B Calculate the car’s average acceleration (in ) between 2.1 and 20.0 . = 0.57 = 2.3 = 12.8 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 4 of 16 3/23/2015 11:12 AM ANSWER: Correct Part C Calculate the car’s average acceleration (in ) between 20.0 and 53 . ANSWER: Correct Exercise 2.19 An antelope moving with constant acceleration covers the distance 79.0 between two points in time 7.00 . Its speed as it passes the second point is 14.5 . Part A What is its speed at the first point? ANSWER: Correct Part B What is the acceleration? ANSWER: Correct Exercise 2.22 In the fastest measured tennis serve, the ball left the racquet at 73.14 . A served tennis ball is typically in contact with = 3.62 = 0.731 = 8.07 = 0.918 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 5 of 16 3/23/2015 11:12 AM the racquet for 27.0 and starts from rest. Assume constant acceleration. Part A What was the ball’s acceleration during this serve? ANSWER: Correct Part B How far did the ball travel during the serve? ANSWER: Correct Exercise 2.30 A cat walks in a straight line, which we shall call the x-axis with the positive direction to the right. As an observant physicist, you make measurements of this cat’s motion and construct a graph of the feline’s velocity as a function of time (the figure ). Part A Find the cat’s velocity at = 5.0 . Express your answer using two significant figures. ANSWER: = 2710 = 0.987 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 6 of 16 3/23/2015 11:12 AM Correct Part B Find the cat’s velocity at = 8.0 . Express your answer using two significant figures. ANSWER: Correct Part C What is the cat’s acceleration at ? Express your answer using two significant figures. ANSWER: Correct Part D What is the cat’s acceleration at ? Express your answer using two significant figures. ANSWER: Correct Part E What is the cat’s acceleration at ? Express your answer using two significant figures. ANSWER: = 1.3 = -2.7 = -1.3 = -1.3 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 7 of 16 3/23/2015 11:12 AM Correct Part F What distance does the cat move during the first 4.5 ? Express your answer using two significant figures. ANSWER: Correct Part G What distance does the cat move from to ? Express your answer using two significant figures. ANSWER: Correct Part H Sketch clear graph of the cat’s acceleration as function of time, assuming that the cat started at the origin. ANSWER: = -1.3 = 23 = 26 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 8 of 16 3/23/2015 11:12 AM Correct Part I Sketch clear graph of the cat’s position as function of time, assuming that the cat started at the origin. ANSWER: Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 9 of 16 3/23/2015 11:12 AM All attempts used; correct answer displayed Exercise 2.35 Part A If a flea can jump straight up to a height of 0.510 , what is its initial speed as it leaves the ground? ANSWER: Correct Part B How long is it in the air? ANSWER: Correct = 3.16 = 0.645 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 10 of 16 3/23/2015 11:12 AM Exercise 2.36 A small rock is thrown vertically upward with a speed of 18.0 from the edge of the roof of a 39.0 tall building. The rock doesn’t hit the building on its way back down and lands in the street below. Air resistance can be neglected. Part A What is the speed of the rock just before it hits the street? Express your answer with the appropriate units. ANSWER: Correct Part B How much time elapses from when the rock is thrown until it hits the street? Express your answer with the appropriate units. ANSWER: Correct Exercise 2.38 You throw a glob of putty straight up toward the ceiling, which is 3.00 above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.70 . Part A What is the speed of the putty just before it strikes the ceiling? Express your answer with the appropriate units. ANSWER: Correct Part B = 33.0 = 5.20 = 5.94 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 11 of 16 3/23/2015 11:12 AM How much time from when it leaves your hand does it take the putty to reach the ceiling? Express your answer with the appropriate units. ANSWER: Correct Exercise 3.1 A squirrel has x- and y-coordinates ( 1.2 , 3.3 ) at time and coordinates ( 5.3 , -0.80 ) at time = 2.6 . Part A For this time interval, find the x-component of the average velocity. Express your answer using two significant figures. ANSWER: Correct Part B For this time interval, find the y-component of the average velocity. Express your answer using two significant figures. ANSWER: Correct Part C Find the magnitude of the average velocity. Express your answer using two significant figures. ANSWER: = 0.384 = 1.6 = -1.6 = 2.2 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 12 of 16 3/23/2015 11:12 AM Correct Part D Find the direction of the average velocity. Express your answer using two significant figures. ANSWER: Correct Exercise 3.3 A web page designer creates an animation in which a dot on a computer screen has a position of 4.1 2.1 4.7 . Part A Find the average velocity of the dot between and . Give your answer as a pair of components separated by a comma. For example, if you think the x component is 3 and the y component is 4, then you should enter 3,4. Express your answer using two significant figures. ANSWER: Correct Part B Find the instantaneous velocity at . Give your answer as a pair of components separated by a comma. For example, if you think the x component is 3 and the y component is 4, then you should enter 3,4. Express your answer using two significant figures. ANSWER: Correct Part C = 45 below the x-axis = 4.2,4.7 = 0,4.7 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 13 of 16 3/23/2015 11:12 AM Find the instantaneous velocity at . Give your answer as a pair of components separated by a comma. For example, if you think the x component is 3 and the y component is 4, then you should enter 3,4. Express your answer using two significant figures. ANSWER: Correct Part D Find the instantaneous velocity at . Give your answer as a pair of components separated by a comma. For example, if you think the x component is 3 and the y component is 4, then you should enter 3,4. Express your answer using two significant figures. ANSWER: Correct Exercise 3.5 A jet plane is flying at a constant altitude. At time it has components of velocity 89 , 108 . At time 32.5 the components are 165 , 37 . Part A For this time interval calculate the average acceleration. Give your answer as a pair of components separated by a comma. For example, if you think the x component is 3 and the y component is 4, then you should enter 3,4. Express your answer using two significant figures. ANSWER: Correct Part B Find the magnitude of the average acceleration. Express your answer using two significant figures. ANSWER: = 4.2,4.7 = 8.4,4.7 = 2.3,-2.2 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 14 of 16 3/23/2015 11:12 AM Correct Part C Find the direction of the average acceleration (let the direction be the angle that the vector makes with the +x-axis, measured counterclockwise). ANSWER: Correct Exercise 3.4 The position of a squirrel running in a park is given by . Part A What is , the -component of the velocity of the squirrel, as function of time? ANSWER: Correct Part B What is , the y-component of the velocity of the squirrel, as function of time? ANSWER: = 3.2 = -43.1 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 15 of 16 3/23/2015 11:12 AM Correct Part C At 4.51 , how far is the squirrel from its initial position? Express your answer to three significant figures and include the appropriate units. ANSWER: All attempts used; correct answer displayed Part D At 4.51 , what is the magnitude of the squirrel’s velocity? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part E At 4.51 , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel’s velocity? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 90.1%. You received 14.42 out of a possible total of 16 points. = 2.65 = 1.31 = 62.5 Week 2 https://session.masteringphysics.com/myct/assignmentPrintView?assignme… 16 of 16 3/23/2015 11:12 AM

