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Brums [2.3K]
3 years ago
5

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

ame location as Rock X. Both rocks fall for several seconds before landing on the ground directly below the cliff. Frictional forces are considered to be negligible. After Rock Y is released from rest several seconds after Rock X is released from rest, what happens to the separation distance S between the rocks as they fall but before they reach the ground, and why? Take the positive direction to be downward.
A) Sis constant because at the moment Rock Y is released, the only difference between the rocks is their difference in height above the ground.
B) S is constant because the difference in speed between the two rocks stays constant as they fall.
C) S increases because the difference in speed between the two rocks increases as they fall.
D) S increases because at all times Rock X falls with a greater speed than Rock Y
Physics
1 answer:
MArishka [77]3 years ago
5 0

Answer:

<h2>D) S increases because at all times Rock X falls with a greater speed than Rock Y</h2>

Explanation:

As we know that Stone X is dropped first and then after some time stone Y is dropped

Now we have

v_x = gt_1

similarly for Y

v_y = g(t_1 - \Delta t)

now the relative acceleration of two stones while they are in air is ZERO

also we know from above equations

v_x > v_y

so the separation between two stones will keep on increasing as the speed of X is more than the speed of Y

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tamaranim1 [39]

the friction force provided by the brakes is 30000 N.

<h3>What is friction force?</h3>

Friction force is the force that opposes the motion between two bodies in contact.

To calculate the average friction force provided by the brakes, we apply the formula below.

Formula:

  • K.E = F'd............. Equation 1

Where:

  • K.E = Kinetic energy of the train
  • F' = Friction force provided by the brakes
  • d = distance

Make F' the subject of the equation

  • F' = K.E/d............ Equation 2

From the question,

Given:

  • K.E = 8.1×10⁶
  • d = 270 m

Substitute these values into equation 2

  • F' = (8.1 ×10⁶)/270
  • F' = 30000 N

Hence, the friction force provided by the brakes is 30000 N

Learn more about friction force here: brainly.com/question/13680415

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3 years ago
30. A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pu
Mnenie [13.5K]

Answer:

84.05

Explanation:

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