The solubility of Ar in water at 25°C is 1.6 × 10-3 M when the pressure of the Ar above the solution is 1.0 atm. The solubility of Ar at a pressure of 2.5 atm is __________ M. A) 1.6 × 103 B) 6.4 × 10-4 C) 4.0 × 10-3 D) 7.5 × 10-2 E) 1.6 × 10-3

The solubility of Ar in water at 25°C is 1.6 × 10-3 M when the pressure of the Ar above the solution is 1.0 atm. The solubility of Ar at a pressure of 2.5 atm is __________ M. A) 1.6 × 103 B) 6.4 × 10-4 C) 4.0 × 10-3 D) 7.5 × 10-2 E) 1.6 × 10-3

Determine the changes in enthalpy , internal energy and entropy when 2.7 kg of water taken at P1 = 1.0133 X 10^5 Pa and T1 = 293 K evaporate at P2 = 0.50665 X 10^5 Pa and T2 = 373 K . Cp(l) ~ Cv(l) = 4.187 X 10^3 J/kg/K the specific heat of evaporation being 2260.98 X 10^3 J/kg

Determine the changes in enthalpy , internal energy and entropy when 2.7 kg of water taken at P1 = 1.0133 X 10^5 Pa and T1 = 293 K evaporate at P2 = 0.50665 X 10^5 Pa and T2 = 373 K . Cp(l) ~ Cv(l) = 4.187 X 10^3 J/kg/K the specific heat of evaporation being 2260.98 X 10^3 J/kg

