3. How might a lesson plan differ between elementary earth science and high school computer science?

3. How might a lesson plan differ between elementary earth science and high school computer science?

Elementary earth science The basic concept are more important here, … Read More...
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 N m2 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%

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 N m2 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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1. How might a lesson plan differ between elementary earth science and high school computer science?

1. How might a lesson plan differ between elementary earth science and high school computer science?

Elementary earth science The basic concept are more important here, … Read More...
Chapter 11 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Understanding Work and Kinetic Energy Learning Goal: To learn about the Work-Energy Theorem and its basic applications. In this problem, you will learn about the relationship between the work done on an object and the kinetic energy of that object. The kinetic energy of an object of mass moving at a speed is defined as . It seems reasonable to say that the speed of an object–and, therefore, its kinetic energy–can be changed by performing work on the object. In this problem, we will explore the mathematical relationship between the work done on an object and the change in the kinetic energy of that object. First, let us consider a sled of mass being pulled by a constant, horizontal force of magnitude along a rough, horizontal surface. The sled is speeding up. Part A How many forces are acting on the sled? ANSWER: Part B This question will be shown after you complete previous question(s). Part C K m v K = (1/2)mv2 m F one two three four

Chapter 11 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Understanding Work and Kinetic Energy Learning Goal: To learn about the Work-Energy Theorem and its basic applications. In this problem, you will learn about the relationship between the work done on an object and the kinetic energy of that object. The kinetic energy of an object of mass moving at a speed is defined as . It seems reasonable to say that the speed of an object–and, therefore, its kinetic energy–can be changed by performing work on the object. In this problem, we will explore the mathematical relationship between the work done on an object and the change in the kinetic energy of that object. First, let us consider a sled of mass being pulled by a constant, horizontal force of magnitude along a rough, horizontal surface. The sled is speeding up. Part A How many forces are acting on the sled? ANSWER: Part B This question will be shown after you complete previous question(s). Part C K m v K = (1/2)mv2 m F one two three four

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What is a décimas? Using the article in the reader on the décima as a reference, provide an explanation of what this is, and make mention of some of its structural characteristics

What is a décimas? Using the article in the reader on the décima as a reference, provide an explanation of what this is, and make mention of some of its structural characteristics

The term décimas is a term indication to a lone … Read More...
CE 309 Fluid Mechanics Laboratory 2015 Assignment: ABET Criterion b You are tasked by SMU to design laboratory equipment for accurately determining discharge coefficients of an orifice in a reservoir discharging into the atmosphere (free jet). The equipment will be used in an undergraduate fluid mechanics laboratory class. You are not allowed to recommend an over-the-shelf system sold by manufacturers but must begin with basic materials. Your design must include the following; • Neat sketches and drawing illustrating your design. Sketches must be to scale. All sections of the sketch must be labeled in detail. As an example, a proposed motor must show the type, horsepower as well as any details necessary for the acquisition of the motor. • Statement of cost of individual items as well as the gross. It must also include installation costs where applicable. You are encouraged to recommend modern instrumentation in you design however costs must be kept as reasonable as possible. An esoteric system with no regard to the cost is of little value. Justify all your choices. • Develop a procedure for students operating the system to achieve the laboratory objectives. Indicate the advantages of your design over the current. • Keep your report to 3 pages maximum.

CE 309 Fluid Mechanics Laboratory 2015 Assignment: ABET Criterion b You are tasked by SMU to design laboratory equipment for accurately determining discharge coefficients of an orifice in a reservoir discharging into the atmosphere (free jet). The equipment will be used in an undergraduate fluid mechanics laboratory class. You are not allowed to recommend an over-the-shelf system sold by manufacturers but must begin with basic materials. Your design must include the following; • Neat sketches and drawing illustrating your design. Sketches must be to scale. All sections of the sketch must be labeled in detail. As an example, a proposed motor must show the type, horsepower as well as any details necessary for the acquisition of the motor. • Statement of cost of individual items as well as the gross. It must also include installation costs where applicable. You are encouraged to recommend modern instrumentation in you design however costs must be kept as reasonable as possible. An esoteric system with no regard to the cost is of little value. Justify all your choices. • Develop a procedure for students operating the system to achieve the laboratory objectives. Indicate the advantages of your design over the current. • Keep your report to 3 pages maximum.

