A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.

D) Both perform no work

Explanation:

The work accomplished by a force is defined as:

where

F represents the applied force

d denotes the displacement

is the angle highlighted between the applied force and the displacement vector

From this formula, it is clear that work is only done when displacement occurs, meaning the object has to move.

In this instance, as the wall is unmoving, the displacement is zero: d = 0, thus no work is performed.

Explanation: I don't know, sorry.

Response:

C. vx

F. ax

G. ay

Clarification:

The projectile follows a curved trajectory toward the ground, causing changes in x and y positions.

Since there is no external force acting in the x-direction, the acceleration in x remains at zero. Consequently, ax and vx remain unchanged.

The projectile is subject to the force of gravity, directed downwards, leading to an increase in its velocity due to the rise in its y-component.

Meanwhile, the y-component of acceleration remains constant due to gravitational acceleration.

Answer:

the maximum static friction force of the wall acting on the book (Increasing)

the normal force of the wall acting on the book (Decreasing)

the weight of the book (Constant)

Explanation:

According to Newton's third law of motion:

"Every action has an equal and opposite reaction"

In the scenario provided, Albert is pressing the book against the wall and subsequently decreases the force applied against the wall.

Let's evaluate all forces influencing the book in this situation.

1. Weight of the book acting downwards (y-axis)

2. Friction from the book against the wall acting upwards (y-axis)

3. Albert’s force exerted on the book against the wall (x-axis)

4. Normal force of the wall reacting to Albert’s applied force (x-axis)

As Albert eases off his force, the new scenario reads:

1. The weight remains constant as represented by W = mg

Since neither mass nor gravitational acceleration has changed, the weight exerted on the book remains the same.

2. As Albert reduces his force, the wall’s normal reaction force decreases correspondingly, following Newton's third law of motion.

3. Friction operates in response to the force applied to it. With a box resting on the floor, no friction acts upon it until it is dragged, at which point friction begins to manifest and rise until it reaches its maximum. Therefore, when Albert diminishes his force, the weight's pull will influence the book and the maximum static friction will rise to resist the book’s downward movement.

It should be noted that the maximum static friction is working to prevent movement of the book. With Albert's force reduced, but the weight of the book unchanged, maximum static friction increases to prevent downward movement.

For motion in a circle.

Centripetal acceleration is calculated as mv²/r = mω²r

where v represents linear velocity, r equals radius which is diameter/2 equating to 1/2 or 0.5m

. Here, m is the mass of the object, which is 175g or 0.175kg.

The angular speed, ω, is derived from Angle covered / time

                         = 2 revolutions per 1 second

                         = 2 * 2π  radians for each second

                         = 4π  radians per second

Thus, Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5. Utilize a calculator

                                                         ≈13.817  m/s²

. The acceleration's magnitude is approximately 13.817  m/s² and it is oriented towards the center of the circular path.

The tension in the string equates to m*a

                                   = 0.175*13.817

                                   = 2.418 N