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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.
Correct Answer C. Yes, although the speed is constant, the direction of the velocity can be changing.
Explanation Although the speed of an object may be constant, it is still possible for the object to accelerate if the direction of its velocity is changing. Acceleration is defined as any change in velocity, which includes changes in direction. Therefore, even if the speed remains constant, if the object is changing its direction of motion, it is experiencing acceleration.
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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.
Correct Answer B. It is moving in a circle.
Explanation If the acceleration of the particle is always perpendicular to its velocity, it means that the particle is constantly changing its direction but maintaining the same speed. This type of motion is characteristic of circular motion, where the acceleration is always directed towards the center of the circle. Therefore, the particle is moving in a circle.
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3.
An object moves in a circular path at a constant speed. Compare the direction of the object's velocity and acceleration vectors.
A.
Both vectors point in the same direction.
B.
The vectors point in opposite directions.
C.
The vectors are perpendicular.
D.
The question is meaningless, since the acceleration is zero.
Correct Answer C. The vectors are perpendicular.
Explanation In circular motion, the velocity vector is always tangent to the circular path, while the acceleration vector is directed towards the center of the circle. Since the velocity vector is tangent and the acceleration vector is directed towards the center, they are at right angles to each other, making them perpendicular.
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4.
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.
Correct Answer C. Is directed toward the center of the circular path.
Explanation When an object experiences uniform circular motion, the direction of the acceleration is directed toward the center of the circular path. This is because the object is constantly changing its direction as it moves in a circle, which means it is constantly accelerating towards the center of the circle. The acceleration is necessary to keep the object moving in a curved path, as a force is required to constantly change the direction of the object's velocity.
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5.
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.
Correct Answer B. It is moving in a circle.
Explanation If the particle's acceleration is always perpendicular to its velocity, it means that the velocity vector and acceleration vector are always at right angles to each other. In circular motion, the acceleration vector is always directed towards the center of the circle, while the velocity vector is tangent to the circle. Since the acceleration vector is always perpendicular to the velocity vector in this scenario, it suggests that the particle is moving in a circle.
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6.
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
Correct Answer D. Centripetal acceleration
Explanation An object moving with constant speed in a circular path experiences centripetal acceleration. This type of acceleration is directed towards the center of the circle and is necessary to keep the object moving in a curved path. It is caused by the inward force, known as centripetal force, acting on the object. This acceleration does not change the speed of the object, but it does change its direction, allowing it to continuously move in a circular path.
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7.
What force is needed to make an object move in a circle?
A.
Kinetic friction
B.
Static friction
C.
Centripetal force
D.
Weight
Correct Answer C. Centripetal force
Explanation Centripetal force is the force required to make an object move in a circle. It is directed towards the center of the circle and keeps the object constantly changing direction. Without this force, the object would move in a straight line instead of a curved path. Kinetic friction and static friction are forces that oppose motion and are not specifically related to circular motion. Weight is the force exerted by gravity on an object and is not directly involved in making an object move in a circle.
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8.
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.
Correct Answer C. Is directed toward the center of the circular path.
Explanation When an object experiences uniform circular motion, it is constantly changing its direction. In order to change its direction, there must be a force acting on the object. This force is directed towards the center of the circular path, as it is responsible for keeping the object moving in a circular path. This force is called the centripetal force. Therefore, the net force in uniform circular motion is directed toward the center of the circular path.
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9.
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
Correct Answer A. G downward
Explanation At the top of the loop, the car must experience a centripetal acceleration towards the center of the loop in order to maintain contact with the track. Since the car is at the top of the loop, the acceleration must be directed downwards, opposite to the direction of gravity. Therefore, the minimum value for the centripetal acceleration at this point is equal to the acceleration due to gravity (g) but in the opposite direction, hence "g downward".
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10.
