An overhang hollow shaft carries a 900 mm diameter pulley, whose centre is 250 mm from the centre of the nearest bearing. The we

Response: The spring constant is 25 N/m.

Details:

The body’s mass is 25 g, which converts to 0.025 kg (since 1 kg = 1000 g).

The total oscillations are 20 in 4 seconds.

Oscillations per second =

Spring's frequency of vibration is =

The spring constant 'k' can be derived from the relationship involving frequency, mass, and spring constant.

The spring constant is 25 N/m.

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.

Answer:

Explanation:

Consider the following:

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field measured at point P

Thus,

Now, by applying integration to the equation above

Answer:

C) True. The distance S increases over time, with v₁ = gt and v₂ = g (t-t₀), illustrating that v₁> v₂ for the same t.

Explanation:

We have a set of statements to evaluate for correctness. The most effective approach is to examine the problem in detail.

Using the equation for vertical launch, we acknowledge that the positive direction signifies downward movement.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Stone 2

Released shortly thereafter, let's assume a delay of one second, we can utilize the same timing mechanism

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

We can now calculate the separation distance between the two stones, which is applicable for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t² - 2 t t₀ + t₀²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t - t₀)

This represents the distance between the two stones over time, with the coefficient outside the parentheses being constant.

For t < to, the first stone remains stationary while the distance grows.

For t > = to, the expression (2t/to-1) yields a value greater than 1, indicating that the distance expands as time progresses.

We can now analyze the different statements

A) false. The height difference increases over time.

B) False S increases.

C) It is true that S increases over time, with v₁ = gt and V₂ = g (t-t₀) indicating v₁> v₂ at the same t.

Answer:

Δx=(v+v0/2)t

Explanation:

We can determine which kinematic equation to apply by selecting the one that encompasses the known variables as well as the unknown we aim to solve for.

In this scenario, the unknown we wish to determine is the initial velocity v_0v

0

​  v, start subscript, 0, end subscript of the roller coaster.