info@checkyourstudy.com
Assignment 8 Due: 11:59pm on Friday, April 4, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 10.3 Part A If a particle’s speed increases by a factor of 5, by what factor does its kinetic energy change? ANSWER: Correct Conceptual Question 10.11 A spring is compressed 1.5 . Part A How far must you compress a spring with twice the spring constant to store the same amount of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct = 25 K2 K1 cm x = 1.1 cm Problem 10.2 The lowest point in Death Valley is below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 . Part A What is the change in potential energy of an energetic 80 hiker who makes it from the floor of Death Valley to the top of Mt.Whitney? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.3 Part A At what speed does a 1800 compact car have the same kinetic energy as a 1.80×104 truck going 21.0 ? Express your answer with the appropriate units. ANSWER: Correct Problem 10.5 A boy reaches out of a window and tosses a ball straight up with a speed of 13 . The ball is 21 above the ground as he releases it. 85m m kg U = 3.5×106 J kg kg km/hr vc = 66.4 km hr m/s m Part A Use energy to find the ball’s maximum height above the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Use energy to find the ball’s speed as it passes the window on its way down. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C Use energy to find the speed of impact on the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Hmax = 30 m v = 13 ms v = 24 ms Problem 10.8 A 59.0 skateboarder wants to just make it to the upper edge of a “quarter pipe,” a track that is one-quarter of a circle with a radius of 2.30 . Part A What speed does he need at the bottom? Express your answer with the appropriate units. ANSWER: Correct Problem 10.12 A 1500 car traveling at 12 suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll. Part A kg m 6.71 ms kg m/s What will be the car’s speed as it coasts into the gas station on the other side of the valley? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Ups and Downs Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy transformations that take place involve the object’s kinetic energy and its gravitational potential energy . The law of conservation of energy for such cases implies that the sum of the object’s kinetic energy and potential energy does not change with time. This idea can be expressed by the equation , where “i” denotes the “initial” moment and “f” denotes the “final” moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. First, let us consider an object launched vertically upward with an initial speed . Neglect air resistance. Part A As the projectile goes upward, what energy changes take place? ANSWER: v = 6.8 ms K = (1/2)mv2 U = mgh Ki + Ui = Kf + Uf v Correct Part B At the top point of the flight, what can be said about the projectile’s kinetic and potential energy? ANSWER: Correct Strictly speaking, it is not the ball that possesses potential energy; rather, it is the system “Earth-ball.” Although we will often talk about “the gravitational potential energy of an elevated object,” it is useful to keep in mind that the energy, in fact, is associated with the interactions between the earth and the elevated object. Part C The potential energy of the object at the moment of launch __________. ANSWER: Both kinetic and potential energy decrease. Both kinetic and potential energy increase. Kinetic energy decreases; potential energy increases. Kinetic energy increases; potential energy decreases. Both kinetic and potential energy are at their maximum values. Both kinetic and potential energy are at their minimum values. Kinetic energy is at a maximum; potential energy is at a minimum. Kinetic energy is at a minimum; potential energy is at a maximum. Correct Usually, the zero level is chosen so as to make the relevant calculations simpler. In this case, it makes good sense to assume that at the ground level–but this is not, by any means, the only choice! Part D Using conservation of energy, find the maximum height to which the object will rise. Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: Correct You may remember this result from kinematics. It is comforting to know that our new approach yields the same answer. Part E At what height above the ground does the projectile have a speed of ? Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: is negative is positive is zero depends on the choice of the “zero level” of potential energy U = 0 hmax v g hmax = v2 2g h 0.5v v g h = 3 v2 8g Correct Part F What is the speed of the object at the height of ? Express your answer in terms of and . Use three significant figures in the numeric coefficient. Hint 1. How to approach the problem You are being asked for the speed at half of the maximum height. You know that at the initial height ( ), the speed is . All of the energy is kinetic energy, and so, the total energy is . At the maximum height, all of the energy is potential energy. Since the gravitational potential energy is proportional to , half of the initial kinetic energy must have been converted to potential energy when the projectile is at . Thus, the kinetic energy must