Unit 4 Multiple Choice Test
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Is it possible for an object moving with a constant speed to accelerate? Explain. A) No, if the speed is constant then the acceleration is equal to zero. B) No, an object can accelerate only if there is a net force acting on it. C) Yes, although the speed is constant, the direction of the velocity can be changing. D) Yes, if an object is moving it can experience acceleration
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About This Quiz
Unit 4 Multiple Choice Test (Chapters 5 and 8)
~Please enter the letter of the best choice in the space available.
(110).jpg "Unit 4 Multiple Choice Test - Quiz")
Quiz Preview
- 2.
Consider a particle moving with constant speed such that its acceleration of constant magnitude is always perpendicular to its velocity. A) It is moving in a straight line. B) It is moving in a circle. C) It is moving in a parabola. D) None of the above is definitely true all of the time.
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Correct Answer
A. BExplanation
The given scenario describes a particle with constant speed and acceleration always perpendicular to its velocity. This is the characteristic behavior of an object moving in a circle. In a circular motion, the velocity vector is always tangent to the circle, and the acceleration vector points towards the center of the circle. Therefore, the correct answer is B) It is moving in a circle.Rate this question:
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- 3.
When an object experiences uniform circular motion, the direction of the acceleration is A) in the same direction as the velocity vector. B) in the opposite direction of the velocity vector. C) is directed toward the center of the circular path. D) is directed away from the center of the circular path.
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Correct Answer
A. CExplanation
When an object experiences uniform circular motion, the direction of the acceleration is directed toward the center of the circular path. This is because in uniform circular motion, the object is constantly changing its direction, which means it is constantly accelerating towards the center of the circle. This centripetal acceleration is necessary to keep the object moving in a circular path.Rate this question:
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- 4.
Consider a particle moving with constant speed such that its acceleration of constant magnitude is always perpendicular to its velocity. A) It is moving in a straight line. B) It is moving in a circle. C) It is moving in a parabola. D) None of the above is definitely true all of the time.
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Correct Answer
A. BExplanation
The given scenario describes a particle with constant speed and an acceleration that is always perpendicular to its velocity. This type of motion is characteristic of circular motion, where the acceleration acts as a centripetal force that continuously changes the direction of the velocity vector. Therefore, the correct answer is B) It is moving in a circle.Rate this question:
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- 5.
What type of acceleration does an object moving with constant speed in a circular path experience? A) free fall B) constant acceleration C) linear acceleration D) centripetal acceleration
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Correct Answer
A. DExplanation
An object moving with constant speed in a circular path experiences centripetal acceleration. This acceleration is directed towards the center of the circle and is responsible for keeping the object in its circular path. It is always perpendicular to the object's velocity and is directly proportional to the square of its speed and inversely proportional to the radius of the circular path.Rate this question:
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- 6.
What force is needed to make an object move in a circle? A) kinetic friction B) static friction C) centripetal force D) weight
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Correct Answer
A. CExplanation
The correct answer is C) centripetal force. When an object moves in a circle, it experiences a force directed towards the center of the circle, which is known as the centripetal force. This force is necessary to keep the object moving in a circular path and is responsible for continuously changing the object's direction. Kinetic friction and static friction are types of forces that oppose motion and are not specifically related to circular motion. Weight is the force exerted on an object due to gravity and is not directly related to circular motion.Rate this question:
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- 7.
When an object experiences uniform circular motion, the direction of the net force is A) in the same direction as the motion of the object. B) in the opposite direction of the motion of the object. C) is directed toward the center of the circular path. D) is directed away from the center of the circular path.
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Correct Answer
A. CExplanation
When an object experiences uniform circular motion, it is constantly changing direction, which means it is accelerating. According to Newton's second law of motion, there must be a net force acting on the object in the direction of its acceleration. In uniform circular motion, this net force is directed toward the center of the circular path, which is why the correct answer is C.Rate this question:
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- 8.
