1.
The diagram below represents a 0.40-kilogram stone attached to a string. The stone is moving at a constant speed of 4.0 meters per second in a horizontal circle having a radius of 0.80 meter.
The magnitude of the centripetal acceleration of the stone is
Correct Answer
D. 20 m/s2
Explanation
The magnitude of the centripetal acceleration of the stone is 20 m/s2 because centripetal acceleration 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 stone is moving at a constant speed of 4.0 m/s and the radius of the circle is 0.80 m. Plugging these values into the formula, we get a = (4.0 m/s)^2 / 0.80 m = 20 m/s2.
2.
A 1000 kilogram car is driven clockwise around a flat circular track of radius 25.0 meters. The speed of the car is a constant 5.00 meters per second.
At the instant shown in the diagram, the car’s centripetal acceleration is directed
Correct Answer
A. Toward E
Explanation
The centripetal acceleration of an object moving in a circular path is always directed towards the center of the circle. In this case, the car is moving clockwise around the circular track, so the center of the circle is towards the east (E). Therefore, the car's centripetal acceleration is directed towards E.
3.
A 1000 kilogram car is driven clockwise around a flat circular track of radius 25.0 meters. The speed of the car is a constant 5.00 meters per second.
If the circular track were to suddenly become frictionless at the instant shown in the diagram, the car’s direction of travel would be
Correct Answer
C. Toward N
Explanation
If the circular track were to suddenly become frictionless, the car would continue to move in the direction it was moving before. Since the car was initially moving in a clockwise direction, it would continue moving in that direction. Therefore, the car's direction of travel would be toward the north (N).
4.
A 1000 kilogram car is driven clockwise around a flat circular track of radius 25.0 meters. The speed of the car is a constant 5.00 meters per second.
At the instant shown in the diagram, the centripetal force acting on the car is directed
Correct Answer
A. Toward E
Explanation
At the instant shown in the diagram, the car is moving in a clockwise direction around the circular track. The centripetal force is the force that keeps an object moving in a circular path. According to the right-hand rule, the centripetal force always points towards the center of the circle. Since the car is moving clockwise, the center of the circle is towards the east (E). Therefore, the centripetal force acting on the car is directed towards E.
5.
A 1000 kilogram car is driven clockwise around a flat circular track of radius 25.0 meters. The speed of the car is a constant 5.00 meters per second.
What centripetal force is a passenger in the car experiencing?
Correct Answer
D. 1000 N
Explanation
The centripetal force experienced by a passenger in the car can be calculated using the formula F = (mv^2)/r, where F is the centripetal force, m is the mass of the car, v is the speed of the car, and r is the radius of the circular track. In this case, the mass of the car is 1000 kg, the speed of the car is 5.00 m/s, and the radius of the track is 25.0 meters. Plugging these values into the formula, we get F = (1000 kg * (5.00 m/s)^2) / 25.0 m = 1000 N. Therefore, the passenger in the car is experiencing a centripetal force of 1000 N.
6.
When a satellite is a distance R from the center of Earth, the force due to gravity on the satellite is F. What is the force due to gravity on the satellite when its distance from the center of Earth is 3R?
Correct Answer
A. 1/9 F
Explanation
The force due to gravity on an object is inversely proportional to the square of its distance from the center of the Earth. This means that as the distance from the center of the Earth increases, the force due to gravity decreases. In this case, when the distance from the center of the Earth is 3R, it is three times farther away than when it was at distance R. Since the force due to gravity is inversely proportional to the square of the distance, the force will be 1/9 of the original force, which is represented by 1/9 F.
7.
The diagram below represents two satellites of equal mass, A and B, in circular orbits around a planet.
Compared to the magnitude of the gravitational force of attraction between satellite A and the planet, the magnitude of the gravitational force of attraction between satellite B and the planet is
Correct Answer
C. One-fourth as great
Explanation
The magnitude of the gravitational force of attraction between satellite B and the planet is one-fourth as great compared to the magnitude of the gravitational force of attraction between satellite A and the planet. This is because the gravitational force is inversely proportional to the square of the distance between two objects. Since both satellites are in circular orbits around the planet, they have the same distance from the planet. Therefore, the force of attraction between satellite B and the planet is one-fourth of the force of attraction between satellite A and the planet.
8.
As an astronaut travels from the surface of Earth to a position that is four times as far away from the center of Earth, the astronaut’s
A. mass decreases B. mass remains the same C. weight increases D. weight remains the same
Correct Answer
B.
Explanation
As the astronaut travels further away from the center of the Earth, the gravitational force acting on them decreases. This is because the gravitational force is inversely proportional to the square of the distance between two objects. Since weight is the force experienced by an object due to gravity, and weight is directly proportional to mass, the weight of the astronaut will decrease as they travel further away. Therefore, the correct answer is D. weight remains the same.
9.
The gravitational force of attraction between two objects would be increased by
A. doubling the mass of both objects, only
B. doubling the distance between the objects, only
C. doubling the mass of both objects and doubling the distance between the objects
D. doubling the mass of one object and doubling the distance between the objects
Correct Answer
A.
Explanation
Doubling the mass of both objects would increase the gravitational force of attraction between them. According to Newton's law of universal gravitation, the force of attraction between two objects is directly proportional to the product of their masses. Therefore, increasing the mass of both objects would result in a stronger gravitational force. Doubling the distance between the objects, on the other hand, would decrease the gravitational force of attraction. This is because the force of attraction is inversely proportional to the square of the distance between the objects. Therefore, option A is the correct answer.
10.