be half of its original value (i.e., when ). You need to determine the speed, as a multiple of , that corresponds to such a kinetic energy. ANSWER: Correct Let us now consider objects launched at an angle. For such situations, using conservation of energy leads to a quicker solution than can be produced by kinematics. Part G A ball is launched as a projectile with initial speed at an angle above the horizontal. Using conservation of energy, find the maximum height of the ball’s flight. Express your answer in terms of , , and . Hint 1. Find the final kinetic energy Find the final kinetic energy of the ball. Here, the best choice of “final” moment is the point at which the ball reaches its maximum height, since this is the point we are interested in. u (1/2)hmax v g h = 0 v (1/2)mv2 h (1/2)hmax (1/4)mv2 h = (1/2)hmax v u = 0.707v v hmax v g Kf Express your answer in terms of , , and . Hint 1. Find the speed at the maximum height The speed of the ball at the maximum height is __________. ANSWER: ANSWER: ANSWER: Correct Part H A ball is launched with initial speed from ground level up a frictionless slope. The slope makes an angle with the horizontal. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of , , and . You may or may not use all of these quantities. v m 0 v v cos v sin v tan Kf = 0.5m(vcos( ))2 hmax = (vsin( ))2 2g v hmax v g ANSWER: Correct Interestingly, the answer does not depend on . The difference between this situation and the projectile case is that the ball moving up a slope has no kinetic energy at the top of its trajectory whereas the projectile launched at an angle does. Part I A ball is launched with initial speed from the ground level up a frictionless hill. The hill becomes steeper as the ball slides up; however, the ball remains in contact with the hill at all times. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of and . ANSWER: Correct The profile of the hill does not matter; the equation would have the same terms regardless of the steepness of the hill. Problem 10.14 A 12- -long spring is attached to the ceiling. When a 2.2 mass is hung from it, the spring stretches to a length of 17 . Part A What is the spring constant ? Express your answer to two significant figures and include the appropriate units. hmax = v2 2g v hmax v g hmax = v2 2g Ki + Ui = Kf + Uf cm kg cm k ANSWER: Correct Part B How long is the spring when a 3.0 mass is suspended from it? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.17 A 6.2 mass hanging from a spring scale is slowly lowered onto a vertical spring, as shown in . You may want to review ( pages 255 – 257) . For help with math skills, you may want to review: Solving Algebraic Equations = 430 k Nm kg y = 19 cm kg Part A What does the spring scale read just before the mass touches the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass before it touches the scale. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? ANSWER: Correct Part B The scale reads 22 when the lower spring has been compressed by 2.7 . What is the value of the spring constant for the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? Use these to determine the force on the mass by the spring, taking note of the directions from your picture. How is the spring constant related to the force by the spring and the compression of the spring? Check your units. ANSWER: F = 61 N N cm k = 1400 k Nm Correct Part C At what compression length will the scale read zero? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces on the mass. When the scale reads zero, what is the force on the mass due to the scale? What is the gravitational force on the mass? What is the force on the mass by the spring? How is the compression length related to the force by the spring and the spring constant? Check your units. ANSWER: Correct Problem 10.18 Part A How far must you stretch a spring with = 800 to store 180 of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: y = 4.2 cm k N/m J Correct Problem 10.22 A 15 runaway grocery cart runs into a spring with spring constant 230 and compresses it by 57 . Part A What was the speed of the cart just before it hit the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Spring Gun A spring-loaded toy gun is used to shoot a ball straight up in the air. The ball reaches a maximum height , measured from the equilibrium position of the spring. s = 0.67 m kg N/m cm v = 2.2 ms H Part A The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to . Hint 1. Potential energy of the spring The potential energy of a spring is proportional to the square of the distance the spring is compressed. The spring was compressed half the distance, so the mass, when launched, has one quarter of the energy as in the first trial. Hint 2. Potential energy of the ball At the highest point in the ball’s trajectory, all of the spring’s potential energy has been converted into gravitational potential energy of the ball. ANSWER: Correct A Bullet Is Fired into a Wooden Block A bullet of mass is fired horizontally with speed at a wooden block of mass resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed . H height = H 4 mb vi mw vf Part A Which of the following best describes this collision? Hint 