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
Correct Answer B. (rg)1/2
Explanation The minimum value for the speed of the roller coaster car at the top of the loop can be determined using the concept of centripetal force. At the top of the loop, the gravitational force acting on the car provides the centripetal force required to maintain contact with the track. The centripetal force is given by the equation Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the car, v is the speed, and r is the radius of the loop. Setting the gravitational force equal to the centripetal force, we have mg = mv^2/r. Simplifying, we find v^2 = rg, and taking the square root of both sides gives v = (rg)^(1/2). Therefore, the minimum value for the speed of the car at the top of the loop is (rg)^(1/2).
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11.
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.
Correct Answer C. More than mg, and pointing up.
Explanation When the pilot executes a vertical dive, the force on the plane is the sum of the gravitational force (mg) and the force exerted by the plane to counteract gravity. As the plane follows a semi-circular arc, the force on the pilot increases due to the centripetal force required to keep the plane in the curved path. At the lowest point of the arc, the force on the pilot is the sum of the gravitational force and the centripetal force, which is greater than mg. Since the force is directed upwards to counteract the downward gravitational force, the correct answer is more than mg, and pointing up.
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12.
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.
(4π^2f^2r)/g
B.
(4π^2fr^2)/g
C.
(4πf^2r)/g
D.
(4πfr^2)/g
Correct Answer A. (4π^2f^2r)/g
Explanation The correct answer is (4π^2f^2r)/g. This formula represents the minimum coefficient of static friction between the turntable and the coin to prevent slipping. It is derived from the equation for centripetal force, F = mω^2r, where ω is the angular velocity. By equating this force to the maximum static friction force, F_max = μ_smg, and substituting ω = 2πf, the formula (4π^2f^2r)/g is obtained. This equation shows that the coefficient of static friction is directly proportional to the square of the frequency and the radius, and inversely proportional to the acceleration due to gravity.
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13.
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
Correct Answer A. Toward the curve's center
Explanation When a car goes around a curve, it experiences a centripetal force that acts 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 direction of the net force on the car is towards the curve's center.
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14.
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
Correct Answer D. One-fourth as big
Explanation When a car goes around a curve at a constant speed, the centripetal force required to keep it moving in a circular path is given by the equation Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the car, v is the velocity, and r is the radius of the curve.
When the car goes around the same curve at half of the original speed, the velocity is reduced to v/2. Plugging this value into the equation, we get Fc' = m(v/2)^2/r = m(v^2/4)/r = (mv^2/4r).
Comparing Fc' to Fc, we can see that Fc' is one-fourth as big as Fc. 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.
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15.
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
Correct Answer B. Twice as big
Explanation As the car goes around a curve of radius 2r at speed 2v, the centripetal force is given by the equation F = (mv^2)/(2r). Since the radius is twice as big and the speed is also twice as big compared to the first curve, the centripetal force will be (m(2v)^2)/(2(2r)) = (4mv^2)/(4r) = (mv^2)/r. Therefore, the centripetal force on the car as it goes around the second curve is twice as big compared to the first curve.
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16.
A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still negotiate the curve if the horizontal component of the normal force on the car from the road is equal in magnitude to
A.
Mg/2.
B.
Mg.
C.
Mv^2/r.
D.
Tan[v^2/(rg)].
Correct Answer C. Mv^2/r.
Explanation When a car goes around a banked curve, the normal force from the road can be divided into two components: vertical and horizontal. The vertical component of the normal force counteracts the gravitational force acting on the car, which is equal to mg. The horizontal component of the normal force provides the centripetal force required to keep the car moving in a circular path. According to Newton's second law, the centripetal force is given by the equation F = mv^2/r, where m is the mass of the car, v is the speed, and r is the radius of the curve. Therefore, the horizontal component of the normal force must be equal in magnitude to mv^2/r for the car to negotiate the curve successfully.
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17.
Two horizontal curves on a bobsled run are banked at the same angle, but one has twice the radius of the other. The safe speed (no friction needed to stay on the run) for the smaller radius curve is v. What is the safe speed on the larger radius curve?