## Phys4A: Practice problems for the 1st midterm test Fall 2015 1 If K has dimensions ML2/T 2, the k in K = kmv 2 must be: Answer: dimensionless 2. A 8.7 hour trip is made at an average speed of 73.0 km/h. If the first third of the trip (chronologically) was driven at 96.5 km/h, what was the average speed for the rest of the journey? Answer: 61 km/h 3. A car travels 95 km to the north at 70.0 km/h, then turns around and travels 21.9 km at 80.0 km/h. What is the difference between the average speed and the average velocity on this trip? Answer: 27 km/h 4. A particle confined to motion along the x axis moves with constant acceleration from x = 2.0 m to x = 8.0 m during a 2.5s time interval. The velocity of the particle at x = 8.0 m is 2.8 m/s. What is the acceleration during this time interval? Answer: 0.32 m/s2 5. A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released? (Disregard air resistance.) Answer: 1000m 6. If vector B is added to vector A, the result is 6i + j. If B is subtracted from A, the result is –4i + 7j. What is the magnitude of A? Answer: 4.1 7. Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second oasis. What is the direction from the first oasis to the second oasis? Answer: 51° W of N 8 A river 1.00 mile wide flows with a constant speed of 1.00 mi/h. A man can row a boat at 2.00 mi/h. He crosses the river in a direction that puts him directly across the river from the starting point, and then he returns in a direction that puts him back at the starting point in the shortest time possible. The travel time for the man is, Answer: 1.15 h 9 An airplane is heading due east. The airspeed indicator shows that the plane is moving at a speed of 370 km/h relative to the air. If the wind is blowing from the south at 92.5 km/h, the velocity of the airplane relative to the ground is: Answer: 381 km/h at 76o east of north 10. A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53° above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. How tall is the building? Answer: 23.6m 11. A boy throws a rock with an initial velocity of 3.13 m/s at 30.0° above the horizontal. How long does it take for the rock to reach the maximum height of its trajectory? Answer: 0.160 s 12. A helicopter is traveling at 54 m/s at a constant altitude of 100 m over a level field. If a wheel falls off the helicopter, with what speed will it hit the ground? Note: air resistance negligible. Answer: 70 m/s 13 A rescue airplane is diving at an angle of 37º below the horizontal with a speed of 250 m/s. It releases a survival package when it is at an altitude of 600 m. If air resistance is ignored, the horizontal distance of the point of impact from the plane at the moment of the package’s release is, Answer: 720 m 14. A hobby rocket reaches a height of 72.3 m and lands 111 m from the launch point. What was the angle of launch? Answer: 69.0° 15. An object moving at a constant speed requires 6.0 s to go once around a circle with a diameter of 4.0 m. What is the magnitude of the instantaneous acceleration of the particle during this time? Answer: 2.2 m/s2 16 A ball is whirled in a horizontal circle of radius r and speed v. The radius is increased to 2r keeping the speed of the ball constant. The period of the ball changes by a factor of Answer: two

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## Chapter 03 Reading Questions Due: 11:59pm on Friday, May 23, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Chapter 3 Reading Quiz Question 1 Part A Isotopes of an element differ from each other by the _____. ANSWER: Correct Chapter 3 Reading Quiz Question 2 Part A Which one of the following statements about pH is correct? ANSWER: Correct Lemon juice is an acid. Chapter 3 Reading Quiz Question 17 Part A In which form are water molecules most closely bonded to each other? ANSWER: number of electrons number of neutrons types of electrons number of protons Stomach acid has more OH- ions than H+ ions. Baking soda has more H+ ions than OH- ions. Lemon juice has more H+ ions than OH- ions. Seawater is slightly acidic. Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 1 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 16 Part A Which one of the following is a molecule but NOT a compound? ANSWER: Correct Oxygen is a molecule made up of just one element. Therefore, it is not a compound. Chapter 3 Reading Quiz Question 3 Part A Which one of the following is a carbohydrate and one of Earth’s most abundant organic molecule? ANSWER: Correct equally closely bonded in water vapor and ice solid ice forming part of an Antarctic sheet liquid water a few degrees above the freezing point water vapor above a boiling pot of water CH4 O2 CO2 H2O oil protein cellulose DNA Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 2 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 4 Part A Which one of the following is a protein that functions as a catalyst? ANSWER: Correct Chapter 3 Reading Quiz Question 18 Part A The process of translation involves the use of _____. ANSWER: Chapter 3 Reading Quiz Question 5 Part A The cooling effect of sweating best represents _____. ANSWER: glucose cellulose enzyme RNA proteins to make lipids lipids to make carbohydrates carbohydrates to make proteins nucleic acids to make proteins latent heat transfer conduction radiation convection Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 3 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 6 Part A When plants use sunlight in photosynthesis, the plants are using a form of _____. ANSWER: Correct Chapter 3 Reading Quiz Question 8 Part A Which of the following converts mass to energy? ANSWER: Correct Chapter 3 Reading Quiz Question 19 Part A When a windmill turns to generate electricity, the amount of kinetic energy input _____. ANSWER: chemical energy in sunlight nuclear fission electromagnetic radiation conduction conduction the breaking of chemical bonds nuclear fission photosynthesis Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 4 of 9 5/21/2014 7:58 PM Correct Chapter 3 Reading Quiz Question 20 Part A Which of the following best represents kinetic energy? ANSWER: Correct Chapter 3 Reading Quiz Question 21 Part A Which of the following processes reduces entropy? ANSWER: Correct Chapter 3 Reading Quiz Question 9 is unrelated to the amount of electrical energy produced is more than the amount of electrical energy produced equals the amount of electrical energy produced is less than the amount of electrical energy produced a charged battery gunpowder in a bullet the energy in the wax molecules of a candle a hot burner on a stove burning gasoline in an automobile engine photosynthesis in a leaf a person walking up a flight of stairs cell respiration in a leaf Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 5 of 9 5/21/2014 7:58 PM Part A Which one of the following planets is a gas giant? ANSWER: Correct Chapter 3 Reading Quiz Question 10 Part A What is the main driving force that causes Earth’s tectonic plates to drift? ANSWER: Correct Chapter 3 Reading Quiz Question 23 Part A In which of the following locations would you expect to find large quantities of young rocks? ANSWER: Venus Jupiter Mars Mercury Heat from Earth’s core causes the mantle rock to circulate. The weight of the tectonic plates causes them to sink and melt. Currents of magma from the core of Earth circulate just beneath the tectonic plates. Electromagnetic radiation from the sun heats the tectonic plates, causing them to expand. the Appalachian Mountains the Himalayas deep in the central parts of India the Mid-Atlantic Ridge Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 6 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 12 Part A The oxygen-rich atmosphere of Earth is mainly the result of _____. ANSWER: Correct Chapter 3 Reading Quiz Question 13 Part A A scientist working on the chemical reactions in the ozone layer is studying the _____. ANSWER: Correct Chapter 3 Reading Quiz Question 24 Part A The total amount of moisture in the air is highest when relative humidity is _____. ANSWER: volcanic activity chemical reactions between the early Earth atmosphere and iron photosynthetic organisms erosion of rocks into soil troposphere thermosphere stratosphere mesosphere Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 7 of 9 5/21/2014 7:58 PM Chapter 3 Reading Quiz Question 15 Part A You are enjoying a spring day but expect a storm to arrive soon . As the storm arrives and the rain begins to fall, you notice that the temperature drops dramatically. Most likely, you have just experienced the arrival of a _____. ANSWER: Correct Chapter 3 Reading Quiz Question 25 Part A Every day tremendous amounts of the sun’s energy strikes Earth. Why doesn’t Earth overheat? ANSWER: Correct Earth’s energy budget is balanced. Over the course of a year, the energy input is equal to the energy output. Chapter 3 Reading Quiz Question 7 low and temperatures are low high and temperatures are high high and temperatures are low low and temperatures are high cold front Hadley cell intertropical convergence stratospheric event The energy is ultimately radiated back to space. Much of the heat melts rocks, forming lava deep inside of Earth. Most of the energy is used in photosynthesis to help plants grow and survive. The energy mostly is absorbed in various weather systems. Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 8 of 9 5/21/2014 7:58 PM Part A How many calories are required to heat up 1,000 grams of liquid water (about 1 liter) from 20 °C to 70 °C? ANSWER: Correct Chapter 3 Reading Quiz Question 14 Part A Hadley cells near the Equator consist of _____. ANSWER: Correct Score Summary: Your score on this assignment is 85.5%. You received 19.67 out of a possible total of 23 points. 100 1,000 5,000 50,000 rising dry air associated with deserts and falling moist air that produces precipitation and rainforests rising moist air that produces precipitation and rainforests, and falling dry air associated with deserts warm, moist air rising up the sides of mountains and cool, dry air descending on the leeward sides cool, dry air rising up the sides of mountains and warm, moist air descending on the leeward sides Chapter 03 Reading Questions http://session.masteringenvironmentalscience.com/myct/assignmentPrintV… 9 of 9 5/21/2014 7:58 PM

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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.