No expert has answered this question yet. You can browse … Read More...
Question 1 In order to properly manage expenses, the company investigates the amount of money spent by its sales office. The below numbers are related to six randomly selected receipts provided by the staff. $147 $124 $93 $158 $164 $171 a) Calculate ̅ , s2 and s for the expense data. b) Assume that the distribution of expenses is approximately normally distributed. Calculate estimates of tolerance intervals containing 68.26 percent, 95.44 percent, and 99.73 percent of all expenses by the sales office. c) If a member of the sales office submits a receipt with the amount of $190, should this expense be considered unusually high? Explain your answer. d) Compute and interpret the z-score for each of the six expenses. Question 2 A survey presents the results of a concept study for the taste of new food. Three hundred consumers between 18 and 49 years old were randomly selected. After sampling the new cuisine, each was asked to rate the quality of food. The rating was made on a scale from 1 to 5, with 5 representing “extremely agree with the quality” and with 1 representing “not at all agree with the new food.” The results obtained are given in Table 1. Estimate the probability that a randomly selected 18- to 49-year-old consumer a) Would give the phrase a rating of 4. b) Would give the phrase a rating of 3 or higher. c) Is in the 18–26 age group; the 27–35 age group; the 36–49 age group. d) Is a male who gives the phrase a rating of 5. e) Is a 36- to 49-year-old who gives the phrase a rating of 2. f) Estimate the probability that a randomly selected 18- to 49-year-old consumer is a 27- to 49-year-old who gives the phrase a rating of 3. g) Estimate the probability that a randomly selected 18- to 49-year-old consumer would 1) give the phrase a rating of 2 or 4 given that the consumer is male; 2) give the phrase a rating of 4 or 5 given that the consumer is female. Based on the results of parts 1 and 2, is the appeal of the phrase among males much different from the appeal of the phrase among females? Explain. h) Give the phrase a rating of 4 or 5, 1) given that the consumer is in the 18–26 age group; 2) given that the consumer is in the 27–35 age group; 3) given that the consumer is in the 36–49 age group. Table 1. Gender Age Group Rating Total Male Female 18-26 27-35 36-49 Extremely Appealing (5) 151 68 83 48 66 37 (4) 91 51 40 36 36 19 (3) 36 21 15 9 12 15 (2) 13 7 6 4 6 3 Not at all appealing(1) 9 3 6 4 3 2 Question 3 Based on the reports provided by the brokers, it is concluded that the annual returns on common stocks are approximately normally distributed with a mean of 17.8 percent and a standard deviation of 29.3 percent. On the other hand, the company reports that the annual returns on tax-free municipal bonds are approximately normally distributed with a mean return of 4.7 percent and a standard deviation of 10.2 percent. Find the probability that a randomly selected a) Common stock will give a positive yearly return. b) Tax-free municipal bond will give a positive yearly return. c) Common stock will give more than a 13 percent return. d) Tax-free municipal bond will give more than a 11.5 percent return. e) Common stock will give a loss of at least 7 percent. f) Tax-free municipal bond will give a loss of at least 10 percent. Question 4 Based on a sample of 176 workers, it is estimated that the mean amount of paid time lost during a three-month period was 1.4 days per employee with a standard deviation of 1.3 days. It is also estimated that the mean amount of unpaid time lost during a three-month period was 1.0 day per employee with a standard deviation of 1.8 days. We randomly select a sample of 100 workers. a) What is the probability that the average amount of paid time lost during a three-month period for the 100 blue-collar workers will exceed 1.5 days? Assume σ equals 1.3 days. b) What is the probability that the average amount of unpaid time lost during a three-month period for the 100 workers will exceed 1.5 days? Assume σ equals 1.8 days. c) A sample of 100 workers is randomly selected. Suppose the sample mean amount of unpaid time lost during a three-month period actually exceeds 1.5 days. Would it be reasonable to conclude that the mean amount of unpaid time lost has increased above the previously estimated 1.0 day? Explain. Assume σ still equals 1.8 days.

