1. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Two kinds of moving in circles Rotary (rotational)= axis of motion at center of movement (sit and spin) Circular motion=axis of rotation outside of object

2. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Angular quantities Always use radians=r=57.3˚ 1 revolution=360 ˚ = 2 π radians –Counterclockwise is + –Clockwise -

3. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Holt book p247 #4 Θ(theta)= angular displacement in rad Θ =Δs/ r s= arc length in meters. R= radius

4. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Holt p 248 # 2 ω(lower case omega)=angular speed measured in rad/s or rev/s ω = Θ/ Δ t

5. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Holt p 250 #2 α ( lower case alpha)= angular acceleration Rad/s/s Δ ω/ Δ t

6. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Tangential Speed tangential speed (v t ) =speed along an imaginary line drawn tangent to the circular path. Larger radius= greater v t uniform circular motion= speed is constant. Section 1 Circular Motion

7. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Holt p255 #4 V t = r ω Where V t = tangential velocity (linear velocity) a t =rα Where a t = tangential acceleration (Linear acceleration)

8. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Centripetal Acceleration Section 1 Circular Motion

9. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Centripetal Acceleration centripetal acceleration= acceleration in a circular path at constant speed is due to a change in direction. Section 1 Circular Motion

10. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 (a. Section 1 Circular Motion

11. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Centripetal force =name given to the net force on an object in uniform circular motion. Any type of force or combination of forces can provide this net force. – friction – gravitational force. –Tension Section 1 Circular Motion

12. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Centripetal Force, Centripetal force is always directed to the center of the circular path. Section 1 Circular Motion

13. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Section 1 Circular Motion

14. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Newton’s Law of Universal Gravitation Chapter 7 Objectives Explain how Newton’s law of universal gravitation accounts for various phenomena, including satellite and planetary orbits, falling objects, and the tides. Apply Newton’s law of universal gravitation to solve problems.

15. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Gravitational Force, continued The centripetal force that holds the planets in orbit is the same force that pulls an apple toward the ground—gravitational force. Gravitational force is the mutual force of attraction between particles of matter. Gravitational force depends on the masses and on the distance between them. Section 2 Newton’s Law of Universal Gravitation

16. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Gravitational Force, continued Section 2 Newton’s Law of Universal Gravitation The constant G, called the constant of universal gravitation, equals 6.673  10 –11 Nm 2 /kg.

17. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Gravitational Force, continued The gravitational forces always equal and opposite. Action reaction pair. Section 2 Newton’s Law of Universal Gravitation

18. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Newton’s Law of Universal Gravitation Section 2 Newton’s Law of Universal Gravitation

19. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Newton’s law of gravitation accounts for ocean tides. High and low tides are due to the gravitational force exerted on Earth by its moon and the sun. Semi diurnal tidal pattern (two high and low tides) Section 2 Newton’s Law of Universal Gravitation

20. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu

21. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Applying the Law of Gravitation, continued weight = mass  gravitational field strength Because it depends on gravitational field strength, weight changes with location: Section 2 Newton’s Law of Universal Gravitation On the surface of any planet, the value of g, as well as your weight, will depend on the planet’s mass and radius.

22. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Universal gravitation What is the gravitational force between you and the earth given your mass of 70 kg and the earths mass of 5.98 X10 24 kg and a radius of 6.38 X10 6 m?

23. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Motion in Space Chapter 7 Objectives Describe Kepler’s laws of planetary motion. Relate Newton’s mathematical analysis of gravitational force to the elliptical planetary orbits proposed by Kepler. Solve problems involving orbital speed and period.

24. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Kepler’s Laws Kepler’s laws describe the motion of the planets. First Law planets travel in elliptical orbits around the sun, the sun is at one of the focal points. Section 3 Motion in Space

25. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Keplers 2 nd law the planet travels faster when it is closer to the sun and slower when it is farther away. Section 3 Motion in Space

26. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Kepler’s third law states that T 2  r 3. T= period of planet (time for 1 cycle or revolution. The constant of proportionality is 4  2 /Gm, where m is the mass of the object being orbited. Thus Section 3 Motion in Space

27. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Kepler’s Laws, continued Kepler’s third law the period of an object in a circular orbit depends on the same factors as speed of an object in a circular orbit : Also Vt=2πr/T hat m is the mass of the central object The mean radius (r) =distance between the centers of the two bodies. Section 3 Motion in Space

28. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Planetary Data Section 3 Motion in Space

29. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Sample Problem Period and Speed of an Orbiting Object Magellan was the first planetary spacecraft to be launched from a space shuttle. During the spacecraft’s fifth orbit around Venus, Magellan traveled at a mean altitude of 361km. If the orbit had been circular, what would Magellan’s period and speed have been? Section 3 Motion in Space

30. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Sample Problem, continued 1. Define Given: r 1 = 361 km = 3.61  10 5 m Unknown: T = ?v t = ? 2. Plan Choose an equation or situation: Use the equations for the period and speed of an object in a circular orbit. Section 3 Motion in Space

31. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Sample Problem, continued Use Table 1 in the textbook to find the values for the radius (r 2 ) and mass (m) of Venus. r 2 = 6.05  10 6 mm = 4.87  10 24 kg Find r by adding the distance between the spacecraft and Venus’s surface (r 1 ) to Venus’s radius (r 2 ). r = r 1 + r 2 r = 3.61  10 5 m + 6.05  10 6 m = 6.41  10 6 m Section 3 Motion in Space

32. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Sample Problem, continued 3. Calculate 4. Evaluate Magellan takes (5.66  10 3 s)(1 min/60 s)  94 min to complete one orbit. Section 3 Motion in Space

33. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Weight and Weightlessness To learn about apparent weightlessness, imagine that you are in an elevator: –When the elevator is at rest, the magnitude of the normal force acting on you equals your weight. –If the elevator were to accelerate downward at 9.81 m/s 2, you and the elevator would both be in free fall. You have the same weight, but there is no normal force acting on you. –This situation is called apparent weightlessness. –Astronauts in orbit experience apparent weightlessness. Section 3 Motion in Space

34. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Weight and Weightlessness Section 3 Motion in Space

35. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Torque and Simple Machines Chapter 7 Objectives Distinguish between torque and force. Calculate the magnitude of a torque on an object. Identify the six types of simple machines. Calculate the mechanical advantage of a simple machine.

36. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Torque and Simple Machines Chapter 7 Rotational Motion Rotational and translational motion can be analyzed separately. –For example, when a bowling ball strikes the pins, the pins may spin in the air as they fly backward. –These pins have both rotational and translational motion. In this section, we will isolate rotational motion. In particular, we will explore how to measure the ability of a force to rotate an object.

37. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Torque and Simple Machines Chapter 7 The Magnitude of a Torque Torque is a quantity that measures the ability of a force to rotate an object around some axis. How easily an object rotates on both how much force is applied and on where the force is applied. The perpendicular distance from the axis of rotation to a line drawn along the direction of the force is equal to d sin  and is called the lever arm.  = Fd sin  torque = force  lever arm

38. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Torque and Simple Machines Chapter 7 The Magnitude of a Torque, continued The applied force may act at an angle. However, the direction of the lever arm (d sin  ) is always perpendicular to the direction of the applied force, as shown here.

39. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Torque Section 4 Torque and Simple Machines

40. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Torque and the Lever Arm Section 4 Torque and Simple Machines In each example, the cat is pushing on the door at the same distance from the axis. To produce the same torque, the cat must apply greater force for smaller angles.

41. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Torque and Simple Machines Chapter 7 The Sign of a Torque Torque is a vector quantity. In this textbook, we will assign each torque a positive or negative sign, depending on the direction the force tends to rotate an object. We will use the convention that the sign of the torque is positive if the rotation is counterclockwise and negative if the rotation is clockwise. Tip: To determine the sign of a torque, imagine that the torque is the only one acting on the object and that the object is free to rotate. Visualize the direction that the object would rotate. If more than one force is acting, treat each force separately.

42. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 The Sign of a Torque Section 4 Torque and Simple Machines

43. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Sample Problem Torque A basketball is being pushed by two players during tip- off. One player exerts an upward force of 15 N at a perpendicular distance of 14 cm from the axis of rotation.The second player applies a downward force of 11 N at a distance of 7.0 cm from the axis of rotation. Find the net torque acting on the ball about its center of mass. Section 4 Torque and Simple Machines

44. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Sample Problem, continued Section 4 Torque and Simple Machines 1. Define Given: F 1 = 15 N F 2 = 11 N d 1 = 0.14 m d 2 = 0.070 m Diagram: Unknown:  net = ?

45. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Sample Problem, continued Section 4 Torque and Simple Machines 2. Plan Choose an equation or situation: Apply the definition of torque to each force,and add up the individual torques. Tip: The factor sin  is not included in the torque equation because each given distance is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force. In other words, each given distance is the lever arm.  = Fd  net =  1 +  2 = F 1 d 1 + F 2 d 2

46. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Sample Problem, continued Section 4 Torque and Simple Machines 4. Evaluate The net torque is negative,so the ball rotates in a clockwise direction. 3.Calculate Substitute the values into the equation and solve: First,determine the torque produced by each force.Use the standard convention for signs.  1 = F 1 d 1 = (15 N)(–0.14 m) = –2.1 Nm  2 = F 2 d 2 = (–11 N)(0.070 m) = –0.77 Nm  net =  1 +  2 = –2.1 Nm – 0.77 Nm  net = –2.9 Nm

47. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Torque and Simple Machines Chapter 7 Simple Machines A machine is any device that transmits or modifies force, usually by changing the force applied to an object. All machines are combinations or modifications of six fundamental types of machines, called simple machines. These six simple machines are the lever, pulley, inclined plane, wheel and axle, wedge, and screw.

48. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Simple Machines Section 4 Torque and Simple Machines

49. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Torque and Simple Machines Chapter 7 Simple Machines, continued Because the purpose of a simple machine is to change the direction or magnitude of an input force, a useful way of characterizing a simple machine is to compare the output and input force. This ratio is called mechanical advantage. If friction is disregarded, mechanical advantage can also be expressed in terms of input and output distance.

50. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Torque and Simple Machines Chapter 7 Simple Machines, continued The diagrams show two examples of a trunk being loaded onto a truck. In the first example, a force (F 1 ) of 360 N moves the trunk through a distance (d 1 ) of 1.0 m. This requires 360 Nm of work. In the second example, a lesser force (F 2 ) of only 120 N would be needed (ignoring friction), but the trunk must be pushed a greater distance (d 2 ) of 3.0 m. This also requires 360 Nm of work.

51. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Torque and Simple Machines Chapter 7 Simple Machines, continued The simple machines we have considered so far are ideal, frictionless machines. Real machines, however, are not frictionless. Some of the input energy is dissipated as sound or heat. The efficiency of a machine is the ratio of useful work output to work input. –The efficiency of an ideal (frictionless) machine is 1, or 100 percent. –The efficiency of real machines is always less than 1.

52. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Mechanical Efficiency Section 4 Torque and Simple Machines

53. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 1. An object moves in a circle at a constant speed. Which of the following is not true of the object? A. Its acceleration is constant. B. Its tangential speed is constant. C. Its velocity is constant. D. A centripetal force acts on the object. Standardized Test Prep Chapter 7

54. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice 1. An object moves in a circle at a constant speed. Which of the following is not true of the object? A. Its acceleration is constant. B. Its tangential speed is constant. C. Its velocity is constant. D. A centripetal force acts on the object. Standardized Test Prep Chapter 7

55. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the passage below to answer questions 2–3. A car traveling at 15 m/s on a flat surface turns in a circle with a radius of 25 m. 2. What is the centripetal acceleration of the car? F. 2.4  10 -2 m/s 2 G. 0.60 m/s 2 H. 9.0 m/s 2 J. zero Standardized Test Prep Chapter 7

56. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the passage below to answer questions 2–3. A car traveling at 15 m/s on a flat surface turns in a circle with a radius of 25 m. 2. What is the centripetal acceleration of the car? F. 2.4  10 -2 m/s 2 G. 0.60 m/s 2 H. 9.0 m/s 2 J. zero Standardized Test Prep Chapter 7

57. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the passage below to answer questions 2–3. A car traveling at 15 m/s on a flat surface turns in a circle with a radius of 25 m. 3. What is the most direct cause of the car’s centripetal acceleration? A. the torque on the steering wheel B. the torque on the tires of the car C. the force of friction between the tires and the road D. the normal force between the tires and the road Standardized Test Prep Chapter 7

58. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the passage below to answer questions 2–3. A car traveling at 15 m/s on a flat surface turns in a circle with a radius of 25 m. 3. What is the most direct cause of the car’s centripetal acceleration? A. the torque on the steering wheel B. the torque on the tires of the car C. the force of friction between the tires and the road D. the normal force between the tires and the road Standardized Test Prep Chapter 7

59. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 4. Earth (m = 5.97  10 24 kg) orbits the sun (m = 1.99  10 30 kg) at a mean distance of 1.50  10 11 m. What is the gravitational force of the sun on Earth? (G = 6.673  10 -11 Nm 2 /kg 2 ) F. 5.29  10 32 N G. 3.52  10 22 N H. 5.90  10 –2 N J. 1.77  10 –8 N Standardized Test Prep Chapter 7

60. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 4. Earth (m = 5.97  10 24 kg) orbits the sun (m = 1.99  10 30 kg) at a mean distance of 1.50  10 11 m. What is the gravitational force of the sun on Earth? (G = 6.673  10 -11 Nm 2 /kg 2 ) F. 5.29  10 32 N G. 3.52  10 22 N H. 5.90  10 –2 N J. 1.77  10 –8 N Standardized Test Prep Chapter 7

61. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 5. Which of the following is a correct interpretation of the expression ? A. Gravitational field strength changes with an object’s distance from Earth. B. Free-fall acceleration changes with an object’s distance from Earth. C. Free-fall acceleration is independent of the falling object’s mass. D. All of the above are correct interpretations. Standardized Test Prep Chapter 7

62. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 5. Which of the following is a correct interpretation of the expression ? A. Gravitational field strength changes with an object’s distance from Earth. B. Free-fall acceleration changes with an object’s distance from Earth. C. Free-fall acceleration is independent of the falling object’s mass. D. All of the above are correct interpretations. Standardized Test Prep Chapter 7

63. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 6. What data do you need to calculate the orbital speed of a satellite? F. mass of satellite, mass of planet, radius of orbit G. mass of satellite, radius of planet, area of orbit H. mass of satellite and radius of orbit only J. mass of planet and radius of orbit only Standardized Test Prep Chapter 7

64. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 6. What data do you need to calculate the orbital speed of a satellite? F. mass of satellite, mass of planet, radius of orbit G. mass of satellite, radius of planet, area of orbit H. mass of satellite and radius of orbit only J. mass of planet and radius of orbit only Standardized Test Prep Chapter 7

65. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 7. Which of the following choices correctly describes the orbital relationship between Earth and the sun? A. The sun orbits Earth in a perfect circle. B. Earth orbits the sun in a perfect circle. C. The sun orbits Earth in an ellipse, with Earth at one focus. D. Earth orbits the sun in an ellipse, with the sun at one focus. Standardized Test Prep Chapter 7

66. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 7. Which of the following choices correctly describes the orbital relationship between Earth and the sun? A. The sun orbits Earth in a perfect circle. B. Earth orbits the sun in a perfect circle. C. The sun orbits Earth in an ellipse, with Earth at one focus. D. Earth orbits the sun in an ellipse, with the sun at one focus. Standardized Test Prep Chapter 7

67. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the diagram to answer questions 8–9. Standardized Test Prep Chapter 7 8. The three forces acting on the wheel have equal magnitudes. Which force will produce the greatest torque on the wheel? F. F 1 G. F 2 H. F 3 J. Each force will produce the same torque.

68. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the diagram to answer questions 8–9. Standardized Test Prep Chapter 7 8. The three forces acting on the wheel have equal magnitudes. Which force will produce the greatest torque on the wheel? F. F 1 G. F 2 H. F 3 J. Each force will produce the same torque.

69. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the diagram to answer questions 8–9. Standardized Test Prep Chapter 7 9. If each force is 6.0 N, the angle between F 1 and F 2 is 60.0°, and the radius of the wheel is 1.0 m, what is the resultant torque on the wheel? A. –18 NmC. 9.0 Nm B. –9.0 NmD. 18 Nm

70. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued Use the diagram to answer questions 8–9. Standardized Test Prep Chapter 7 9. If each force is 6.0 N, the angle between F 1 and F 2 is 60.0°, and the radius of the wheel is 1.0 m, what is the resultant torque on the wheel? A. –18 NmC. 9.0 Nm B. –9.0 NmD. 18 Nm

71. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 10. A force of 75 N is applied to a lever. This force lifts a load weighing 225 N. What is the mechanical advantage of the lever? F. 1/3 G. 3 H. 150 J. 300 Standardized Test Prep Chapter 7

72. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 10. A force of 75 N is applied to a lever. This force lifts a load weighing 225 N. What is the mechanical advantage of the lever? F. 1/3 G. 3 H. 150 J. 300 Standardized Test Prep Chapter 7

73. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 11. A pulley system has an efficiency of 87.5 percent. How much work must you do to lift a desk weighing 1320 N to a height of 1.50 m? A. 1510 J B. 1730 J C. 1980 J D. 2260 J Standardized Test Prep Chapter 7

74. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 11. A pulley system has an efficiency of 87.5 percent. How much work must you do to lift a desk weighing 1320 N to a height of 1.50 m? A. 1510 J B. 1730 J C. 1980 J D. 2260 J Standardized Test Prep Chapter 7

75. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 12. Which of the following statements is correct? F. Mass and weight both vary with location. G. Mass varies with location, but weight does not. H. Weight varies with location, but mass does not. J. Neither mass nor weight varies with location. Standardized Test Prep Chapter 7

76. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 12. Which of the following statements is correct? F. Mass and weight both vary with location. G. Mass varies with location, but weight does not. H. Weight varies with location, but mass does not. J. Neither mass nor weight varies with location. Standardized Test Prep Chapter 7

77. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 13. Which astronomer discovered that planets travel in elliptical rather than circular orbits? A. Johannes Kepler B. Nicolaus Copernicus C. Tycho Brahe D. Claudius Ptolemy Standardized Test Prep Chapter 7

78. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Multiple Choice, continued 13. Which astronomer discovered that planets travel in elliptical rather than circular orbits? A. Johannes Kepler B. Nicolaus Copernicus C. Tycho Brahe D. Claudius Ptolemy Standardized Test Prep Chapter 7

79. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response 14. Explain how it is possible for all the water to remain in a pail that is whirled in a vertical path, as shown below. Standardized Test Prep Chapter 7

80. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response 14. Explain how it is possible for all the water to remain in a pail that is whirled in a vertical path, as shown below. Standardized Test Prep Chapter 7 Answer: The water remains in the pail even when the pail is upside down because the water tends to move in a straight path due to inertia.

81. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response, continued 15. Explain why approximately two high tides take place every day at a given location on Earth. Standardized Test Prep Chapter 7

82. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response, continued 15. Explain why approximately two high tides take place every day at a given location on Earth. Answer: The moon’s tidal forces create two bulges on Earth. As Earth rotates on its axis once per day, any given point on Earth passes through both bulges. Standardized Test Prep Chapter 7

83. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response, continued 16. If you used a machine to increase the output force, what factor would have to be sacrificed? Give an example. Standardized Test Prep Chapter 7

84. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Short Response, continued 16. If you used a machine to increase the output force, what factor would have to be sacrificed? Give an example. Answer: You would have to apply the input force over a greater distance. Examples may include any machines that increase output force at the expense of input distance. Standardized Test Prep Chapter 7

85. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Extended Response Standardized Test Prep Chapter 7 17. Mars orbits the sun (m = 1.99  10 30 kg) at a mean distance of 2.28  10 11 m. Calculate the length of the Martian year in Earth days. Show all of your work. (G = 6.673  10 –11 Nm 2 /kg 2 )

86. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Extended Response 17. Mars orbits the sun (m = 1.99  10 30 kg) at a mean distance of 2.28  10 11 m. Calculate the length of the Martian year in Earth days. Show all of your work. (G = 6.673  10 –11 Nm 2 /kg 2 ) Standardized Test Prep Chapter 7 Answer: 687 days

87. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Centripetal Acceleration Section 1 Circular Motion

88. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Centripetal Force Section 1 Circular Motion

89. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 7 Kepler’s Laws Section 3 Motion in Space

90. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Torque and Simple Machines Chapter 7 The Magnitude of a Torque

91. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 4 Torque and Simple Machines Chapter 7 Simple Machines