Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of t

Answer:

The books are displaced to the left due to inertia and ultimately halt when impacted by the car door.

Explanation:

The movement of the books can be understood through Newton's first two laws:

- The first law (Law of Inertia): an object will remain at rest or continue moving in a straight line unless an unbalanced force acts upon it.

- The second law: if unbalanced forces act on an object, it experiences an acceleration that can be described by the formula

where F is the object's net force, m its mass, and a its acceleration.

Now let's relate this to the scenario:

- When Argelia makes a sharp right turn, the books, which are not secured in the car, maintain their straight-line motion due to inertia so they appear to move left as the car shifts right.

- Upon contacting the car door, the books cease moving due to the second law: the door exerts an unbalanced force, causing the books to decelerate and ultimately come to rest.

Response:

3.5 N

Reasoning:

Taking the 0 cm position as the pivot point, to achieve balance, the total moment calculated around this point must equal zero. We will analyze the moment generated at each point, moving from 0 to 100 cm:

- Tension from the string at the 0 cm mark is 0, since the moment arm is nonexistent.

- A 2 N weight at the 10 cm point creates a moment of 20 Ncm moving clockwise.

- Another 2 N weight at the 50 cm position produces a moment of 100 Ncm also clockwise.

- The weight of the 1 N stick located at its center of mass (50 cm) has a moment of 50 Ncm, clockwise.

- A 3 N weight at the 60 cm position generates a moment of 180 Ncm, clockwise.

- The tension T (N) from the string at the 100 cm end contributes a moment of 100T Ncm, moving counter-clockwise.

Total clockwise moments = 20 + 100 + 50 + 180 = 350 Ncm.

Total counter-clockwise moment = 100T.

To achieve balance, we set 100 T = 350, leading to T = 350 / 100 = 3.5 N.

A. a<span> = 1.3 m/s^2</span><span>; </span>FN<span> = 63.1 N</span>

Refer to the diagram below.

Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².

The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²

Thus, the deceleration magnitude is 82 m/s².

Answer:

The duration, t = 3.53 seconds

Explanation:

The following information is provided:

The equation to calculate the time t is expressed as:

...... (1)

Where

s denotes the distance in feet

We are to determine the duration taken by the stone to fall a distance of 200 feet, where s = 200 feet

Substituting the value of s into equation (1) yields:

t = 3.53 seconds

Thus, the time taken by the object is 3.53 seconds, which provides the required answer.