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

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## Assignment 12 Due: 11:59pm on Friday, May 9, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 15.6 A 0.600 -diameter vat of liquid is 2.30 deep. The pressure at the bottom of the vat is 1.30 . Part A What is the mass of the liquid in the vat? Express your answer with the appropriate units. ANSWER: Correct Problem 15.8 A 90-cm-thick layer of oil floats on a 160-cm-thick layer of water. Part A What is the pressure at the bottom of the water layer? Express your answer with the appropriate units. ANSWER: Correct m m atm 876 kg p = 1.25×105 Pa Problem 15.9 A research submarine has a 40.0 -diameter window 9.00 thick. The manufacturer says the window can withstand forces up to 1.20×106 . What is the submarine’s maximum safe depth? Part A The pressure inside the submarine is maintained at 1.0 atm. Express your answer with the appropriate units. ANSWER: Correct Problem 15.13 Part A What is the minimum hose diameter of an ideal vacuum cleaner that could lift a 12 dog off the floor? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 15.40 The 78 student in the figure balances a 1100 elephant on a hydraulic lift. cm cm N 947 m kg d = 3.8 cm kg kg You may want to review ( pages 415 – 419) . For help with math skills, you may want to review: Rearrangement of Equations Involving Multiplication and Division Part A What is the diameter of the piston the student is standing on? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Given that the height of the fluid on the two sides is the same in the figure, how is the pressure of the fluid on the two sides related? What is the definition of pressure? What is the area of the right cylinder? What is the force exerted by the elephant on the right cylinder? What is the additional pressure above atmospheric pressure in the fluid under the elephant? What is the additional pressure above atmospheric pressure under the student in the left cylinder? What is the force exerted by the student on the left cylinder? What is the area of the left cylinder? ANSWER: Correct Part B d = 0.53 m When a second student joins the first, the piston sinks 40 . What is the second student’s mass? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the definition of pressure? How is the height difference between the left and right cylinders related to the pressure difference in the two cylinders? What is the standard value for the density of the oil given in the text? What is the force due to the elephant on the right cylinder? What is the additional pressure above atmospheric pressure in the fluid under the elephant? Given the height difference between the two cylinders and the pressure in the right cylinder, what is the pressure above atmospheric pressure in the left cylinder? What is the force due to both students on the left cylinder? What is the sum of the masses of the students? What is the mass of the second student? ANSWER: Correct Enhanced EOC: Problem 15.17 A 6.80 rock whose density is 4900 is suspended by a string such that half of the rock’s volume is under water. You may want to review ( pages 419 – 423) . For help with math skills, you may want to review: Conversion Factors Part A What is the tension in the string? Express your answer with the appropriate units. Hint 1. How to approach the problem cm m = 80 kg kg kg/m3 What are the three forces acting on the rock? Draw a picture indicating the direction of the forces on the rock and an appropriate coordinate system indicating the positive direction. How is volume related to mass and density? What is the volume of the rock? What is the buoyant force on the rock given that half of the rock is underwater? What is the gravitational force on the rock? Given that the rock is suspended, what is the net force on the rock? Now, determine the tension in the string. ANSWER: Correct Problem 15.15 A block floats in water with its long axis vertical. The length of the block above water is 1.0 . Part A What is the block’s mass density? Express your answer with the appropriate units. ANSWER: Correct Crown of Gold? According to legend, the following challenge led Archimedes to the discovery of his famous principle: Hieron, king of Syracuse, was suspicious that a new crown that he had received from the royal goldsmith was not pure gold, as claimed. Archimedes was ordered to determine whether the crown was in fact made of pure gold, with the condition that only a nondestructive test would be allowed. Rather than answer the problem in the politically most expedient way (or perhaps extract a bribe from the goldsmith), Archimedes thought about the problem scientifically. The legend relates that when 59.8 N 2.0 cm × 2.0 cm × 7.0 cm cm 857 kg m3 Archimedes stepped into his bath and caused it to overflow, he realized that he could answer the challenge by comparing the volume of water displaced by the crown with the volume of water displaced by an amount of pure gold equal in weight to the crown. If the crown was made of pure gold, the two volumes would be equal. If some other (less dense) metal had been substituted for some of the gold, then the crown would displace more water than the pure gold. A similar method of answering the challenge, based on the same physical principle, is to compute the ratio of the actual weight of the crown, , and the apparent weight of the crown when it is submerged in water, . See whether you can follow in Archimedes’ footsteps. The figure shows what is meant by weighing the crown while it is submerged in water. Part A Take the density of the crown to be . What is the ratio of the crown’s apparent weight (in water) to its actual weight ? Express your answer in terms of the density of the crown and the density of water . Hint 1. Find an expression for the actual weight of the crown Assume that the crown has volume . Find the actual weight of the crown. Express in terms of , (the magnitude of the acceleration due to gravity), and . ANSWER: Wactual Wapparent c Wapparent Wactual c w V Wactual Wactual V g c Wactual = cV g Hint 2. Find an expression for the apparent weight of the crown Assume that the crown has volume , and take the density of water to be . Find the apparent weight of the crown submerged in water. Express your answer in terms of , (the magnitude of the acceleration due to gravity), , and . Hint 1. How to approach the problem The apparent weight of the crown when it is submerged in water will be less than its actual weight (weight in air) due to the buoyant force, which opposes gravity. Hint 2. Find an algebraic expression for the buoyant force. Find the magnitude of the buoyant force on the crown when it is completely submerged in water. Express your answer in terms of , , and the gravitational acceleration . ANSWER: ANSWER: ANSWER: Correct Part B Imagine that the apparent weight of the crown in water is , and the actual weight is . Is the crown made of pure (100%) gold? The density of water is V w Wapparent V g w c Fbuoyant w V g Fbuoyant = wV g Wapparent = (c − w)gV = Wapparent Wactual 1 − w c Wapparent = 4.50 N Wactual = 5.00 N grams per cubic centimeter. The density of gold is grams per cubic centimeter. Hint 1. Find the ratio of weights for a crown of pure gold Given the expression you obtained for , what should the ratio of weights be if the crown is made of pure gold? Express your answer numerically, to two decimal places. ANSWER: ANSWER: Correct For the values given, , whereas for pure gold, . Thus, you can conclude that the the crown is not pure gold but contains some less-dense metal. The goldsmith made sure that the crown’s (true) weight was the same as that of the amount of gold he was allocated, but in so doing was forced to make the volume of the crown slightly larger than it would otherwise have been. Problem 15.23 A 1.0-cm-diameter pipe widens to 2.0 cm, then narrows to 0.5 cm. Liquid flows through the first segment at a speed of 9.0 . Part A What is the speed in the second segment? Express your answer with the appropriate units. w = 1.00 g = 19.32 Wapparent Wactual = 0.95 Wapparent Wactual Yes No = 4.50/5.00 = 0.90 Wapparent Wactual = 1 − = 0.95 Wapparent Wactual w g m/s ANSWER: Correct Part B What is the speed in the third segment? Express your answer with the appropriate units. ANSWER: Correct Part C What is the volume flow rate through the pipe? Express your answer with the appropriate units. ANSWER: Correct Understanding Bernoulli’s Equation Bernoulli’s equation is a simple relation that can give useful insight into the balance among fluid pressure, flow speed, and elevation. It applies exclusively to ideal fluids with steady flow, that is, fluids with a constant density and no internal friction forces, whose flow patterns do not change with time. Despite its limitations, however, Bernoulli’s equation is an essential tool in understanding the behavior of fluids in many practical applications, from plumbing systems to the flight of airplanes. 