Question 1 In order to properly manage expenses, the company investigates the amount of money spent by its sales office. The below numbers are related to six randomly selected receipts provided by the staff. $147 $124 $93 $158 $164 $171 a) Calculate ̅ , s2 and s for the expense data. b) Assume that the distribution of expenses is approximately normally distributed. Calculate estimates of tolerance intervals containing 68.26 percent, 95.44 percent, and 99.73 percent of all expenses by the sales office. c) If a member of the sales office submits a receipt with the amount of $190, should this expense be considered unusually high? Explain your answer. d) Compute and interpret the z-score for each of the six expenses. Question 2 A survey presents the results of a concept study for the taste of new food. Three hundred consumers between 18 and 49 years old were randomly selected. After sampling the new cuisine, each was asked to rate the quality of food. The rating was made on a scale from 1 to 5, with 5 representing “extremely agree with the quality” and with 1 representing “not at all agree with the new food.” The results obtained are given in Table 1. Estimate the probability that a randomly selected 18- to 49-year-old consumer a) Would give the phrase a rating of 4. b) Would give the phrase a rating of 3 or higher. c) Is in the 18–26 age group; the 27–35 age group; the 36–49 age group. d) Is a male who gives the phrase a rating of 5. e) Is a 36- to 49-year-old who gives the phrase a rating of 2. f) Estimate the probability that a randomly selected 18- to 49-year-old consumer is a 27- to 49-year-old who gives the phrase a rating of 3. g) Estimate the probability that a randomly selected 18- to 49-year-old consumer would 1) give the phrase a rating of 2 or 4 given that the consumer is male; 2) give the phrase a rating of 4 or 5 given that the consumer is female. Based on the results of parts 1 and 2, is the appeal of the phrase among males much different from the appeal of the phrase among females? Explain. h) Give the phrase a rating of 4 or 5, 1) given that the consumer is in the 18–26 age group; 2) given that the consumer is in the 27–35 age group; 3) given that the consumer is in the 36–49 age group. Table 1. Gender Age Group Rating Total Male Female 18-26 27-35 36-49 Extremely Appealing (5) 151 68 83 48 66 37 (4) 91 51 40 36 36 19 (3) 36 21 15 9 12 15 (2) 13 7 6 4 6 3 Not at all appealing(1) 9 3 6 4 3 2 Question 3 Based on the reports provided by the brokers, it is concluded that the annual returns on common stocks are approximately normally distributed with a mean of 17.8 percent and a standard deviation of 29.3 percent. On the other hand, the company reports that the annual returns on tax-free municipal bonds are approximately normally distributed with a mean return of 4.7 percent and a standard deviation of 10.2 percent. Find the probability that a randomly selected a) Common stock will give a positive yearly return. b) Tax-free municipal bond will give a positive yearly return. c) Common stock will give more than a 13 percent return. d) Tax-free municipal bond will give more than a 11.5 percent return. e) Common stock will give a loss of at least 7 percent. f) Tax-free municipal bond will give a loss of at least 10 percent. Question 4 Based on a sample of 176 workers, it is estimated that the mean amount of paid time lost during a three-month period was 1.4 days per employee with a standard deviation of 1.3 days. It is also estimated that the mean amount of unpaid time lost during a three-month period was 1.0 day per employee with a standard deviation of 1.8 days. We randomly select a sample of 100 workers. a) What is the probability that the average amount of paid time lost during a three-month period for the 100 blue-collar workers will exceed 1.5 days? Assume σ equals 1.3 days. b) What is the probability that the average amount of unpaid time lost during a three-month period for the 100 workers will exceed 1.5 days? Assume σ equals 1.8 days. c) A sample of 100 workers is randomly selected. Suppose the sample mean amount of unpaid time lost during a three-month period actually exceeds 1.5 days. Would it be reasonable to conclude that the mean amount of unpaid time lost has increased above the previously estimated 1.0 day? Explain. Assume σ still equals 1.8 days.

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