A roller coaster car is on a track that forms a circular loop in the vertical plane. If the car is to just maintain contact with track at the top of the loop, what is the minimum value for its centripetal acceleration at this point? A) g downward B) 0.5g downward C) g upward D) 2g upward
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Correct Answer
A. AExplanation
At the top of the loop, the car is in a state of circular motion. In order to maintain contact with the track, the car must experience a centripetal force directed towards the center of the loop. This force is provided by the gravitational force acting on the car. Since the car is at the top of the loop, the centripetal force must be equal to the weight of the car. Therefore, the minimum value for the centripetal acceleration is equal to the acceleration due to gravity, which is directed downward. Therefore, the correct answer is A) g downward.Rate this question:
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- 9.
A roller coaster car (mass = M) is on a track that forms a circular loop (radius = r) in the vertical plane. If the car is to just maintain contact with the track at the top of the loop, what is the minimum value for its speed at that point? A) rg B) (rg)1/2 C) (2rg)1/2 D) (0.5rg)1/2
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Correct Answer
A. BExplanation
At the top of the loop, the car is in a state of circular motion. In order for the car to maintain contact with the track, the centripetal force acting on the car must be equal to the gravitational force. The centripetal force is given by mv^2/r, where m is the mass of the car and v is its speed. The gravitational force is given by mg, where g is the acceleration due to gravity. Equating the two forces, we have mv^2/r = mg. Solving for v, we get v = (rg)^(1/2). Therefore, the minimum value for the car's speed at the top of the loop is (rg)^(1/2), which is option B.Rate this question:
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- 10.
A pilot executes a vertical dive then follows a semi-circular arc until it is going straight up. Just as the plane is at its lowest point, the force on him is A) less than mg, and pointing up. B) less than mg, and pointing down. C) more than mg, and pointing up. D) more than mg, and pointing down.
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Correct Answer
A. CExplanation
As the pilot executes a vertical dive and follows a semi-circular arc, the plane experiences a change in direction and velocity. At the lowest point of the arc, the plane is still moving upwards, which means there must be a net force acting in the upward direction to counteract the downward force of gravity (mg). Therefore, the force on the pilot is more than mg and pointing up.Rate this question:
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- 11.
A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What is the minimum coefficient of static friction between the turntable and the coin if the coin is not to slip? A) (4Ï€2f2r)/g B) (4Ï€2fr2)/g C) (4Ï€f2r)/g D) (4Ï€fr2)/g
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Correct Answer
A. AExplanation
The minimum coefficient of static friction between the turntable and the coin can be determined by considering the forces acting on the coin. The gravitational force acting on the coin is mg, where m is the mass of the coin and g is the acceleration due to gravity. The centripetal force acting on the coin is given by (m)(4Ï€^2)(f^2)(r), where f is the frequency of rotation and r is the distance from the axis of rotation. In order for the coin not to slip, the static friction force between the coin and the turntable must be equal to or greater than the centripetal force. Therefore, the minimum coefficient of static friction is given by (4Ï€^2)(f^2)(r)/g, which corresponds to option A.Rate this question:
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- 12.
A car goes around a curve of radius r at a constant speed v. What is the direction of the net force on the car? A) toward the curve's center B) away from the curve's center C) toward the front of the car D) toward the back of the car
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Correct Answer
A. AExplanation
The direction of the net force on the car is toward the curve's center. This is because when a car goes around a curve, there is a centripetal force acting towards the center of the curve. This force is necessary to keep the car moving in a curved path and prevent it from moving in a straight line. Therefore, the net force on the car is directed towards the center of the curve.Rate this question:
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- 13.
A car goes around a curve of radius r at a constant speed v. Then it goes around the same curve at half of the original speed. What is the centripetal force on the car as it goes around the curve for the second time, compared to the first time? A) twice as big B) four times as big C) half as big D) one-fourth as big
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Correct Answer
A. DExplanation
When a car goes around a curve at a constant speed, the centripetal force acting on it is given by the equation F = mv^2 / r, where m is the mass of the car, v is the velocity, and r is the radius of the curve.
If the car goes around the same curve at half of the original speed, the velocity v is halved. Plugging this new velocity into the equation, we get F' = m(v/2)^2 / r = mv^2 / (4r).
Comparing F' to the original force F, we can see that F' is one-fourth (1/4) as big as F. Therefore, the centripetal force on the car as it goes around the curve for the second time is one-fourth as big as the first time, which corresponds to option D.Rate this question:
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- 14.