The centers of two 15.0-kilogram spheres are separated by 3.00 meters. The magnitude of the gravitational force between the two spheres is approximately
A. 1.1 E 10^{–10}N B. 1.67 E 10^{–9} N C. 3.34 E 10^{–10} N D. 5.00 E 10^{–9} N
Correct Answer
B.
Explanation
The magnitude of the gravitational force between two objects can be calculated using the formula F = G * (m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between their centers. In this case, the masses of the spheres are given as 15.0 kilograms each, and the distance between their centers is given as 3.00 meters. Plugging these values into the formula, we can calculate the magnitude of the gravitational force.
11.
On the surface of Earth, a spacecraft has a mass of 2.00 × 10^{4} kilograms. What is the mass of the
spacecraft at a distance of one Earth radius above Earth’s surface?
A. 5.00 × 10^{3} kg B. 4.90 × 10^{4} kg C. 2.00 × 10^{4} kg D. 1.96 × 10^{5} kg
Correct Answer
C.
Explanation
The mass of the spacecraft remains the same at a distance of one Earth radius above Earth's surface. Therefore, the correct answer is C. 2.00 × 104 kg.
12.
When Earth and the Moon are separated by a distance of 3.84 × 10^{8} meters, the magnitude of
the gravitational force of attraction between them is 2.0 × 10^{20} newtons. What would be the magnitude of this gravitational force of attraction if Earth and the Moon were separated by a distance of 1.92 × 10^{8} meters?
A. 5.0 × 10^{19} N B. 4.0 × 10^{20} N C. 2.0 × 10^{20} N D. 8.0 × 10^{20} N
Correct Answer
D.
Explanation
The magnitude of the gravitational force of attraction between two objects is inversely proportional to the square of the distance between them. Therefore, if the distance between Earth and the Moon is halved from 3.84 × 108 meters to 1.92 × 108 meters, the magnitude of the gravitational force of attraction would be four times greater. So, the correct answer is B. 4.0 × 1020 N.
13.
As a meteor moves from a distance of 16 Earth radii to a distance of 2 Earth radii from the center of Earth, the magnitude of the gravitational force between the meteor and Earth becomes
A. 1/8 as great B. 64 times as great C. 8 times as great D. 4 times as great
Correct Answer
B.
Explanation
As the distance between the meteor and the center of the Earth decreases from 16 Earth radii to 2 Earth radii, the magnitude of the gravitational force between them becomes 64 times as great. This is because the gravitational force is inversely proportional to the square of the distance between two objects. Therefore, when the distance is reduced by a factor of 8 (from 16 Earth radii to 2 Earth radii), the gravitational force increases by a factor of 8^2, which is 64.
14.
A 0.50-kilogram object moves in a horizontal circular path with a radius of 0.25 meter at a constant speed of 4.0 meters per second. What is the magnitude of the object’s acceleration?
A. 8.0 m/s^{2} B. 32 m/s^{2} C. 16 m/s^{2} D. 64 m/s^{2}
Correct Answer
D.
Explanation
The magnitude of the object's acceleration can be found using the formula for centripetal acceleration, which is given by a = v^2/r, where v is the velocity and r is the radius. Plugging in the values, we have a = (4.0 m/s)^2 / 0.25 m = 16 m/s^2. Therefore, the correct answer is C. 16 m/s^2.
15.
In the diagram below, S is a point on a car tire rotating at a constant rate.
Which graph best represents the magnitude of the centripetal acceleration of point S as a function of time?
A. 1 B. 2 C. 3 D. 4
Correct Answer
B. B. 2
16.
A 1.0 × 10^{3}-kilogram car travels at a constant speed of 20 meters per second around a horizontal circular
track. Which diagram correctly represents the direction of the car’s velocity (v) and the direction of the centripetal force (Fc) acting on the car at one particular moment?
Correct Answer
A. A. 1
17.
The radius of Mars is approximately one-half the radius of Earth, and the mass of Mars is approximately
one-tenth the mass of Earth. Compared to the acceleration due to gravity on the surface of Earth, the acceleration due to gravity on the surface of Mars is
A. smaller B. larger C. the same
Correct Answer
A.
Explanation
A. smaller. The acceleration due to gravity is directly proportional to the mass of the planet and inversely proportional to the square of the radius. Since Mars has a smaller mass and radius compared to Earth, the acceleration due to gravity on the surface of Mars is smaller than that on Earth.
18.
A 1200-kilogram car traveling at a constant speed of 9.0 meters per second turns at an intersection. The car follows a horizontal circular path with a radius of 25 meters to point P.
The magnitude of the centripetal force acting on the car as it travels around the circular path is approximately
A. 1.1 X 10^{4 }N B. 3.9 X 10^{3} N C. 1.2 X 10^{4} N D. 4.3 X 10^{2} N
Correct Answer
B. B. 3.9 X 10^{3} N
Explanation
The magnitude of the centripetal force acting on the car as it travels around the circular path is approximately 1.1 X 104 N. This can be calculated using the formula for centripetal force, F = (m * v^2) / r, where m is the mass of the car, v is the velocity of the car, and r is the radius of the circular path. Plugging in the given values, we get F = (1200 kg * (9.0 m/s)^2) / 25 m = 1.1 X 104 N.
19.
A 1200-kilogram car traveling at a constant speed of 9.0 meters per second turns at an intersection. The car follows a horizontal circular path with a radius of 25 meters to point P.
At point P, the car hits an area of ice and loses all frictional force on its tires. Which path does the
car follow on the ice?
Correct Answer
B. B
Explanation
The car will continue to move in a circular path with a radius of 25 meters on the ice. The absence of frictional force on the tires does not affect the car's ability to turn, as the centripetal force required for circular motion is provided by the car's inertia.