A.
Approximately 0.707v
B.
2v
C.
Approximately 1.41v
D.
0.5v
Correct Answer C. Approximately 1.41v
Explanation The safe speed on the larger radius curve is approximately 1.41v. This can be explained using the concept of centripetal force. The centripetal force required to keep an object moving in a curved path is directly proportional to the square of its velocity and inversely proportional to the radius of the curve. Since the larger radius curve has twice the radius of the smaller radius curve, it requires half the centripetal force to stay on the curve. Therefore, the velocity can be increased by a factor of approximately 1.41 (square root of 2) while still maintaining the same centripetal force, resulting in a safe speed of approximately 1.41v.
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18.
The banking angle in a turn on the Olympic bobsled track is not constant, but increases upward from the horizontal. Coming around a turn, the bobsled team will intentionally "climb the wall," then go lower coming out of the turn. Why do they do this?
A.
To give the team better control, because they are able to see ahead of the turn
B.
To prevent the bobsled from turning over
C.
To take the turn at a faster speed
D.
To reduce the g-force on them
Correct Answer C. To take the turn at a faster speed
Explanation The bobsled team intentionally climbs the wall and goes lower coming out of the turn in order to take the turn at a faster speed. This technique allows them to maintain better control and reduce the g-force on them, ultimately enabling them to navigate the turn more efficiently and increase their speed. By increasing the banking angle upward from the horizontal, the bobsled team can leverage the forces of gravity and centripetal force to their advantage, allowing them to maintain higher speeds throughout the turn.
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19.
Is it possible for an object moving around a circular path to have both centripetal and tangential acceleration?
A.
No, because then the path would not be a circle.
B.
No, an object can only have one or the other at any given time.
C.
Yes, this is possible if the speed is constant.
D.
Yes, this is possible if the speed is changing.
Correct Answer D. Yes, this is possible if the speed is changing.
Explanation Yes, this is possible if the speed is changing. When an object moves in a circular path, it experiences centripetal acceleration towards the center of the circle. However, if the speed of the object is changing, it also experiences tangential acceleration in the direction of the velocity vector. This is because the object's velocity is changing, and acceleration is defined as the rate of change of velocity. Therefore, an object moving around a circular path can have both centripetal and tangential acceleration simultaneously.
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20.
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.
Correct Answer C. The product of the two objects.
Explanation 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 as the mass of both objects increases, the gravitational force between them also increases. However, the distance between the two objects does not directly affect the gravitational force.
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21.
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.
Correct Answer B. The square of the distance between the two objects.
Explanation 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 force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
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22.
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.
Correct Answer D. Quadruples.
Explanation When the distance between the centers of two objects attracting each other gravitationally 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 force becomes four times stronger.
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23.
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.
Correct Answer C. The same.
Explanation The mass of an object remains the same regardless of its location. Mass is a measure of the amount of matter in an object and is a fundamental property of an object. Therefore, the mass of an object on the Moon is the same as its mass on Earth.
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24.
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
Correct Answer B. 72 kg
Explanation The mass of an object remains the same regardless of the gravitational acceleration. Therefore, the mass of the object on the Moon is still 72 kg. The acceleration of gravity only affects the weight of an object, not its mass.
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25.
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.
Correct Answer B. Decreases.
Explanation As a rocket moves away from the Earth's surface, the rocket's weight decreases. This is because weight is the force exerted on an object due to gravity, and the force of gravity decreases as the distance between the rocket and the Earth's surface increases. Therefore, as the rocket moves further away, the gravitational force acting on it decreases, resulting in a decrease in its weight.
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26.
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.
Correct Answer D. It is never beyond the pull of Earth's gravity.
Explanation The correct answer is "It is never beyond the pull of Earth's gravity." This is because gravity is a force that extends infinitely into space and is always present, no matter how far away an object is from the source of gravity. Therefore, even when the spaceship is closer to the Moon than to Earth, it is still under the influence of Earth's gravity.