A car goes around a curve of radius r at a constant speed v. Then it goes around a curve of radius 2r at speed 2v. What is the centripetal force on the car as it goes around the second curve, compared to the first? A) four times as big B) twice as big C) one-half as big D) one-fourth as big
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Correct Answer
A. BExplanation
The centripetal force on an object moving in a circular path is directly proportional to the square of its speed and inversely proportional to the radius of the curve. In this case, the car's speed is doubled and the radius is doubled. Therefore, the centripetal force on the car as it goes around the second curve is four times bigger than the first curve.Rate this question:
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- 15.
The gravitational force between two objects is proportional to A) the distance between the two objects. B) the square of the distance between the two objects. C) the product of the two objects. D) the square of the product of the two objects.
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Correct Answer
A. CExplanation
The gravitational force between two objects is proportional to the product of the two objects. This means that as the mass of one object increases or the mass of the other object increases, the gravitational force between them will also increase. The distance between the two objects does not directly affect the gravitational force, as stated in option A. The square of the distance between the two objects is also not directly proportional to the gravitational force, as stated in option B. The square of the product of the two objects, as stated in option D, is not a correct representation of the gravitational force. Therefore, the correct answer is option C.Rate this question:
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- 16.
The gravitational force between two objects is inversely proportional to A) the distance between the two objects. B) the square of the distance between the two objects. C) the product of the two objects. D) the square of the product of the two objects.
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Correct Answer
A. BExplanation
The correct answer is B. The gravitational force between two objects is inversely proportional to the square of the distance between the two objects. This means that as the distance between the objects increases, the gravitational force decreases. Conversely, as the distance decreases, the gravitational force increases. This relationship is described by Newton's law of universal gravitation, which states that the gravitational force is directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance between them.Rate this question:
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- 17.
Two objects attract each other gravitationally. If the distance between their centers is cut in half, the gravitational force A) is cut to one fourth. B) is cut in half. C) doubles. D) quadruples
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Correct Answer
A. DExplanation
When the distance between the centers of two objects is cut in half, the gravitational force between them quadruples. This is because the gravitational force is inversely proportional to the square of the distance between the objects. So, when the distance is halved, the square of that distance becomes one-fourth, resulting in the gravitational force increasing by a factor of four.Rate this question:
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- 18.
Two objects, with masses m1 and m2, are originally a distance r apart. The gravitational force between them has magnitude F. The second object has its mass changed to 2m2, and the distance is changed to r/4. What is the magnitude of the new gravitational force? A) F/32 B) F/16 C) 16F D) 32F
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Correct Answer
A. DExplanation
When the mass of the second object is changed to 2m2, and the distance is changed to r/4, the gravitational force between the two objects can be calculated using the formula for gravitational force: F = G(m1*m2)/r^2.
Substituting the new values into the formula, we get F' = G(m1*(2m2))/(r/4)^2 = 16G(m1*m2)/r^2 = 16F.
Therefore, the magnitude of the new gravitational force is 16F, which corresponds to option C.Rate this question:
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- 19.
Two objects, with masses m1 and m2, are originally a distance r apart. The magnitude of the gravitational force between them is F. The masses are changed to 2m1 and 2m2, and the distance is changed to 4r. What is the magnitude of the new gravitational force? A) F/16 B) F/4 C) 16F D) 4F
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Correct Answer
A. BExplanation
When the masses are changed to 2m1 and 2m2, and the distance is changed to 4r, the magnitude of the new gravitational force can be determined using the formula for gravitational force: F = G(m1 * m2) / r^2.
Substituting the new masses and distance into the formula, we get:
New force = G((2m1) * (2m2)) / (4r)^2
= G(4m1m2) / 16r^2
= (1/4) * (G(m1 * m2) / r^2)
= (1/4) * F
Therefore, the magnitude of the new gravitational force is F/4, which corresponds to option B.Rate this question:
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- 20.
Compared to its mass on the Earth, the mass of an object on the Moon is A) less. B) more. C) the same. D) half as much.
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Correct Answer
A. CExplanation
The mass of an object remains the same regardless of its location. Therefore, the mass of an object on the Moon is the same as its mass on Earth.Rate this question:
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- 21.