2.25 ms 36.0 ms 7.07×10−4 m3 s For a fluid element of density that flows along a streamline, Bernoulli’s equation states that , where is the pressure, is the flow speed, is the height, is the acceleration due to gravity, and subscripts 1 and 2 refer to any two points along the streamline. The physical interpretation of Bernoulli’s equation becomes clearer if we rearrange the terms of the equation as follows: . The term on the left-hand side represents the total work done on a unit volume of fluid by the pressure forces of the surrounding fluid to move that volume of fluid from point 1 to point 2. The two terms on the right-hand side represent, respectively, the change in potential energy, , and the change in kinetic energy, , of the unit volume during its flow from point 1 to point 2. In other words, Bernoulli’s equation states that the work done on a unit volume of fluid by the surrounding fluid is equal to the sum of the change in potential and kinetic energy per unit volume that occurs during the flow. This is nothing more than the statement of conservation of mechanical energy for an ideal fluid flowing along a streamline. Part A Consider the portion of a flow tube shown in the figure. Point 1 and point 2 are at the same height. An ideal fluid enters the flow tube at point 1 and moves steadily toward point 2. If the cross section of the flow tube at point 1 is greater than that at point 2, what can you say about the pressure at point 2? Hint 1. How to approach the problem Apply Bernoulli’s equation to point 1 and to point 2. Since the points are both at the same height, their elevations cancel out in the equation and you are left with a relation between pressure and flow speeds. Even though the problem does not give direct information on the flow speed along the flow tube, it does tell you that the cross section of the flow tube decreases as the fluid flows toward point 2. Apply the continuity equation to points 1 and 2 and determine whether the flow speed at point 2 is greater than or smaller than the flow speed at point 1. With that information and Bernoulli’s equation, you will be able to determine the pressure at point 2 with respect to the pressure at point 1. Hint 2. Apply Bernoulli’s equation p1 +gh1 + = +g + 1 2 v21 p2 h2 1 2 v22 p v h g p1 − p2 = g(h2 −h1)+ ( − ) 1 2 v22 v21 p1 − p2 g(h2 − h1) 1 ( − ) 2 v22 v21 Apply Bernoulli’s equation to point 1 and to point 2 to complete the expression below. Here and are the pressure and flow speed, respectively, and subscripts 1 and 2 refer to point 1 and point 2. Also, use for elevation with the appropriate subscript, and use for the density of the fluid. Express your answer in terms of some or all of the variables , , , , , , and . Hint 1. Flow along a horizontal streamline Along a horizontal streamline, the change in potential energy of the flowing fluid is zero. In other words, when applying Bernoulli’s equation to any two points of the streamline, and they cancel out. ANSWER: Hint 3. Determine with respect to By applying the continuity equation, determine which of the following is true. Hint 1. The continuity equation The continuity equation expresses conservation of mass for incompressible fluids flowing in a tube. It says that the amount of fluid flowing through a cross section of the tube in a time interval must be the same for all cross sections, or . Therefore, the flow speed must increase when the cross section of the flow tube decreases, and vice versa. ANSWER: p v h p1 v1 h1 p2 v2 h2 h1 = h2 p1 + = 1 2 v21 p2 + v2 2 2 v2 v1 $V A $t $V = = $t A1v1 A2v2 v2 > v1 v2 = v1 v2 < v1 ANSWER: Correct Thus, by combining the continuity equation and Bernoulli's equation, one can characterize the flow of an ideal fluid.When the cross section of the flow tube decreases, the flow speed increases, and therefore the pressure decreases. In other words, if , then and . Part B As you found out in the previous part, Bernoulli's equation tells us that a fluid element that flows through a flow tube with decreasing cross section moves toward a region of lower pressure. Physically, the pressure drop experienced by the fluid element between points 1 and 2 acts on the fluid element as a net force that causes the fluid to __________. Hint 1. Effects from conservation of mass Recall that, if the cross section of the flow tube varies, the flow speed must change to conserve mass. This means that there is a nonzero net force acting on the fluid that causes the fluid to increase or decrease speed depending on whether the fluid is flowing through a portion of the tube with a smaller or larger cross section. ANSWER: Correct Part C Now assume that point 2 is at height with respect to point 1, as shown in the figure. The ends of the flow tube have the same areas as The pressure at point 2 is lower than the pressure at point 1. equal to the pressure at point 1. higher than the pressure at point 1. A2 < A1 v2 > v1 p2 < p1 A v decrease in speed increase in speed remain in equilibrium h the ends of the horizontal flow tube shown in Part A. Since the cross section of the flow tube is decreasing, Bernoulli's equation tells us that a fluid element flowing toward point 2 from point 1 moves toward a region of lower pressure. In this case, what is the pressure drop experienced by the fluid element? Hint 1. How to approach the problem Apply Bernoulli's equation to point 1 and to point 2, as you did in Part A. Note that this time you must take into account the difference in elevation between points 1 and 2. Do you need to add this additional term to the other term representing the pressure drop between the two ends of the flow tube or do you subtract it? ANSWER: Correct Part D From a physical point of view, how do you explain the fact that the pressure drop at the ends of the elevated flow tube from Part C is larger than the pressure drop occurring in the similar but purely horizontal flow from Part A? Hint 1. Physical meaning of the pressure drop in a tube As explained in the introduction, the difference in pressure between the ends of a flow tube represents the total work done on a unit volume of fluid by the pressure forces of the The pressure drop is smaller than the pressure drop occurring in a purely horizontal flow. equal to the pressure drop occurring in a purely horizontal flow. larger than the pressure drop occurring in a purely horizontal flow. p1 − p2 surrounding fluid to move that volume of fluid from one end to the other end of the flow tube. ANSWER: Correct In the case of purely horizontal flow, the difference in pressure between the two ends of the flow tube had to balance only the increase in kinetic energy resulting from the acceleration of the fluid. In an elevated flow tube, the difference in pressure must also balance the increase in potential energy of the fluid; therefore a higher pressure is needed for the flow to occur. Venturi Meter with Two Tubes A pair of vertical, open-ended glass tubes inserted into a horizontal pipe are often used together to measure flow velocity in the pipe, a configuration called a Venturi meter. Consider such an arrangement with a horizontal pipe carrying fluid of density . The fluid rises to heights and in the two open-ended tubes (see figure). The cross-sectional area of the pipe is at the position of tube 1, and at the position of tube 2. A greater amount of work is needed to balance the increase in potential energy from the elevation change. decrease in potential energy from the elevation change. larger increase in kinetic energy. larger decrease in kinetic energy. h1 h2 A1 A2 Part A Find , the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.) Express your answer in terms of quantities given in the problem introduction and , the magnitude of the acceleration due to gravity. Hint 1. How to approach the problem Use Bernoulli's law to compute the difference in pressure between the top and bottom of tube 1. The pressure at the top of the tube is defined to be atmospheric pressure. Note: Inside the tube, since the fluid is not flowing, the terms involving velocity in Bernoulli's equation can be ignored. Thus, Bernoulli's equation reduces to the formula for pressure as a function of depth in a fluid of uniform density. Hint 2. Simplified Bernoulli's equation For a fluid of uniform density that is not flowing, the pressure at a depth below the surface is given by , where is the pressure at the surface and is the magnitude of the acceleration due to gravity. ANSWER: Correct The fluid is pushed up tube 1 by the pressure of the fluid at the base of the tube, and not by its kinetic energy, since there is no streamline around the sharp edge of the tube. Thus energy is not conserved (there is turbulence at the edge of the tube) at the entrance of the tube. Since Bernoulli's law is essentially a statement of energy conservation, it must be applied separately to the fluid in the tube and the fluid flowing in the main pipe. However, the pressure in the fluid is the same just inside and just outside the tube. Part B Find , the speed of the fluid in the left end of the main pipe. Express your answer in terms of , , , and either and or , which is equal to . Hint 1. How to approach the problem Energy is conserved along the streamlines in the main flow. This means that Bernoulli's law can be applied to obtain a relationship between the fluid pressure and velocity at the bottom of p1 g p h p = p0 + gh p0 g p1 = gh1 v1 h1 h2 g A1 A2 A1 A2 tube 1, and the fluid pressure and velocity at the bottom of tube 2. Hint 2. Find in terms of What is , the pressure at the bottom of tube 2? Express your answer in terms of , , and any other given quantities. Hint 1. Recall Part A Obtain the solution for this part in the same way that you found an expression for in terms of in Part A of this problem. ANSWER: Hint 3. Find in terms of given quantities Find , the fluid pressure at the bottom of tube 2. Express your answer in terms of , , , , and . Hint 1. Find the pressure at the bottom of tube 2 Find , the fluid pressure at the bottom of tube 2. Express your answer in terms of , , and . ANSWER: Hint 2. Find in terms of The fluid is incompressible, so you can use the continuity equation to relate the fluid velocities and in terms of the geometry of the pipe. Find , the fluid velocity at the bottom p2 h2 p2 h2 g p1 h1 p2 = gh2 p2 p2 p1 v1 A1 A2 p2 p1 v1 v2 p2 = p1 + ( − ) 1 2 v1 2 v2 2 v2 v1 v1 v2 v2 of tube 2, in terms of . Your answer may include and , the cross-sectional areas of the pipe. ANSWER: ANSWER: ANSWER: Correct Note that this result depends on the difference between the heights of the fluid in the tubes, a quantity that is more easily measured than the heights themselves. Problem 15.39 The container shown in the figure is filled with oil. It is open to the atmosphere on the left. v1 A1 A2 v2 = A1 A2 v1 p2 = p1 + ( )(1 − ) 1 2 v1 2 ( ) A1 A2 2 v1 = 2g h1−h2 ( ) −1 A1 A2 2 −−−−−−−−−−−−−− Part A What is the pressure at point A? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part B What is the pressure difference between points A and B? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the pressure difference between points A and C? PA = 106 kPa PB − PA = 4.4 kPa Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 15.48 You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 32.4 . If you then lower the statue into a tub of water, so that it is completely submerged, the scale reads 17 . Part A What is the density? Express your answer with the appropriate units. ANSWER: Correct Problem 15.60 Water flows from the pipe shown in the figure with a speed of 7.0 . PC − PA = 4.4 kPa N N statue = 2100 kg m3 m/s Part A What is the water pressure as it exits into the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the height of the standing column of water? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.9%. You received 93.92 out of a possible total of 94 points. P = 1.0×105 Pa h h = 5.9 m

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## Chapter 13 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 A Matter of Some Gravity Learning Goal: To understand Newton’s law of gravitation and the distinction between inertial and gravitational masses. In this problem, you will practice using Newton’s law of gravitation. According to that law, the magnitude of the gravitational force between two small particles of masses and , separated by a distance , is given by , where is the universal gravitational constant, whose numerical value (in SI units) is . This formula applies not only to small particles, but also to spherical objects. In fact, the gravitational force between two uniform spheres is the same as if we concentrated all the mass of each sphere at its center. Thus, by modeling the Earth and the Moon as uniform spheres, you can use the particle approximation when calculating the force of gravity between them. Be careful in using Newton’s law to choose the correct value for . To calculate the force of gravitational attraction between two uniform spheres, the distance in the equation for Newton’s law of gravitation is the distance between the centers of the spheres. For instance, if a small object such as an elephant is located on the surface of the Earth, the radius of the Earth would be used in the equation. Note that the force of gravity acting on an object located near the surface of a planet is often called weight. Also note that in situations involving satellites, you are often given the altitude of the satellite, that is, the distance from the satellite to the surface of the planet; this is not the distance to be used in the formula for the law of gravitation. There is a potentially confusing issue involving mass. Mass is defined as a measure of an object’s inertia, that is, its ability to resist acceleration. Newton’s second law demonstrates the relationship between mass, acceleration, and the net force acting on an object: . We can now refer to this measure of inertia more precisely as the inertial mass. On the other hand, the masses of the particles that appear in the expression for the law of gravity seem to have nothing to do with inertia: Rather, they serve as a measure of the strength of gravitational interactions. It would be reasonable to call such a property gravitational mass. Does this mean that every object has two different masses? Generally speaking, yes. However, the good news is that according to the latest, highly precise, measurements, the inertial and the gravitational mass of an object are, in fact, equal to each other; it is an established consensus among physicists that there is only one mass after all, which is a measure of both the object’s inertia and its ability to engage in gravitational interactions. Note that this consensus, like everything else in science, is open to possible amendments in the future. In this problem, you will answer several questions that require the use of Newton’s law of gravitation. Part A Two particles are separated by a certain distance. The force of gravitational interaction between them is . Now the separation between the particles is tripled. Find the new force of gravitational Fg m1 m2 r Fg = G m1m2 r2 G 6.67 × 10−11 Nm2 kg2 r r rEarth F = m net a F0 interaction . Express your answer in terms of . ANSWER: Part B A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is . Then the satellite moves to a different orbit, so that its altitude is tripled. Find the new force of gravitational interaction . Express your answer in terms of . You did not open hints for this part. ANSWER: Part C A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is . Then the satellite is brought back to the surface of the planet. Find the new force of gravitational interaction . Express your answer in terms of . ANSWER: F1 F0 F1 = F0 F2 F0 F2 = F0 F4 F0 Typesetting math: 81% Part D Two satellites revolve around the Earth. Satellite A has mass and has an orbit of radius . Satellite B has mass and an orbit of unknown radius . The forces of gravitational attraction between each satellite and the Earth is the same. Find . Express your answer in terms of . ANSWER: Part E An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far from the center of the Earth should an elephant be placed so that its weight equals that of the elephant shrew on the surface of the Earth? The radius of the Earth is 6400 . ( .) Express your answer in kilometers. ANSWER: The table below gives the masses of the Earth, the Moon, and the Sun. Name Mass (kg) Earth Moon Sun F4 = m r 6m rb rb r rb = r km 1 ton = 103 kg r = km 5.97 × 1024 7.35 × 1022 1.99 × 1030 Typesetting math: 81% The average distance between the Earth and the Moon is . The average distance between the Earth and the Sun is . Use this information to answer the following questions. Part F Find the net gravitational force acting on the Earth in the Sun-Earth-Moon system during the new moon (when the moon is located directly between the Earth and the Sun). Express your answer in newtons to three significant figures. You did not open hints for this part. ANSWER: Part G Find the net gravitational force acting on the Earth in the Sun-Earth-Moon system during the full moon (when the Earth is located directly between the moon and the sun). Express your answer in newtons to three significant figures. ANSWER: ± Understanding Newton’s Law of Universal Gravitation Learning Goal: To understand Newton’s law of universal gravitation and be able to apply it in two-object situations and (collinear) three-object situations; to distinguish between the use of and . 3.84 × 108 m 1.50 × 1011 m Fnet Fnet = N Fnet Fnet = N Typesetting math: 81% G g In the late 1600s, Isaac Newton proposed a rule to quantify the attractive force known as gravity between objects that have mass, such as those shown in the figure. Newton’s law of universal gravitation describes the magnitude of the attractive gravitational force between two objects with masses and as , where is the distance between the centers of the two objects and is the gravitational constant. The gravitational force is attractive, so in the figure it pulls to the right on (toward ) and toward the left on (toward ). The gravitational force acting on is equal in size to, but exactly opposite in direction from, the gravitational force acting on , as required by Newton’s third law. The magnitude of both forces is calculated with the equation given above. The gravitational constant has the value and should not be confused with the magnitude of the gravitational free-fall acceleration constant, denoted by , which equals 9.80 near the surface of the earth. The size of in SI units is tiny. This means that gravitational forces are sizeable only in the vicinity of very massive objects, such as the earth. You are in fact gravitationally attracted toward all the objects around you, such as the computer you are using, but the size of that force is too small to be noticed without extremely sensitive equipment. Consider the earth following its nearly circular orbit (dashed curve) about the sun. The earth has mass and the sun has mass . They are separated, center to center, by . Part A What is the size of the gravitational force acting on the earth due to the sun? Express your answer in newtons. F g m1 m2 Fg = G( ) m1m2 r2 r G m1 m2 m2 m1 m1 m2 G G = 6.67 × 10−11 N m2/kg2 g m/s2 G mearth = 5.98 × 1024 kg msun = 1.99 × 1030 kg r = 93 million miles = 150 million km Typesetting math: 81% 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). Part E This question will be shown after you complete previous question(s). Part F N Typesetting math: 81% This question will be shown after you complete previous question(s). Understanding Mass and Weight Learning Goal: To understand the distinction between mass and weight and to be able to calculate the weight of an object from its mass and Newton’s law of gravitation. The concepts of mass and weight are often confused. In fact, in everyday conversations, the word “weight” often replaces “mass,” as in “My weight is seventy-five kilograms” or “I need to lose some weight.” Of course, mass and weight are related; however, they