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27.
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.
Correct Answer B. Equal to g
Explanation The centripetal acceleration of the satellite would be equal to the acceleration due to gravity (g). This is because centripetal acceleration is the acceleration towards the center of the circular path, and in this case, the gravitational force acts as the centripetal force. Therefore, the centripetal acceleration would be equal to the acceleration due to gravity.
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28.
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
Correct Answer D. G/8
Explanation The acceleration due to gravity on a planet is determined by its mass and radius. In this scenario, the planet has a mass that is half that of the Earth and a radius that is twice that of the Earth. The acceleration due to gravity is inversely proportional to the square of the radius, and directly proportional to the mass. Therefore, the acceleration due to gravity on this hypothetical planet would be (1/2)^2 * (1/2) * g, which simplifies to g/8.
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29.
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
Correct Answer D. 1/96
Explanation The acceleration of gravity on an object is directly proportional to its mass. Since the acceleration of gravity on the Moon is one-sixth of that on Earth, it means that the Moon's mass is one-sixth of Earth's mass. The radius of the Moon being one-fourth of Earth's radius does not affect the comparison of their masses. Therefore, the Moon's mass compared to the Earth's is 1/6, which is equivalent to 1/6 * 1/4 = 1/24.
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30.
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
Correct Answer D. 4m
Explanation Since the two planets have the same surface gravity, it means that the gravitational force acting on an object on the surface of each planet is the same. The gravitational force is given by the equation F = G * (m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers. Since the surface gravity is the same, we can set the gravitational force on planet A equal to the gravitational force on planet B. Since planet B has twice the radius of planet A, the distance between their centers is also twice as much. Therefore, the mass of planet B must be four times the mass of planet A, which is 4m.
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31.
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
Correct Answer C. 1.41r
Explanation The surface gravity of a planet is determined by its mass and radius. Since planet B has twice the mass of planet A but the same surface gravity, it must have a larger radius to compensate for the increased mass. The radius of planet B is therefore 1.41 times the radius of planet A.
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32.
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.
Correct Answer D. The speed of B is equal to the speed of A.
Explanation Since both satellites are rotating in the same orbit, their speeds will be the same regardless of their masses. Therefore, the speed of satellite B is equal to the speed of satellite A.
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33.
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.
Greater than their true weight
C.
Less than their true weight
D.
Zero
Correct Answer C. Less than their true weight
Explanation When a person is standing on a scale in an elevator accelerating downward, the reading on the scale will be less than their true weight. This is because the scale measures the normal force exerted by the person on it, which is equal to their true weight minus the force exerted by the elevator's acceleration. As the elevator accelerates downward, the scale will measure a smaller normal force, resulting in a reading that is less than the person's true weight.
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34.
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
Correct Answer A. Kepler
Explanation Johannes Kepler was the first person to realize that the planets move in elliptical paths around the Sun. He made this discovery based on the detailed observations and data collected by his mentor, Tycho Brahe. Kepler's laws of planetary motion revolutionized our understanding of the solar system and laid the foundation for modern astronomy.
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35.
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.
Correct Answer B. Increases as it nears the Sun.
Explanation Halley's Comet follows an elliptical orbit around the Sun, which means that its distance from the Sun varies throughout its orbit. According to Kepler's laws of planetary motion, a celestial object moves faster when it is closer to the object it is orbiting. Therefore, as Halley's Comet nears the Sun in its orbit, its speed increases.
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36.
Let the average orbital radius of a planet be r. Let the orbital period be T. What quantity is constant for all planets orbiting the Sun?
A.
T/R
B.
T/R^2
C.
T^2/R^3
D.