The acceleration of gravity on the Moon is one-sixth what it is on Earth. An object of mass 72 kg is taken to the Moon. What is its mass there? A) 12 kg B) 72 kg C) 72 N D) 12 N
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Correct Answer
A. BExplanation
The mass of an object remains the same regardless of the location, so the mass of the object on the Moon would still be 72 kg. The acceleration of gravity may be different on the Moon, but it does not affect the mass of the object.Rate this question:
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- 22.
As a rocket moves away from the Earth's surface, the rocket's weight A) increases. B) decreases. C) remains the same. D) depends on how fast it is moving.
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Correct Answer
A. BExplanation
As a rocket moves away from the Earth's surface, the force of gravity acting on it decreases. Weight is the force of gravity acting on an object, so as the force of gravity decreases, the rocket's weight also decreases. Therefore, the correct answer is B) decreases.Rate this question:
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- 23.
A spaceship is traveling to the Moon. At what point is it beyond the pull of Earth's gravity? A) when it gets above the atmosphere B) when it is half-way there C) when it is closer to the Moon than it is to Earth D) It is never beyond the pull of Earth's gravity.
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Correct Answer
A. D -
- 24.
Suppose a satellite were orbiting the Earth just above the surface. What is its centripetal acceleration? A) smaller than g B) equal to g C) larger than g D) Impossible to say without knowing the mass.
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Correct Answer
A. BExplanation
The centripetal acceleration of a satellite orbiting the Earth just above the surface would be equal to g. This is because the centripetal acceleration is given by the equation a = v^2/r, where v is the velocity of the satellite and r is the radius of the orbit. In this case, the satellite is in a circular orbit just above the surface of the Earth, so the radius of the orbit is equal to the radius of the Earth. The velocity of the satellite is such that it maintains a stable orbit, which means it is moving at a constant speed. Therefore, the centripetal acceleration is equal to g, the acceleration due to gravity.Rate this question:
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- 25.
A hypothetical planet has a mass of half that of the Earth and a radius of twice that of the Earth. What is the acceleration due to gravity on the planet in terms of g, the acceleration due to gravity at the Earth? A) g B) g/2 C) g/4 D) g/8
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Correct Answer
A. DExplanation
The acceleration due to gravity on a planet is determined by its mass and radius. In this hypothetical planet, the mass is half that of the Earth and the radius is twice that of the Earth. According to the formula for gravitational acceleration (g = G * M / R^2), where G is the gravitational constant, M is the mass, and R is the radius, we can see that the acceleration due to gravity is inversely proportional to the radius squared. Since the radius of this planet is twice that of the Earth, the acceleration due to gravity would be (1/2)^2 = 1/4 or g/8. Therefore, the correct answer is D) g/8.Rate this question:
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- 26.
The acceleration of gravity on the Moon is one-sixth what it is on Earth. The radius of the Moon is one-fourth that of the Earth. What is the Moon's mass compared to the Earth's? A) 1/6 B) 1/16 C) 1/24 D) 1/96
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Correct Answer
A. DExplanation
The acceleration of gravity on the Moon is one-sixth of that on Earth, which means that the Moon's gravitational force is weaker than Earth's. The radius of the Moon is one-fourth that of the Earth, which means that the Moon is smaller in size compared to Earth. The mass of an object is directly proportional to its gravitational force and inversely proportional to the radius squared. Therefore, since the Moon has a smaller radius and weaker gravitational force compared to Earth, its mass must be significantly smaller. Therefore, the Moon's mass compared to Earth's is 1/96.Rate this question:
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- 27.
Two planets have the same surface gravity, but planet B has twice the radius of planet A. If planet A has mass m, what is the mass of planet B? A) 0.707m B) m C) 1.41m D) 4m
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Correct Answer
A. DExplanation
Planet B has twice the radius of planet A, but both planets have the same surface gravity. This means that planet B must have a greater mass than planet A in order to generate the same surface gravity with a larger radius. Therefore, the mass of planet B is 4 times the mass of planet A, which corresponds to answer choice D.Rate this question:
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- 28.