are also very different. Mass, as you recall, is a measure of an object’s inertia (ability to resist acceleration). Newton’s 2nd law demonstrates the relationship among an object’s mass, its acceleration, and the net force acting on it: . Mass is an intrinsic property of an object and is independent of the object’s location. Weight, in contrast, is defined as the force due to gravity acting on the object. That force depends on the strength of the gravitational field of the planet: , where is the weight of an object, is the mass of that object, and is the local acceleration due to gravity (in other words, the strength of the gravitational field at the location of the object). Weight, unlike mass, is not an intrinsic property of the object; it is determined by both the object and its location. Part A Which of the following quantities represent mass? Check all that apply. ANSWER: Fnet = ma w = mg w m g 12.0 lbs 0.34 g 120 kg 1600 kN 0.34 m 411 cm 899 MN Typesetting math: 81% Part B This question will be shown after you complete previous question(s). Using the universal law of gravity, we can find the weight of an object feeling the gravitational pull of a nearby planet. We can write an expression , where is the weight of the object, is the gravitational constant, is the mass of that object, is mass of the planet, and is the distance from the center of the planet to the object. If the object is on the surface of the planet, is simply the radius of the planet. Part C The gravitational field on the surface of the earth is stronger than that on the surface of the moon. If a rock is transported from the moon to the earth, which properties of the rock change? ANSWER: Part D This question will be shown after you complete previous question(s). Part E If acceleration due to gravity on the earth is , which formula gives the acceleration due to gravity on Loput? You did not open hints for this part. ANSWER: w = GmM/r2 w G m M r r mass only weight only both mass and weight neither mass nor weight g Typesetting math: 81% Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). ± Weight on a Neutron Star Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but a much smaller diameter. g 1.7 5.6 g 1.72 5.6 g 1.72 5.62 g 5.6 1.7 g 5.62 1.72 g 5.6 1.72 Typesetting math: 81% Part A If you weigh 655 on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 19.0 ? Take the mass of the sun to be = 1.99×1030 , the gravitational constant to be = 6.67×10−11 , and the acceleration due to gravity at the earth’s surface to be = 9.810 . Express your weight in newtons. You did not open hints for this part. ANSWER: ± Escape Velocity Learning Goal: To introduce you to the concept of escape velocity for a rocket. The escape velocity is defined to be the minimum speed with which an object of mass must move to escape from the gravitational attraction of a much larger body, such as a planet of total mass . The escape velocity is a function of the distance of the object from the center of the planet , but unless otherwise specified this distance is taken to be the radius of the planet because it addresses the question “How fast does my rocket have to go to escape from the surface of the planet?” Part A The key to making a concise mathematical definition of escape velocity is to consider the energy. If an object is launched at its escape velocity, what is the total mechanical energy of the object at a very large (i.e., infinite) distance from the planet? Follow the usual convention and take the gravitational potential energy to be zero at very large distances. You did not open hints for this part. ANSWER: N km ms kg G N m2/kg2 g m/s2 wstar wstar = N m M R Etotal Typesetting math: 81% Consider the motion of an object between a point close to the planet and a point very very far from the planet. Indicate whether the following statements are true or false. Part B Angular momentum about the center of the planet is conserved. ANSWER: Part C Total mechanical energy is conserved. ANSWER: Part D Kinetic energy is conserved. ANSWER: Etotal = true false true false Typesetting math: 81% 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 Satellite in a Circular Orbit Consider a satellite of mass that orbits a planet of mass in a circle a distance from the center of the planet. The satellite’s mass is negligible compared with that of the planet. Indicate whether each of the statements in this problem is true or false. Part A The information given is sufficient to uniquely specify the speed, potential energy, and angular momentum of the satellite. You did not open hints for this part. ANSWER: true false m1 m2 r true false Typesetting math: 81% Part B The total mechanical energy of the satellite is conserved. You did not open hints for this part. ANSWER: Part C The linear momentum vector of the satellite is conserved. You did not open hints for this part. ANSWER: Part D The angular momentum of the satellite about the center of the planet is conserved. You did not open hints for this part. ANSWER: true false true false Typesetting math: 81% Part E The equations that express the conservation laws of total mechanical energy and linear momentum are sufficient to solve for the speed necessary to maintain a circular orbit at without using . You did not open hints for this part. ANSWER: At the Galaxy’s Core Astronomers have observed a small, massive object at the center of our Milky Way galaxy. A ring of material orbits this massive object; the ring has a diameter of about 15 light years and an orbital speed of about 200 . Part A Determine the mass of the massive object at the center of the Milky Way galaxy. Take the distance of one light year to be . Express your answer in kilograms. You did not open hints for this part. true false R F = ma true false km/s M 9.461 × 1015 m Typesetting math: 81% 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). Part E This question will be shown after you complete previous question(s). Properties of Circular Orbits Learning Goal: To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth. M = kg Typesetting math: 81% The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were observed by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the speed of the space shuttle and the height of geostationary satellites) is the simplest orbit–a circular one. This problem concerns the properties of circular orbits for a satellite orbiting a planet of mass . For all parts of this problem, where appropriate, use for the universal gravitational constant. Part A Find the orbital speed for a satellite in a circular orbit of radius . Express the orbital speed in terms of , , and . You did not open hints for this part. ANSWER: Part B Find the kinetic energy of a satellite with mass in a circular orbit with radius . Express your answer in terms of \texttip{m}{m}, \texttip{M}{M}, \texttip{G}{G}, and \texttip{R}{R}. ANSWER: Part C M G v R G M R v = K m R \texttip{K}{K} = Typesetting math: 81% This question will be shown after you complete previous question(s). Part D Find the orbital period \texttip{T}{T}. Express your answer in terms of \texttip{G}{G}, \texttip{M}{M}, \texttip{R}{R}, and \texttip{\pi }{pi}. You did not open hints for this part. ANSWER: Part E This question will be shown after you complete previous question(s). Part F Find \texttip{L}{L}, the magnitude of the angular momentum of the satellite with respect to the center of the planet. Express your answer in terms of \texttip{m}{m}, \texttip{M}{M}, \texttip{G}{G}, and \texttip{R}{R}. You did not open hints for this part. ANSWER: \texttip{T}{T} = Typesetting math: 81% Part G The quantities \texttip{v}{v}, \texttip{K}{K}, \texttip{U}{U}, and \texttip{L}{L} all represent physical quantities characterizing the orbit that depend on radius \texttip{R}{R}. Indicate the exponent (power) of the radial dependence of the absolute value of each. Express your answer as a comma-separated list of exponents corresponding to \texttip{v}{v}, \texttip{K}{K}, \texttip{U}{U}, and \texttip{L}{L}, in that order. For example, -1,-1/2,-0.5,-3/2 would mean v \propto R^{-1}, K \propto R^{-1/2}, and so forth. You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. \texttip{L}{L} = Typesetting math: 81%

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## Assignment 12 Due: 11:59pm on Friday, May 9, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 15.6 A 2.00 -diameter vat of liquid is 2.90 deep. The pressure at the bottom of the vat is 1.20 . Part A What is the mass of the liquid in the vat? Express your answer with the appropriate units. ANSWER: Correct Problem 15.8 A 120-cm-thick layer of oil floats on a 130-cm-thick layer of water. Part A What is the pressure at the bottom of the water layer? Express your answer with the appropriate units. ANSWER: Correct m m atm 6490 kg p = 1.25×105 Pa Problem 15.9 A research submarine has a 40.0 -diameter window 8.00 thick. The manufacturer says the window can withstand forces up to 1.20×106 . What is the submarine’s maximum safe depth? Part A The pressure inside the submarine is maintained at 1.0 atm. Express your answer with the appropriate units. ANSWER: Correct Problem 15.13 Part A What is the minimum hose diameter of an ideal vacuum cleaner that could lift a 13 dog off the floor? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 15.40 The 80 student in the figure balances a 1300 elephant on a hydraulic lift. cm cm N 947 m kg d = 4.0 cm kg kg Typesetting math: 100% You may want to review ( pages 415 – 419) . For help with math skills, you may want to review: Rearrangement of Equations Involving Multiplication and Division Part A What is the diameter of the piston the student is standing on? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Given that the height of the fluid on the two sides is the same in the figure, how is the pressure of the fluid on the two sides related? What is the definition of pressure? What is the area of the right cylinder? What is the force exerted by the elephant on the right cylinder? What is the additional pressure above atmospheric pressure in the fluid under the elephant? What is the additional pressure above atmospheric pressure under the student in the left cylinder? What is the force exerted by the student on the left cylinder? What is the area of the left cylinder? ANSWER: Correct Part B d = 0.50 m Typesetting math: 100% When a second student joins the first, the piston sinks 30 . What is the second student’s mass? Express your answer to two significant figures and include the appropriate units. You did not open hints for this part. ANSWER: Enhanced EOC: Problem 15.17 A 4.70 rock whose density is 4300 is suspended by a string such that half of the rock’s volume is under water. You may want to review ( pages 419 – 423) . For help with math skills, you may want to review: Conversion Factors Part A What is the tension in the string? Express your answer with the appropriate units. Hint 1. How to approach the problem What are the three forces acting on the rock? Draw a picture indicating the direction of the forces on the rock and an appropriate coordinate system indicating the positive direction. How is volume related to mass and density? What is the volume of the rock? What is the buoyant force on the rock given that half of the rock is underwater? What is the gravitational force on the rock? Given that the rock is suspended, what is the net force on the rock? Now, determine the tension in the string. cm m = kg kg/m3 Typesetting math: 100% ANSWER: Correct Problem 15.15 A block floats in water with its long axis vertical. The length of the block above water is 1.0 . Part A What is the block’s mass density? Express your answer with the appropriate units. ANSWER: Correct Crown of Gold? According to legend, the following challenge led Archimedes to the discovery of his famous principle: Hieron, king of Syracuse, was suspicious that a new crown that he had received from the royal goldsmith was not pure gold, as claimed. Archimedes was ordered to determine whether the crown was in fact made of pure gold, with the condition that only a nondestructive test would be allowed. Rather than answer the problem in the politically most expedient way (or perhaps extract a bribe from the goldsmith), Archimedes thought about the problem scientifically. The legend relates that when Archimedes stepped into his bath and caused it to overflow, he realized that he could answer the challenge by comparing the volume of water displaced by the crown with the volume of water displaced by an amount of pure gold equal in weight to the crown. If the crown was made of pure gold, the two volumes would be equal. If some other (less dense) metal had been substituted for some of the gold, then the crown would displace more water than the pure gold. A similar method of answering the challenge, based on the same physical principle, is to compute the ratio of the actual weight of the crown, , and the apparent weight of the crown when it is submerged in water, . See whether you can follow in Archimedes’ footsteps. The figure shows what is meant by weighing the crown while it is submerged in water. 40.7 N 2.0 cm × 2.0 cm × 8.0 cm cm 875 kg m3 Wactual Wapparent Typesetting math: 100% Part A Take the density of the crown to be . What is the ratio of the crown’s apparent weight (in water) to its actual weight ? Express your answer in terms of the density of the crown and the density of water . Hint 1. Find an expression for the actual weight of the crown Assume that the crown has volume . Find the actual weight of the crown. Express in terms of , (the magnitude of the acceleration due to gravity), and . ANSWER: Hint 2. Find an expression for the apparent weight of the crown Assume that the crown has volume , and take the density of water to be . Find the apparent weight of the crown submerged in water. Express your answer in terms of , (the magnitude of the acceleration due to gravity), , and . Hint 1. How to approach the problem c Wapparent Wactual c w V Wactual Wactual V g c Wactual = cV g V w Wapparent V g w c Typesetting math: 100% The apparent weight of the crown when it is submerged in water will be less than its actual weight (weight in air) due to the buoyant force, which opposes gravity. Hint 2. Find an algebraic expression for the buoyant force. Find the magnitude of the buoyant force on the crown when it is completely submerged in water. Express your answer in terms of , , and the gravitational acceleration . ANSWER: ANSWER: ANSWER: Correct Part B Imagine that the apparent weight of the crown in water is , and the actual weight is . Is the crown made of pure (100%) gold? The density of water is grams per cubic centimeter. The density of gold is grams per cubic centimeter. Hint 1. Find the ratio of weights for a crown of pure gold Given the expression you obtained for , what should the ratio of weights be if the crown is made of pure gold? Express your answer numerically, to two decimal places. Fbuoyant w V g Fbuoyant = wV g Wapparent = (c − w)gV = Wapparent Wactual 1 − w c Wapparent = 4.50 N Wactual = 5.00 N w = 1.00 g = 19.32 Wapparent Wactual Typesetting math: 100% ANSWER: ANSWER: Correct For the values given, , whereas for pure gold, . Thus, you can conclude that the the crown is not pure gold but contains some less-dense metal. The goldsmith made sure that the crown’s (true) weight was the same as that of the amount of gold he was allocated, but in so doing was forced to make the volume of the crown slightly larger than it would otherwise have been. Problem 15.23 A 1.0-cm-diameter pipe widens to 2.0 cm, then narrows to 0.5 cm. Liquid flows through the first segment at a speed of 2.0 . Part A What is the speed in the second segment? Express your answer with the appropriate units. ANSWER: Correct = 0.95 Wapparent Wactual Yes No = 4.50/5.00 = 0.90 Wapparent Wactual = 1 − = 0.95 Wapparent Wactual w g m/s 0.500 ms Typesetting math: 100% Part B What is the speed in the third segment? Express your answer with the appropriate units. ANSWER: Correct Part C What is the volume flow rate through the pipe? Express your answer with the appropriate units. ANSWER: Correct Understanding Bernoulli’s Equation Bernoulli’s equation is a simple relation that can give useful insight into the balance among fluid pressure, flow speed, and elevation. It applies exclusively to ideal fluids with steady flow, that is, fluids with a constant density and no internal friction forces, whose flow patterns do not change with time. Despite its limitations, however, Bernoulli’s equation is an essential tool in understanding the behavior of fluids in many practical applications, from plumbing systems to the flight of airplanes. For a fluid element of density that flows along a streamline, Bernoulli’s equation states that , where is the pressure, is the flow speed, is the height, is the acceleration due to gravity, and subscripts 1 and 2 refer to any two points along the streamline. The physical interpretation of Bernoulli’s equation becomes clearer if we rearrange the terms of the equation as follows: . 8.00 ms 1.57×10−4 m3 s p1 +gh1 + = +g + 1 2 v21 p2 h2 1 2 v22 p v h g p1 − p2 = g(h2 −h1)+ ( − ) 1 2 v22 v21 Typesetting math: 100% The term on the left-hand side represents the total work done on a unit volume of fluid by the pressure forces of the surrounding fluid to move that volume of fluid from point 1 to point 2. The two terms on the right-hand side represent, respectively, the change in potential energy, , and the change in kinetic energy, , of the unit volume during its flow from point 1 to point 2. In other words, Bernoulli’s equation states that the work done on a unit volume of fluid by the surrounding fluid is equal to the sum of the change in potential and kinetic energy per unit volume that occurs during the flow. This is nothing more than the statement of conservation of mechanical energy for an ideal fluid flowing along a streamline. Part A Consider the portion of a flow tube shown in the figure. Point 1 and point 2 are at the same height. An ideal fluid enters the flow tube at point 1 and moves steadily toward point 2. If the cross section of the flow tube at point 1 is greater than that at point 2, what can you say about the pressure at point 2? Hint 1. How to approach the problem Apply Bernoulli’s equation to point 1 and to point 2. Since the points are both at the same height, their elevations cancel out in the equation and you are left with a relation between pressure and flow speeds. Even though the problem does not give direct information on the flow speed along the flow tube, it does tell you that the cross section of the flow tube decreases as the fluid flows toward point 2. Apply the continuity equation to points 1 and 2 and determine whether the flow speed at point 2 is greater than or smaller than the flow speed at point 1. With that information and Bernoulli’s equation, you will be able to determine the pressure at point 2 with respect to the pressure at point 1. Hint 2. Apply Bernoulli’s equation Apply Bernoulli’s equation to point 1 and to point 2 to complete the expression below. Here and are the pressure and flow speed, respectively, and subscripts 1 and 2 refer to point 1 and point 2. Also, use for elevation with the appropriate subscript, and use for the density of the fluid. Express your answer in terms of some or all of the variables , , , , , , and . Hint 1. Flow along a horizontal streamline p1 − p2 g(h2 − h1) 1 ( − ) 2 v22 v21 p v h p1 v1 h1 p2 v2 h2 Typesetting math: 100% Along a horizontal streamline, the change in potential energy of the flowing fluid is zero. In other words, when applying Bernoulli’s equation to any two points of the streamline, and they cancel out. ANSWER: Hint 3. Determine with respect to By applying the continuity equation, determine which of the following is true. Hint 1. The continuity equation The continuity equation expresses conservation of mass for incompressible fluids flowing in a tube. It says that the amount of fluid flowing through a cross section of the tube in a time interval must be the same for all cross sections, or . Therefore, the flow speed must increase when the cross section of the flow tube decreases, and vice versa. ANSWER: ANSWER: h1 = h2 p1 + = 1 2 v21 p2 + v2 2 2 v2 v1 $V A $t $V = = $t A1v1 A2v2 v2 > v1 v2 = v1 v2 < v1 Typesetting math: 100% Correct Thus, by combining the continuity equation and Bernoulli's equation, one can characterize the flow of an ideal fluid.When the cross section of the flow tube decreases, the flow speed increases, and therefore the pressure decreases. In other words, if , then and . Part B As you found out in the previous part, Bernoulli's equation tells us that a fluid element that flows through a flow tube with decreasing cross section moves toward a region of lower pressure. Physically, the pressure drop experienced by the fluid element between points 1 and 2 acts on the fluid element as a net force that causes the fluid to __________. Hint 1. Effects from conservation of mass Recall that, if the cross section of the flow tube varies, the flow speed must change to conserve mass. This means that there is a nonzero net force acting on the fluid that causes the fluid to increase or decrease speed depending on whether the fluid is flowing through a portion of the tube with a smaller or larger cross section. ANSWER: Correct Part C Now assume that point 2 is at height with respect to point 1, as shown in the figure. The ends of the flow tube have the same areas as the ends of the horizontal flow tube shown in Part A. Since the cross section of the flow tube is decreasing, Bernoulli's equation tells us that a fluid element flowing toward point 2 from point 1 moves toward a region of lower pressure. In this case, what is the pressure drop The pressure at point 2 is lower than the pressure at point 1. equal to the pressure at point 1. higher than the pressure at point 1. A2 < A1 v2 > v1 p2 < p1 A v decrease in speed increase in speed remain in equilibrium h Typesetting math: 100% experienced by the fluid element? Hint 1. How to approach the problem Apply Bernoulli's equation to point 1 and to point 2, as you did in Part A. Note that this time you must take into account the difference in elevation between points 1 and 2. Do you need to add this additional term to the other term representing the pressure drop between the two ends of the flow tube or do you subtract it? ANSWER: Correct Part D From a physical point of view, how do you explain the fact that the pressure drop at the ends of the elevated flow tube from Part C is larger than the pressure drop occurring in the similar but purely horizontal flow from Part A? The pressure drop is smaller than the pressure drop occurring in a purely horizontal flow. equal to the pressure drop occurring in a purely horizontal flow. larger than the pressure drop occurring in a purely horizontal flow. Typesetting math: 100% Hint 1. Physical meaning of the pressure drop in a tube As explained in the introduction, the difference in pressure between the ends of a flow tube represents the total work done on a unit volume of fluid by the pressure forces of the surrounding fluid to move that volume of fluid from one end to the other end of the flow tube. ANSWER: Correct In the case of purely horizontal flow, the difference in pressure between the two ends of the flow tube had to balance only the increase in kinetic energy resulting from the acceleration of the fluid. In an elevated flow tube, the difference in pressure must also balance the increase in potential energy of the fluid; therefore a higher pressure is needed for the flow to occur. Venturi Meter with Two Tubes A pair of vertical, open-ended glass tubes inserted into a horizontal pipe are often used together to measure flow velocity in the pipe, a configuration called a Venturi meter. Consider such an arrangement with a horizontal pipe carrying fluid of density . The fluid rises to heights and in the two open-ended tubes (see figure). The cross-sectional area of the pipe is at the position of tube 1, and at the position of tube 2. p1 − p2 A greater amount of work is needed to balance the increase in potential energy from the elevation change. decrease in potential energy from the elevation change. larger increase in kinetic energy. larger decrease in kinetic energy. h1 h2 A1 A2 Typesetting math: 100% Part A Find , the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.) Express your answer in terms of quantities given in the problem introduction and , the magnitude of the acceleration due to gravity. Hint 1. How to approach the problem Use Bernoulli's law to compute the difference in pressure between the top and bottom of tube 1. The pressure at the top of the tube is defined to be atmospheric pressure. Note: Inside the tube, since the fluid is not flowing, the terms involving velocity in Bernoulli's equation can be ignored. Thus, Bernoulli's equation reduces to the formula for pressure as a function of depth in a fluid of uniform density. Hint 2. Simplified Bernoulli's equation For a fluid of uniform density that is not flowing, the pressure at a depth below the surface is given by , where is the pressure at the surface and is the magnitude of the acceleration due to gravity. ANSWER: Correct The fluid is pushed up tube 1 by the pressure of the fluid at the base of the tube, and not by its kinetic energy, since there is no streamline around the sharp edge of the tube. Thus energy is not conserved (there is turbulence at the edge of the tube) at the entrance of the tube. Since Bernoulli's law is essentially a statement of energy conservation, it must be applied separately to the fluid in the tube and the fluid flowing in the main pipe. However, the pressure in the fluid is the same just inside and just outside the tube. Part B Find , the speed of the fluid in the left end of the main pipe. Express your answer in terms of , , , and either and or , which is equal to . p1 g p h p = p0 + gh p0 g p1 = gh1 v1 h1 h2 g A1 A2 A1 A2 Typesetting math: 100% Hint 1. How to approach the problem Energy is conserved along the streamlines in the main flow. This means that Bernoulli's law can be applied to obtain a relationship between the fluid pressure and velocity at the bottom of tube 1, and the fluid pressure and velocity at the bottom of tube 2. Hint 2. Find in terms of What is , the pressure at the bottom of tube 2? Express your answer in terms of , , and any other given quantities. Hint 1. Recall Part A Obtain the solution for this part in the same way that you found an expression for in terms of in Part A of this problem. ANSWER: Hint 3. Find in terms of given quantities Find , the fluid pressure at the bottom of tube 2. Express your answer in terms of , , , , and . Hint 1. Find the pressure at the bottom of tube 2 Find , the fluid pressure at the bottom of tube 2. Express your answer in terms of , , and . ANSWER: p2 h2 p2 h2 g p1 h1 p2 = gh2 p2 p2 p1 v1 A1 A2 p2 p1 v1 v2 p2 = p1 + ( − ) 1 2 v1 2 v2 2 Typesetting math: 100% Hint 2. Find in terms of The fluid is incompressible, so you can use the continuity equation to relate the fluid velocities and in terms of the geometry of the pipe. Find , the fluid velocity at the bottom of tube 2, in terms of . Your answer may include and , the cross-sectional areas of the pipe. ANSWER: ANSWER: ANSWER: Correct Note that this result depends on the difference between the heights of the fluid in the tubes, a quantity that is more easily measured than the heights themselves. Problem 15.39 The container shown in the figure is filled with oil. It is open to the atmosphere on the left. v2 v1 v1 v2 v2 v1 A1 A2 v2 = A1 A2 v1 p2 = p1 + ( )(1 − ) 1 2 v1 2 ( ) A1 A2 2 v1 = 2g h1−h2 ( ) −1 A1 A2 2 −−−−−−−−−−−−−− Typesetting math: 100% Part A What is the pressure at point A? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part B What is the pressure difference between points A and B? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct PA = 106 kPa PB − PA = 4.4 kPa Typesetting math: 100% Part C What is the pressure difference between points A and C? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 15.48 You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 32.4 . If you then lower the statue into a tub of water, so that it is completely submerged, the scale reads 17 . Part A What is the density? Express your answer with the appropriate units. ANSWER: Correct Problem 15.60 Water flows from the pipe shown in the figure with a speed of 2.0 . PC − PA = 4.4 kPa N N statue = 2100 kg m3 m/s Typesetting math: 100% Part A What is the water pressure as it exits into the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the height of the standing column of water? Express your answer to two significant figures and include the appropriate units. ANSWER: Incorrect; Try Again P = 1.0×105 Pa h h = Typesetting math: 100% Score Summary: Your score on this assignment is 83.9%. You received 78.84 out of a possible total of 94 points. Typesetting math: 100%

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