1. Types of collisions An inelastic collision is a collision in which kinetic energy is not conserved. In a partially inelastic collision, kinetic energy is lost, but the objects colliding do not stick together. From this information, you can infer what completely inelastic and elastic collisions are. ANSWER: Correct Part B Which of the following quantities, if any, are conserved during this collision? Hint 1. When is kinetic energy conserved? Kinetic energy is conserved only in perfectly elastic collisions. ANSWER: perfectly elastic partially inelastic perfectly inelastic Correct Part C What is the speed of the block/bullet system after the collision? Express your answer in terms of , , and . Hint 1. Find the momentum after the collision What is the total momentum of the block/bullet system after the collision? Express your answer in terms of and other given quantities. ANSWER: Hint 2. Use conservation of momentum The momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for , this time expressed as the total momentum of the system before the collision. Express your answer in terms of and other given quantities. ANSWER: kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy vi mw mb ptotal vf ptotal = (mw + mb)vf ptotal vi ptotal = mbvi ANSWER: Correct Problem 10.31 Ball 1, with a mass of 150 and traveling at 15.0 , collides head on with ball 2, which has a mass of 340 and is initially at rest. Part A What are the final velocities of each ball if the collision is perfectly elastic? Express your answer with the appropriate units. ANSWER: Correct Part B Express your answer with the appropriate units. ANSWER: Correct Part C vf = mb vi mb+mw g m/s g (vfx) = -5.82 1 ms (vfx) = 9.18 2 ms What are the final velocities of each ball if the collision is perfectly inelastic? Express your answer with the appropriate units. ANSWER: Correct Part D Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.43 A package of mass is released from rest at a warehouse loading dock and slides down the = 2.2 – high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass , from the bottom of the chute. You may want to review ( pages 265 – 269) . For help with math skills, you may want to review: Solving Algebraic Equations (vfx) = 4.59 1 ms (vfx) = 4.59 2 ms m h m 2m Part A Suppose the packages stick together. What is their common speed after the collision? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are two parts to this problem: the block sliding down the frictionless incline and the collision. What conservation laws are valid in each part? In terms of , what are the kinetic and potential energies of the block at the top of the incline? What is the potential energy of the same block at the bottom just before the collision? What are the kinetic energy and velocity of block just before the collision? What is conserved during the collision? What is the total momentum of the two blocks before the collision? What is the momentum of the two blocks stuck together after the collision? What is the velocity of the two blocks after the collision? ANSWER: Correct Part B Suppose the collision between the packages is perfectly elastic. To what height does the package of mass rebound? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are three parts to this problem: the block sliding down the incline, the collision, and mass going back up the incline. What conservation laws are valid in each part? m m v = 2.2 ms m m What is an elastic collision? For an elastic collision, how are the initial and final velocities related when one of the masses is initially at rest? Using the velocity of just before the collision from Part A, what is the velocity of just after the collision in this case? What are the kinetic and potential energies of mass just after the collision? What is the kinetic energy of mass at its maximum rebound height? Using conservation of energy, what is the potential energy of mass at its maximum height? What is the maximum height? ANSWER: Correct Problem 10.35 A cannon tilted up at a 35.0 angle fires a cannon ball at 79.0 from atop a 21.0 -high fortress wall. Part A What is the ball’s impact speed on the ground below? Express your answer with the appropriate units. ANSWER: Correct Problem 10.45 A 1000 safe is 2.5 above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 48 . m m m m m h = 24 cm $ m/s m vf = 81.6 ms kg m cm Part A What is the spring constant of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.49 A 100 block on a frictionless table is firmly attached to one end of a spring with = 21 . The other end of the spring is anchored to the wall. A 30 ball is thrown horizontally toward the block with a speed of 6.0 . Part A If the collision is perfectly elastic, what is the ball’s speed immediately after the collision? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the maximum compression of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: = 2.5×105 k Nm g k N/m g m/s v = 3.2 ms Correct Part C Repeat part A for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D Repeat part B for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.4%. You received 120.28 out of a possible total of 121 points. x = 0.19 m v = 1.4 ms x = 0.11 m

