An axle passes through a pulley. Each end of the axle has a string that is tied to a support. A third string is looped many time

In the first case, once equilibrium is established,
mg = kx
135 = k(0.21) [Converted to SI units]
k = 135/(0.21)

In the second case, similarly when reaching equilibrium,
mg = kx
mg = [(135)(.449)]/(0.21)
mg = weight of the electronic equipment = approximately 289 N.

Answer:

(a) the coefficient of friction is 0.451

This was derived using the energy conservation principle (the total energy in a closed system remains constant).

(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.

Explanation:

For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.

Thank you for your attention; I trust this is beneficial to you.

The answer is

-Small f and large D.

The explanation:

-A car jack acts as a machine, defined as an apparatus that aids individuals in exerting force more easily.

-Hence, by applying a small force to the jack, the height at which the car is elevated increases.

Machines are essential for people to amplify their strength; without them, lifting a car would be impossible.

Employing leverage or hydraulic principles, machines enhance your exerted force.

Utilizing a greater lever allows for extensive movement with minimal force, resulting in the opposite side moving shorter distances with an increased force.

A. A car moving at a constant speed on a flat, straight road. B. A vehicle traveling at a steady speed on a 10-degree incline. An object operates within an inertial reference frame if there is no net force acting upon it. According to Newton's second law, this implies that the object's acceleration also equals zero. Assessing the scenarios yields: A. A car moving at a constant speed on a flat road qualifies as an inertial reference frame, since its velocity and direction remain unchanged; thus, acceleration is zero. B. A car moving steadily up a 10-degree incline still constitutes an inertial reference frame, for similar reasons. C. A car accelerating after departing a stop sign does not represent an inertial frame due to its change in speed. D. A car driving at a steady speed around a curve cannot be considered an inertial reference frame since its direction is changing, resulting in a change in velocity and thus acceleration. Therefore, options A and B are correct.