Classroom Quiz 2 PDF

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Mastering Physics...

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Classroom Quiz 2 Due: 11:59pm on Sunday, September 2, 2018 You will receive no credit for items you complete after the assignment is due. Grading Policy

Prelecture Concept Question 4.02 Part A An object is moving with constant velocity. Which of the following best describes the force(s) acting on the object? ANSWER:

No forces are acting on the object. A constant net force is acting on the object in a direction perpendicular to the direction of motion. A constant net force is acting on the object in a direction opposite of the motion. A constant net force is acting on the object in the direction of the motion. The net force acting on the object is zero.

Correct

Prelecture Concept Question 4.03 Part A A constant net force acts on an object. Which of the following best describes the object's motion? ANSWER:

The object is moving with a constant acceleration. The object is moving with a constant velocity. The object is at rest; its position is constant. The object is moving with an increasing acceleration. The object is moving with a decreasing acceleration.

Correct

Prelecture Concept Question 4.04 Part A The same net force is applied to two different objects. The second object has twice the mass of the first object. Compare the acceleration of the two objects. ANSWER: The acceleration of object 1 is four times the acceleration of object 2. The acceleration of object 1 is twice the acceleration of object 2. The acceleration of object 1 is equal to the acceleration of object 2. The acceleration of object 1 is one-fourth of the acceleration of object 2. The acceleration of object 1 is one-half of the acceleration of object 2.

Correct

Prelecture Concept Question 4.05

Part A An object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg. If this force serves as the action force, what is the reaction force in the action–reaction pair? ANSWER:

the object pushing down on the table Earth pushing upward on the table the object pulling upward on Earth the table pushing up on the object the table pushing down on Earth

Correct

Prelecture Concept Question 4.06 Part A A large truck collides head-on with a small car. The car is severely damaged as a result of the collision. According to Newton's third law, how do the forces acting between the truck and car compare during the collision? ANSWER: The force on the truck is smaller than the force on the car. The force on the truck is larger than the force on the car. The car does not exert a force on the truck during the collision. The force on the truck is equal to the force on the car. The truck does not exert a force on the car during the collision.

Correct

Video Tutor: Cart with Fan and Sail First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question o the right. You can watch the video again at any point.

Part A Which of the force diagrams in the figure correctly displays all of the horizontal forces exerted on the cart by the surrounding air?

Hint 1. How to approach the problem First, what does the acceleration of the cart imply about the net force acting on the cart? (Only two of the choices are compatible with the cart’s behavior.) Next, decide which direction is correct for the force vectors. • When the fan pushes air to the right, in which direction does the air push the fan? (Remember how the cart moved in the video when only the fan was attached to it.) • When air moving to the right strikes the sail, in which direction does the air push the sail?

ANSWER:

A B C D

Correct The net force on the cart is zero, since the air is the only thing acting on the cart in the horizontal direction.

Video Tutor: Tension in String between Hanging Weights First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question a right. You can watch the video again at any point.

Part A Consider the video tutorial you just watched. Suppose that we duplicate this experimental setup in an elevator. What will the spring scale read if the elevator is moving upward a constant speed?

Hint 1. How to approach the problem What does the phrase "at constant speed" imply about the acceleration of the system?

ANSWER:

18 More than 18 0 Less than 18

but greater than 0

Correct Since the elevator is not accelerating, the reading on the scale is the same as in the video.

Video Tutor: Weighing a Hovering Magnet First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions right. You can watch the video again at any point.

Part A Consider the video tutorial you just watched. Suppose that we repeat the experiment, but this time we replace the original 56- magnet with a more powerful magnet of the sam mass. As you know from experience, the more powerful a magnet is, the more strongly it attracts or repels other magnets or magnetic objects. You have also probably noticed that magnetic forces fall off sharply with distance—two magnets that interact strongly across a distance of millimeters interact more weakly at a distance of a centimeter. (In fact, the strength of the force falls off with the square of the distance.) With the magnet hovering above the base, what will the scale read? The scale has been zeroed (tared) to subtract the weight of the base.

Hint 1. How to approach the problem Here’s the key: Because the magnet hovers motionless above the base, its weight exactly counters the upward magnetic repulsion exerted on it by the base. Hence, we can draw free-body diagrams for the magnet and the base:

As we just said, equal magnitudes.

has the same magnitude as

. And because

and

The scale reading reflects the force exerted by the base on the scale, which is equal in magnitude to the normal force forces constitute another third-law pair.)

constitute a Newton’s-third-law pair, they also have

exerted by the scale on the base. (These two

Given the above facts, is it possible to change the scale reading by changing the strength of the magnet?

ANSWER: Less than 56

but greater than 0

More than 56 0 56 –56

Correct In equilibrium, the base must push up on the magnet with a force equal to the weight of the magnet, and the magnet will exert an equal-but-opposite force on the base. The reading on the scale will not change.