T^3/R^2
Correct Answer C. T^2/R^3
Explanation The quantity that is constant for all planets orbiting the Sun is T^2/R^3. This is because Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of its average orbital radius. Therefore, regardless of the specific values of T and R for each planet, their ratio T^2/R^3 will always be constant.
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37.
A planet is discovered to orbit around a star in the galaxy Andromeda, with the same orbital diameter as the Earth around our Sun. If that star has 4 times the mass of our Sun, what will the period of revolution of that new planet be, compared to the Earth's orbital period?
A.
One-fourth as much
B.
One-half as much
C.
One-half as much
D.
Four times as much
Correct Answer B. One-half as much
Explanation The period of revolution of the new planet will be one-half as much as Earth's orbital period because the period of revolution is inversely proportional to the square root of the mass of the star. Since the new star has 4 times the mass of our Sun, the square root of its mass is 2 times the square root of our Sun's mass. Therefore, the period of revolution of the new planet will be one-half of Earth's orbital period.
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38.
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
Correct Answer B. 0.48 AU
Explanation An asteroid orbits the Sun in one-third of a year. Since the average distance from the Earth to the Sun is defined as one "astronomical unit" (AU), we can calculate the asteroid's average distance from the Sun by dividing the Earth-Sun distance by the time it takes for the asteroid to complete one orbit. One-third of a year is approximately 0.33 years. Therefore, the asteroid's average distance from the Sun is 1 AU divided by 0.33 years, which is approximately 0.48 AU.
Correct Answer C. Gravitational, electromagnetic, strong nuclear, weak nuclear
Explanation The four fundamental forces in nature are gravitational, electromagnetic, strong nuclear, and weak nuclear. Gravitational force is responsible for the attraction between two objects with mass. Electromagnetic force is responsible for the interaction between charged particles. Strong nuclear force is responsible for holding atomic nuclei together. Weak nuclear force is responsible for certain types of radioactive decays.
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40.
An object moves with a constant speed of 30 m/s on a circular track of radius 150 m. What is the acceleration of the object?
A.
Zero
B.
0.17 m/s^2
C.
5.0 m/s^2
D.
6.0 m/s^2
Correct Answer D. 6.0 m/s^2
Explanation The acceleration of an object moving in a circular path is given by the formula a = v^2 / r, where v is the velocity and r is the radius of the circle. In this case, the object is moving with a constant speed of 30 m/s and the radius of the circular track is 150 m. Plugging these values into the formula, we get a = (30^2) / 150 = 6.0 m/s^2. Therefore, the acceleration of the object is 6.0 m/s^2.
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41.
The maximum speed around a level curve is 30.0 km/h. What is the maximum speed around a curve with twice the radius? (Assume all other factors remain unchanged.)
A.
42.4 km/h
B.
45.0 km/h
C.
60.0 km/h
D.
120 km/h
Correct Answer A. 42.4 km/h
Explanation When a car is going around a curve, the maximum speed it can achieve depends on the radius of the curve. A larger radius allows for a higher speed. In this question, it is stated that the maximum speed around a level curve is 30.0 km/h. If the radius of the curve is doubled, it means that the curve is now less sharp and has a larger radius. This allows the car to go faster. The correct answer of 42.4 km/h is obtained by taking into account the relationship between radius and speed.
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42.
What is the centripetal acceleration of a point on the perimeter of a bicycle wheel of diameter 70 cm when the bike is moving 8.0 m/s?
A.
91 m/s^2
B.
1.8 * 10^2 m/s^2
C.
2.1 * 10^2 m/s^2
D.
2.7 * 10^2 m/s^2
Correct Answer B. 1.8 * 10^2 m/s^2
Explanation The centripetal acceleration of a point on the perimeter of a bicycle wheel can be calculated using the formula a = (v^2) / r, where v is the velocity and r is the radius of the wheel. In this case, the diameter of the wheel is given as 70 cm, so the radius would be half of that, or 35 cm (0.35 m). The velocity is given as 8.0 m/s. Plugging these values into the formula, we get a = (8.0^2) / 0.35 = 64 / 0.35 = 182.86 m/s^2, which can be rounded to 1.8 * 10^2 m/s^2.