Assignment 8 Due: 11:59pm on Friday, April 4, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 10.3 Part A If a particle’s speed increases by a factor of 5, by what factor does its kinetic energy change? ANSWER: Correct Conceptual Question 10.11 A spring is compressed 1.5 . Part A How far must you compress a spring with twice the spring constant to store the same amount of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct = 25 K2 K1 cm x = 1.1 cm Problem 10.2 The lowest point in Death Valley is below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 . Part A What is the change in potential energy of an energetic 80 hiker who makes it from the floor of Death Valley to the top of Mt.Whitney? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.3 Part A At what speed does a 1800 compact car have the same kinetic energy as a 1.80×104 truck going 21.0 ? Express your answer with the appropriate units. ANSWER: Correct Problem 10.5 A boy reaches out of a window and tosses a ball straight up with a speed of 13 . The ball is 21 above the ground as he releases it. 85m m kg U = 3.5×106 J kg kg km/hr vc = 66.4 km hr m/s m Part A Use energy to find the ball’s maximum height above the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Use energy to find the ball’s speed as it passes the window on its way down. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C Use energy to find the speed of impact on the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Hmax = 30 m v = 13 ms v = 24 ms Problem 10.8 A 59.0 skateboarder wants to just make it to the upper edge of a “quarter pipe,” a track that is one-quarter of a circle with a radius of 2.30 . Part A What speed does he need at the bottom? Express your answer with the appropriate units. ANSWER: Correct Problem 10.12 A 1500 car traveling at 12 suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll. Part A kg m 6.71 ms kg m/s What will be the car’s speed as it coasts into the gas station on the other side of the valley? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Ups and Downs Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy transformations that take place involve the object’s kinetic energy and its gravitational potential energy . The law of conservation of energy for such cases implies that the sum of the object’s kinetic energy and potential energy does not change with time. This idea can be expressed by the equation , where “i” denotes the “initial” moment and “f” denotes the “final” moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. First, let us consider an object launched vertically upward with an initial speed . Neglect air resistance. Part A As the projectile goes upward, what energy changes take place? ANSWER: v = 6.8 ms K = (1/2)mv2 U = mgh Ki + Ui = Kf + Uf v Correct Part B At the top point of the flight, what can be said about the projectile’s kinetic and potential energy? ANSWER: Correct Strictly speaking, it is not the ball that possesses potential energy; rather, it is the system “Earth-ball.” Although we will often talk about “the gravitational potential energy of an elevated object,” it is useful to keep in mind that the energy, in fact, is associated with the interactions between the earth and the elevated object. Part C The potential energy of the object at the moment of launch __________. ANSWER: Both kinetic and potential energy decrease. Both kinetic and potential energy increase. Kinetic energy decreases; potential energy increases. Kinetic energy increases; potential energy decreases. Both kinetic and potential energy are at their maximum values. Both kinetic and potential energy are at their minimum values. Kinetic energy is at a maximum; potential energy is at a minimum. Kinetic energy is at a minimum; potential energy is at a maximum. Correct Usually, the zero level is chosen so as to make the relevant calculations simpler. In this case, it makes good sense to assume that at the ground level–but this is not, by any means, the only choice! Part D Using conservation of energy, find the maximum height to which the object will rise. Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: Correct You may remember this result from kinematics. It is comforting to know that our new approach yields the same answer. Part E At what height above the ground does the projectile have a speed of ? Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: is negative is positive is zero depends on the choice of the “zero level” of potential energy U = 0 hmax v g hmax = v2 2g h 0.5v v g h = 3 v2 8g Correct Part F What is the speed of the object at the height of ? Express your answer in terms of and . Use three significant figures in the numeric coefficient. Hint 1. How to approach the problem You are being asked for the speed at half of the maximum height. You know that at the initial height ( ), the speed is . All of the energy is kinetic energy, and so, the total energy is . At the maximum height, all of the energy is potential energy. Since the gravitational potential energy is proportional to , half of the initial kinetic energy must have been converted to potential energy when the projectile is at . Thus, the kinetic energy must be half of its