Part B Compared to the magnet in the video, the magnet in Part A will hover at a position

Hint 1. How to approach the problem Recall that the forces between the magnet and the base weaken with increasing distance. The magnet hovers at the distance where its weight is balanced by the magnetic repulsion of the base:

The fact that our current magnet is stronger than the one in the video means that the magnetic forces between it and the base will be stronger at any given distance. Where will this force balance the magnet’s weight?

ANSWER: farther above the base. closer to the base. at the same distance above the base.

Correct In equilibrium, the magnetic force exerted by the base on the magnet must equal the magnet's weight (or else the magnet would accelerate). Because this magnet is stronger than the one in the video (and because the magnetic forces weaken with distance), this magnet hovers farther from the base.

Prelecture Concept Question 5.01

Part A A packing crate is sitting at rest on an inclined loading ramp. How does the magnitude of the force of static friction compare to the other forces acting on the crate? ANSWER:

The force of static friction is the only force acting on the crate, and it is responsible for keeping the crate at rest. The magnitude of the force of static friction is equal to the magnitude of the component of the weight of the crate perpendicular to the inclined ramp. The magnitude of the force of static friction is equal to the magnitude of the weight of the crate. The magnitude of the force of static friction is equal to the magnitude of the component of the weight of the crate parallel to the inclined ramp. The magnitude of the force of static friction is equal to the magnitude of the normal force acting on the crate.

Correct

Prelecture Concept Question 5.02 Part A In general, how does the coefficient of static friction compare to the coefficient of kinetic friction for the same two materials? ANSWER: The coefficient of static friction can be greater than or equal to the coefficient of kinetic friction. The coefficient of static friction can be less than or equal to the coefficient of kinetic friction. The coefficient of static friction is less than the coefficient of kinetic friction. The coefficient of static friction is equal to the coefficient of kinetic friction.

Correct

Prelecture Concept Question 5.03 Part A An object is hanging by a string from the ceiling of an elevator. The elevator is moving upward with a constant speed. What is the magnitude of the tension in the string? ANSWER:

The magnitude of the tension in the string is greater than the magnitude of the weight of the object. The magnitude of the tension in the string is equal to the magnitude of the weight of the object. The magnitude of the tension in the string is less than the magnitude of the weight of the object. The tension in the string cannot be determined without knowing the speed of the elevator. The tension in the string is zero.

Correct

Prelecture Concept Question 5.04 Part A An object is hanging by a string from the ceiling of an elevator. The elevator is slowing down while moving upward. What is the magnitude of the tension in the string? ANSWER:

The tension in the string cannot be determined without knowing the speed of the elevator. The magnitude of the tension in the string is equal to the magnitude of the weight of the object. The magnitude of the tension in the string is less than the magnitude of the weight of the object. The magnitude of the tension in the string is greater than the magnitude of the weight of the object. The tension in the string is zero.

Correct

Prelecture Concept Question 5.05 Part A An object moves in a circular path at a constant speed. What is the relationship between the directions of the object's velocity and acceleration vectors? ANSWER: The velocity and acceleration vectors point in opposite directions. The velocity vector points toward the center of the circular path. The acceleration is zero. The velocity vector points in a direction tangent to the circular path. The acceleration is zero. The velocity and acceleration vectors are perpendicular. The velocity and acceleration vectors point in the same direction.

Correct

Prelecture Concept Question 5.06

Part A An object moves in a circular path at a constant speed. What is the direction of the net force acting on the object? ANSWER:

The net force points in the direction opposite to the motion of the object. The net force is directed away from the center of the circular path. The net force is directed toward the center of the circular path. The net force points in the same direction as the motion of the object. The net force is zero because the object is moving with a constant speed.

Correct

Video Tutor: Ball Leaves Circular Track First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions right. You can watch the video again at any point.

Part A

Consider the video demonstration that you just watched. Which of the following changes would make it more likely for the ball to hit both the white can and the green can?

Hint 1. How to approach the problem To answer this question, you first have to decide whether changing the ball’s mass or its speed can change the path it follows after it leaves the track. Newton’s second law says that a net force acting on the ball will change the ball’s motion—that is, its speed and/or direction. Newton’s first law says that, in the absence of a net force, the ball’s motion won’t change. After the ball leaves the track, does a net force act on it? Draw a free-body diagram for the ball if you’re not sure. To hit the green can, the ball must continue following a curved path. What would be needed to make that happen?

ANSWER:

Roll the ball faster. Use a ball that is heavier than the original ball. Use a ball that is lighter than the original ball, but still heavier than an empty can. Roll the ball slower. None of the above

Correct By Newton’s first law, after it has left the circular track, the ball will travel in a straight line until it is subject to a nonzero net force. Thus, the ball can only hit the white can, because that is the only can in the ball’s straight-line path.

Score Summary: Your score on this assignment is 98.3%. You received 73.75 out of a possible total of 75 points....