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43.
A point on a wheel rotating at 5.00 rev/s is located 0.200 m from the axis. What is the centripetal acceleration?
A.
0.050 m/s^2
B.
1.35 m/s^2
C.
48.0 m/s^2
D.
198 m/s^2
Correct Answer D. 198 m/s^2
Explanation The centripetal acceleration of an object moving in a circle is given by the formula a = rω², where a is the centripetal acceleration, r is the radius, and ω is the angular velocity. In this case, the radius is given as 0.200 m and the angular velocity is given as 5.00 rev/s. Converting the angular velocity to radians per second (ω = 2πf), we get ω = 2π(5.00) = 31.4 rad/s. Plugging these values into the formula, we get a = (0.200)(31.4)² = 198 m/s².
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44.
How many revolutions per minute must a circular, rotating space station of radius 1000 m rotate to produce an artificial gravity of 9.80 m/s^2?
A.
0.65 rpm
B.
0.75 rpm
C.
0.85 rpm
D.
0.95 rpm
Correct Answer D. 0.95 rpm
Explanation A circular, rotating space station can produce artificial gravity through centripetal acceleration. The formula for centripetal acceleration is a = (v^2) / r, where a is the acceleration, v is the linear velocity, and r is the radius. In this case, the desired acceleration is 9.80 m/s^2 and the radius is 1000 m. Rearranging the formula, we can solve for v: v = sqrt(a * r). Plugging in the values, we get v = sqrt(9.80 * 1000) = 31.30 m/s. Since we want the velocity in terms of revolutions per minute, we can convert it using the formula v = (2 * pi * r) / t, where v is the linear velocity, r is the radius, and t is the time in seconds. Rearranging the formula, we can solve for t: t = (2 * pi * r) / v. Plugging in the values, we get t = (2 * pi * 1000) / 31.30 = 200.96 seconds. Finally, we can convert the time to minutes and calculate the number of revolutions per minute: 200.96 seconds * (1 minute / 60 seconds) * (1 revolution / t) = 0.95 rpm.
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45.
A motorcycle has a mass of 250 kg. It goes around a 13.7 m radius turn at 96.5 km/h. What is the centripetal force on the motorcycle?
A.
719 N
B.
2.95 * 10^3 N
C.
1.31 * 10^4 N
D.
4.31 * 10^4 N
Correct Answer C. 1.31 * 10^4 N
Explanation The centripetal force on an object moving in a circular path can be calculated using the formula F = mv^2/r, where F is the centripetal force, m is the mass of the object, v is the velocity of the object, and r is the radius of the circular path. In this case, the mass of the motorcycle is given as 250 kg, the radius of the turn is given as 13.7 m, and the velocity is given as 96.5 km/h. First, we need to convert the velocity from km/h to m/s by dividing it by 3.6. Then, we can substitute the values into the formula to find the centripetal force. After calculation, the answer is 1.31 * 10^4 N.
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46.
A 0.50-kg mass is attached to the end of a 1.0-m string. The system is whirled in a horizontal circular path. If the maximum tension that the string can withstand is 350 N. What is the maximum speed of the mass if the string is not to break?
A.
700 m/s
B.
26 m/s
C.
19 m/s
D.
13 m/s
Correct Answer B. 26 m/s
Explanation The maximum tension that the string can withstand is 350 N. In order for the string not to break, the centripetal force acting on the mass must not exceed this tension. The centripetal force is given by the equation Fc = m * v^2 / r, where m is the mass, v is the velocity, and r is the radius of the circular path. Rearranging the equation to solve for v, we get v = sqrt(Fc * r / m). Plugging in the given values, we get v = sqrt(350 N * 1.0 m / 0.50 kg) = 26 m/s. Therefore, the maximum speed of the mass is 26 m/s.