original value (i.e., when ). You need to determine the speed, as a multiple of , that corresponds to such a kinetic energy. ANSWER: Correct Let us now consider objects launched at an angle. For such situations, using conservation of energy leads to a quicker solution than can be produced by kinematics. Part G A ball is launched as a projectile with initial speed at an angle above the horizontal. Using conservation of energy, find the maximum height of the ball’s flight. Express your answer in terms of , , and . Hint 1. Find the final kinetic energy Find the final kinetic energy of the ball. Here, the best choice of “final” moment is the point at which the ball reaches its maximum height, since this is the point we are interested in. u (1/2)hmax v g h = 0 v (1/2)mv2 h (1/2)hmax (1/4)mv2 h = (1/2)hmax v u = 0.707v v hmax v g Kf Express your answer in terms of , , and . Hint 1. Find the speed at the maximum height The speed of the ball at the maximum height is __________. ANSWER: ANSWER: ANSWER: Correct Part H A ball is launched with initial speed from ground level up a frictionless slope. The slope makes an angle with the horizontal. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of , , and . You may or may not use all of these quantities. v m 0 v v cos v sin v tan Kf = 0.5m(vcos( ))2 hmax = (vsin( ))2 2g v hmax v g ANSWER: Correct Interestingly, the answer does not depend on . The difference between this situation and the projectile case is that the ball moving up a slope has no kinetic energy at the top of its trajectory whereas the projectile launched at an angle does. Part I A ball is launched with initial speed from the ground level up a frictionless hill. The hill becomes steeper as the ball slides up; however, the ball remains in contact with the hill at all times. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of and . ANSWER: Correct The profile of the hill does not matter; the equation would have the same terms regardless of the steepness of the hill. Problem 10.14 A 12- -long spring is attached to the ceiling. When a 2.2 mass is hung from it, the spring stretches to a length of 17 . Part A What is the spring constant ? Express your answer to two significant figures and include the appropriate units. hmax = v2 2g v hmax v g hmax = v2 2g Ki + Ui = Kf + Uf cm kg cm k ANSWER: Correct Part B How long is the spring when a 3.0 mass is suspended from it? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.17 A 6.2 mass hanging from a spring scale is slowly lowered onto a vertical spring, as shown in . You may want to review ( pages 255 – 257) . For help with math skills, you may want to review: Solving Algebraic Equations = 430 k Nm kg y = 19 cm kg Part A What does the spring scale read just before the mass touches the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass before it touches the scale. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? ANSWER: Correct Part B The scale reads 22 when the lower spring has been compressed by 2.7 . What is the value of the spring constant for the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? Use these to determine the force on the mass by the spring, taking note of the directions from your picture. How is the spring constant related to the force by the spring and the compression of the spring? Check your units. ANSWER: F = 61 N N cm k = 1400 k Nm Correct Part C At what compression length will the scale read zero? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces on the mass. When the scale reads zero, what is the force on the mass due to the scale? What is the gravitational force on the mass? What is the force on the mass by the spring? How is the compression length related to the force by the spring and the spring constant? Check your units. ANSWER: Correct Problem 10.18 Part A How far must you stretch a spring with = 800 to store 180 of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: y = 4.2 cm k N/m J Correct Problem 10.22 A 15 runaway grocery cart runs into a spring with spring constant 230 and compresses it by 57 . Part A What was the speed of the cart just before it hit the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Spring Gun A spring-loaded toy gun is used to shoot a ball straight up in the air. The ball reaches a maximum height , measured from the equilibrium position of the spring. s = 0.67 m kg N/m cm v = 2.2 ms H Part A The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to . Hint 1. Potential energy of the spring The potential energy of a spring is proportional to the square of the distance the spring is compressed. The spring was compressed half the distance, so the mass, when launched, has one quarter of the energy as in the first trial. Hint 2. Potential energy of the ball At the highest point in the ball’s trajectory, all of the spring’s potential energy has been converted into gravitational potential energy of the ball. ANSWER: Correct A Bullet Is Fired into a Wooden Block A bullet of mass is fired horizontally with speed at a wooden block of mass resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed . H height = H 4 mb vi mw vf Part A Which of the following best describes this collision? Hint 1. Types of