A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 %

(A) velocity = 2.8 m/s (B) Average force = 1.9536 Newtons.

Answer:

Option b. it has the same position and the same acceleration as A

Explanation:

Now, let's examine each statement:

a. it has the same position and the same velocity as A

At the moment B overtakes A, they share the same location. However, their velocities cannot match, or else ball B wouldn't be able to pass ball A. Thus, this statement is false.

b. it has the same position and the same acceleration as A

As established before, their positions coincide, and both balls are subjected to gravitational acceleration, making this statement true.

c. it has the same velocity and the same acceleration as A

While the accelerations are identical, their velocities differ, rendering this statement false.

d. it has the same displacement and the same velocity as A

Though they have traveled the same distance, their velocities are not the same, thus this statement is false.

e. it has the same position, displacement and velocity as A

While their position and displacement match, their velocities do not, so this statement is false.

Therefore, only option b is correct.

Answer:

A. Yes, the ball clears the crossbar by 2.83 meters

Explanation:

This situation pertains to projectile motion.

The horizontal velocity component of the ball is calculated as 26 cos 35 = 21.3 m/s

The vertical velocity component is 26 sin 35 = 14.9 m/s

The time taken to travel the horizontal distance to the goalpost, which is 54.9 m, is:

= distance / horizontal speed

= 54.9 / 21.3

= 2.577 seconds.

The vertical distance achieved during this time is:

h = ut - 1/2 gt², where u is the initial vertical velocity, and t = 2.577 seconds.

h = 14.9 x 2.577 - 0.5 x 9.8 x (2.577)²

= 38.39 - 32.54

= 5.85 m

Thus, the ball surpasses the crossbar by 5.85 - 3.05 = 2.8 m

Speed is defined as distance over time. Hence, to determine the distance, we use d = V * t. Plugging in the values yields d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km. Thus, during this distracted period, a distance of 0.08Km was covered.

Answer:

The primary factors that influence the intensity of shaking during an earthquake include the depth of the quake, its distance from the fault line, the type of soil below, and specifics about the buildings—especially their height. We will focus on the last two (soil quality and structures) and their interaction.