Two planets have the same surface gravity, but planet B has twice the mass of planet A. If planet A has radius r, what is the radius of planet B? A) 0.707r B) r C) 1.41r D) 4r
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Correct Answer
A. CExplanation
Planet B has twice the mass of planet A, but both planets have the same surface gravity. This means that planet B must have a larger radius than planet A in order to maintain the same surface gravity. The correct answer, C) 1.41r, represents a radius that is 1.41 times larger than the radius of planet A.Rate this question:
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- 29.
Consider a small satellite moving in a circular orbit (radius r) about a spherical planet (mass M). Which expression gives this satellite's orbital velocity? A) v = GM/r B) (GM/r)1/2 C) GM/r2 D) (GM/r2)1/2
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Correct Answer
A. BExplanation
The correct answer is B. The expression (GM/r)1/2 gives the satellite's orbital velocity. This is because the gravitational force between the satellite and the planet is given by F = GMm/r^2, where G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, and r is the distance between them. The centripetal force required to keep the satellite in its circular orbit is provided by the gravitational force. Equating the centripetal force to the gravitational force, we get mv^2/r = GMm/r^2. Simplifying this equation gives v = (GM/r)1/2, which is the orbital velocity of the satellite.Rate this question:
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- 30.
Satellite A has twice the mass of satellite B, and rotates in the same orbit. Compare the two satellite's speeds. A) The speed of B is twice the speed of A. B) The speed of B is half the speed of A. C) The speed of B is one-fourth the speed of A. D) The speed of B is equal to the speed of A.
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Correct Answer
A. DExplanation
The speed of a satellite in orbit is determined by the mass of the planet it is orbiting and the radius of its orbit, but not by its own mass. Since both satellites are in the same orbit, they will have the same speed regardless of their masses. Therefore, the speed of B is equal to the speed of A.Rate this question:
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- 31.
A person is standing on a scale in an elevator accelerating downward. Compare the reading on the scale to the person's true weight. A) greater than their true weight B) equal to their true weight C) less than their true weight D) zero
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Correct Answer
A. CExplanation
When the elevator is accelerating downward, the person experiences a greater force due to gravity than when they are at rest. This is because the person and the scale are accelerating downward together. According to Newton's second law, the force experienced by an object is equal to its mass multiplied by its acceleration. Since the person's mass remains constant, the force experienced by the person is greater, resulting in a reading on the scale that is less than their true weight. Therefore, the correct answer is C) less than their true weight.Rate this question:
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- 32.
Who was the first person to realize that the planets move in elliptical paths around the Sun? A) Kepler B) Brahe C) Einstein D) Copernicus
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A. AExplanation
Johannes Kepler was the first person to realize that the planets move in elliptical paths around the Sun. He developed three laws of planetary motion, known as Kepler's Laws, which revolutionized our understanding of the solar system. Kepler's Laws replaced the previous belief in circular orbits and provided a more accurate description of the planetary motion. Kepler's work laid the foundation for Isaac Newton's theory of gravity and had a significant impact on the development of modern astronomy.Rate this question:
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- 33.
The speed of Halley's Comet, while traveling in its elliptical orbit around the Sun, A) is constant. B) increases as it nears the Sun. C) decreases as it nears the Sun. D) is zero at two points in the orbit.
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Correct Answer
A. BExplanation
As Halley's Comet travels in its elliptical orbit around the Sun, its speed increases as it nears the Sun. This is because of the gravitational pull exerted by the Sun, which accelerates the comet as it gets closer. As the comet moves away from the Sun, the gravitational pull decreases, causing its speed to decrease as well. Therefore, the correct answer is B) increases as it nears the Sun.Rate this question:
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- 34.
The average distance from the Earth to the Sun is defined as one "astronomical unit" (AU). An asteroid orbits the Sun in one-third of a year. What is the asteroid's average distance from the Sun? A) 0.19 AU B) 0.48 AU C) 2.1 AU D) 5.2 AU
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Correct Answer
A. BExplanation
The average distance from the Earth to the Sun is defined as one "astronomical unit" (AU). Since the asteroid orbits the Sun in one-third of a year, its average distance from the Sun would be less than one AU. Among the given options, the closest distance is 0.48 AU, so that would be the correct answer.Rate this question:
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Mar 21, 2023Quiz Edited by
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Feb 17, 2009Quiz Created by
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