collisions An inelastic collision is a collision in which kinetic energy is not conserved. In a partially inelastic collision, kinetic energy is lost, but the objects colliding do not stick together. From this information, you can infer what completely inelastic and elastic collisions are. ANSWER: Correct Part B Which of the following quantities, if any, are conserved during this collision? Hint 1. When is kinetic energy conserved? Kinetic energy is conserved only in perfectly elastic collisions. ANSWER: perfectly elastic partially inelastic perfectly inelastic Correct Part C What is the speed of the block/bullet system after the collision? Express your answer in terms of , , and . Hint 1. Find the momentum after the collision What is the total momentum of the block/bullet system after the collision? Express your answer in terms of and other given quantities. ANSWER: Hint 2. Use conservation of momentum The momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for , this time expressed as the total momentum of the system before the collision. Express your answer in terms of and other given quantities. ANSWER: kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy vi mw mb ptotal vf ptotal = (mw + mb)vf ptotal vi ptotal = mbvi ANSWER: Correct Problem 10.31 Ball 1, with a mass of 150 and traveling at 15.0 , collides head on with ball 2, which has a mass of 340 and is initially at rest. Part A What are the final velocities of each ball if the collision is perfectly elastic? Express your answer with the appropriate units. ANSWER: Correct Part B Express your answer with the appropriate units. ANSWER: Correct Part C vf = mb vi mb+mw g m/s g (vfx) = -5.82 1 ms (vfx) = 9.18 2 ms What are the final velocities of each ball if the collision is perfectly inelastic? Express your answer with the appropriate units. ANSWER: Correct Part D Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.43 A package of mass is released from rest at a warehouse loading dock and slides down the = 2.2 – high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass , from the bottom of the chute. You may want to review ( pages 265 – 269) . For help with math skills, you may want to review: Solving Algebraic Equations (vfx) = 4.59 1 ms (vfx) = 4.59 2 ms m h m 2m Part A Suppose the packages stick together. What is their common speed after the collision? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are two parts to this problem: the block sliding down the frictionless incline and the collision. What conservation laws are valid in each part? In terms of , what are the kinetic and potential energies of the block at the top of the incline? What is the potential energy of the same block at the bottom just before the collision? What are the kinetic energy and velocity of block just before the collision? What is conserved during the collision? What is the total momentum of the two blocks before the collision? What is the momentum of the two blocks stuck together after the collision? What is the velocity of the two blocks after the collision? ANSWER: Correct Part B Suppose the collision between the packages is perfectly elastic. To what height does the package of mass rebound? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are three parts to this problem: the block sliding down the incline, the collision, and mass going back up the incline. What conservation laws are valid in each part? m m v = 2.2 ms m m What is an elastic collision? For an elastic collision, how are the initial and final velocities related when one of the masses is initially at rest? Using the velocity of just before the collision from Part A, what is the velocity of just after the collision in this case? What are the kinetic and potential energies of mass just after the collision? What is the kinetic energy of mass at its maximum rebound height? Using conservation of energy, what is the potential energy of mass at its maximum height? What is the maximum height? ANSWER: Correct Problem 10.35 A cannon tilted up at a 35.0 angle fires a cannon ball at 79.0 from atop a 21.0 -high fortress wall. Part A What is the ball’s impact speed on the ground below? Express your answer with the appropriate units. ANSWER: Correct Problem 10.45 A 1000 safe is 2.5 above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 48 . m m m m m h = 24 cm $ m/s m vf = 81.6 ms kg m cm Part A What is the spring constant of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.49 A 100 block on a frictionless table is firmly attached to one end of a spring with = 21 . The other end of the spring is anchored to the wall. A 30 ball is thrown horizontally toward the block with a speed of 6.0 . Part A If the collision is perfectly elastic, what is the ball’s speed immediately after the collision? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the maximum compression of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: = 2.5×105 k Nm g k N/m g m/s v = 3.2 ms Correct Part C Repeat part A for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D Repeat part B for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.4%. You received 120.28 out of a possible total of 121 points. x = 0.19 m v = 1.4 ms x = 0.11 m