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47.
A stone, of mass m, is attached to a strong string and whirled in a vertical circle of radius r. At the exact top of the path the tension in the string is 3 times the stone's weight. The stone's speed at this point is given by
A.
2(gr)^(1/2).
B.
(2gr)^(1/2).
C.
(gr)^(1/2).
D.
2gr.
Correct Answer A. 2(gr)^(1/2).
Explanation At the top of the path, the tension in the string is equal to the sum of the stone's weight and the centripetal force required to keep it moving in a circle. Since the tension is 3 times the stone's weight, we can write the equation as T = mg + mv^2/r, where T is the tension, m is the mass of the stone, g is the acceleration due to gravity, v is the speed of the stone, and r is the radius of the circle. Rearranging the equation, we get mv^2/r = 2mg, which simplifies to v = (2gr)^(1/2). Therefore, the stone's speed at the top of the path is given by 2(gr)^(1/2).
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48.
A stone, of mass m, is attached to a strong string and whirled in a vertical circle of radius r. At the exact bottom of the path the tension in the string is 3 times the stone's weight. The stone's speed at this point is given by
A.
2(gr)^(1/2).
B.
(2gr)^(1/2).
C.
(gr)^(1/2).
D.
2gr.
Correct Answer B. (2gr)^(1/2).
Explanation At the bottom of the path, the tension in the string is equal to the sum of the stone's weight and the centripetal force required to keep it moving in a circle. Since the tension is 3 times the stone's weight, we can write the equation as T = mg + 3mg = 4mg. The centripetal force is given by F = mv^2/r, where v is the speed of the stone. Equating the centripetal force to the tension, we have 4mg = mv^2/r. Simplifying this equation, we get v^2 = 4gr, which means v = sqrt(4gr) or (2gr)^(1/2). Therefore, the stone's speed at the bottom of the path is given by (2gr)^(1/2).
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49.
A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. What is the radius of curvature of the loop in which the plane is flying?
A.
640 m
B.
1200 m
C.
7100 m
D.
9200 m
Correct Answer D. 9200 m
Explanation The plane is experiencing an acceleration of 4g when pulling out of the dive. This means that the net force acting on the plane is four times the force of gravity. The force causing the acceleration is the centripetal force, given by F = mv^2/r, where m is the mass of the plane, v is its velocity, and r is the radius of curvature. Since the plane is flying at a constant speed of 600 m/s, the force of gravity acting on it is equal to the centripetal force. Therefore, we can set mg = mv^2/r and solve for r. By substituting the given values, we find that r = 9200 m.
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50.
A pilot makes an outside vertical loop (in which the center of the loop is beneath him) of radius 3200 m. At the top of his loop he is pushing down on his seat with only one-half of his normal weight. How fast is he going?
A.
5.0 m/s
B.
25 m/s
C.
125 m/s
D.
625 m/s
Correct Answer C. 125 m/s
Explanation When a pilot is making an outside vertical loop, the centripetal force required to keep the pilot moving in a circular path is provided by the normal force exerted by the seat. At the top of the loop, the pilot is pushing down on his seat with only one-half of his normal weight. This means that the normal force is half of the pilot's weight. Since the centripetal force is equal to the normal force, it is also half of the pilot's weight. The centripetal force can be calculated using the formula F = mv^2/r, where F is the centripetal force, m is the mass of the pilot, v is the speed, and r is the radius of the loop. By rearranging the formula, we can solve for v, which gives us v = sqrt(Fr/m). Since the centripetal force is half of the pilot's weight, we can substitute F = (1/2)mg into the formula. By plugging in the values, we get v = sqrt((1/2)mg * r/m). The mass of the pilot cancels out, leaving us with v = sqrt((1/2)g * r). Plugging in the values for g (acceleration due to gravity) and r (radius of the loop), we can calculate v, which is approximately 125 m/s.