please email info@checkyourstudy.com
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%

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

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

please email info@checkyourstudy.com
The value of DH° for the reaction below is -482 kJ. Calculate the heat (kJ) released to the surroundings when 24.0 g of CO (g) reacts completely. 2 2 2CO(g) +O (g)®2CO (g) A) 3 2.89×10 B) 103 C) 207 D) 65.7 E) -482

The value of DH° for the reaction below is -482 kJ. Calculate the heat (kJ) released to the surroundings when 24.0 g of CO (g) reacts completely. 2 2 2CO(g) +O (g)®2CO (g) A) 3 2.89×10 B) 103 C) 207 D) 65.7 E) -482

C) 207
An average sample of coal contains 3.0% sulfur by mass. Calculate the moles of sulfur present in 2.40  103 kg of coal. (Points : 4) 9.4  102 mol 2.2  103 mol 6.0  103 mol 7.2  104 mol 7.5  104 mol

An average sample of coal contains 3.0% sulfur by mass. Calculate the moles of sulfur present in 2.40  103 kg of coal. (Points : 4) 9.4  102 mol 2.2  103 mol 6.0  103 mol 7.2  104 mol 7.5  104 mol

An average sample of coal contains 3.0% sulfur by mass. … Read More...
Biomedical Signal and Image Processing (4800_420_001) Assigned on September 12th, 2017 Assignment 4 – Noise and Correlation 1. If a signal is measured as 2.5 V and the noise is 28 mV (28 × 10−3 V), what is the SNR in dB? 2. A single sinusoidal signal is found with some noise. If the RMS value of the noise is 0.5 V and the SNR is 10 dB, what is the RMS amplitude of the sinusoid? 3. The file signal_noise.mat contains a variable x that consists of a 1.0-V peak sinusoidal signal buried in noise. What is the SNR for this signal and noise? Assume that the noise RMS is much greater than the signal RMS. Note: “signal_noise.mat” and other files used in these assignments can be downloaded from the content area of Brightspace, within the “Data Files for Exercises” folder. These files can be opened in Matlab by copying into the active folder and double-clicking on the file or using the Matlab load command using the format: load(‘signal_noise.mat’). To discover the variables within the files use the Matlab who command. 4. An 8-bit ADC converter that has an input range of ±5 V is used to convert a signal that ranges between ±2 V. What is the SNR of the input if the input noise equals the quantization noise of the converter? Hint: Refer to Equation below to find the quantization noise: 5. The file filter1.mat contains the spectrum of a fourth-order lowpass filter as variable x in dB. The file also contains the corresponding frequencies of x in variable freq. Plot the spectrum of this filter both as dB versus log frequency and as linear amplitude versus linear frequency. The frequency axis should range between 10 and 400 Hz in both plots. Hint: Use Equation below to convert: Biomedical Signal and Image Processing (4800_420_001) Assigned on September 12th, 2017 6. Generate one cycle of the square wave similar to the one shown below in a 500-point MATLAB array. Determine the RMS value of this waveform. [Hint: When you take the square of the data array, be sure to use a period before the up arrow so that MATLAB does the squaring point-by-point (i.e., x.^2).]. 7. A resistor produces 10 μV noise (i.e., 10 × 10−6 V noise) when the room temperature is 310 K and the bandwidth is 1 kHz (i.e., 1000 Hz). What current noise would be produced by this resistor? 8. A 3-ma current flows through both a diode (i.e., a semiconductor) and a 20,000-Ω (i.e., 20-kΩ) resistor. What is the net current noise, in? Assume a bandwidth of 1 kHz (i.e., 1 × 103 Hz). Which of the two components is responsible for producing the most noise? 9. Determine if the two signals, x and y, in file correl1.mat are correlated by checking the angle between them. 10. Modify the approach used in Practice Problem 3 to find the angle between short signals: Do not attempt to plot these vectors as it would require a 6-dimensional plot!

Biomedical Signal and Image Processing (4800_420_001) Assigned on September 12th, 2017 Assignment 4 – Noise and Correlation 1. If a signal is measured as 2.5 V and the noise is 28 mV (28 × 10−3 V), what is the SNR in dB? 2. A single sinusoidal signal is found with some noise. If the RMS value of the noise is 0.5 V and the SNR is 10 dB, what is the RMS amplitude of the sinusoid? 3. The file signal_noise.mat contains a variable x that consists of a 1.0-V peak sinusoidal signal buried in noise. What is the SNR for this signal and noise? Assume that the noise RMS is much greater than the signal RMS. Note: “signal_noise.mat” and other files used in these assignments can be downloaded from the content area of Brightspace, within the “Data Files for Exercises” folder. These files can be opened in Matlab by copying into the active folder and double-clicking on the file or using the Matlab load command using the format: load(‘signal_noise.mat’). To discover the variables within the files use the Matlab who command. 4. An 8-bit ADC converter that has an input range of ±5 V is used to convert a signal that ranges between ±2 V. What is the SNR of the input if the input noise equals the quantization noise of the converter? Hint: Refer to Equation below to find the quantization noise: 5. The file filter1.mat contains the spectrum of a fourth-order lowpass filter as variable x in dB. The file also contains the corresponding frequencies of x in variable freq. Plot the spectrum of this filter both as dB versus log frequency and as linear amplitude versus linear frequency. The frequency axis should range between 10 and 400 Hz in both plots. Hint: Use Equation below to convert: Biomedical Signal and Image Processing (4800_420_001) Assigned on September 12th, 2017 6. Generate one cycle of the square wave similar to the one shown below in a 500-point MATLAB array. Determine the RMS value of this waveform. [Hint: When you take the square of the data array, be sure to use a period before the up arrow so that MATLAB does the squaring point-by-point (i.e., x.^2).]. 7. A resistor produces 10 μV noise (i.e., 10 × 10−6 V noise) when the room temperature is 310 K and the bandwidth is 1 kHz (i.e., 1000 Hz). What current noise would be produced by this resistor? 8. A 3-ma current flows through both a diode (i.e., a semiconductor) and a 20,000-Ω (i.e., 20-kΩ) resistor. What is the net current noise, in? Assume a bandwidth of 1 kHz (i.e., 1 × 103 Hz). Which of the two components is responsible for producing the most noise? 9. Determine if the two signals, x and y, in file correl1.mat are correlated by checking the angle between them. 10. Modify the approach used in Practice Problem 3 to find the angle between short signals: Do not attempt to plot these vectors as it would require a 6-dimensional plot!

Whatsapp +919911743277