2 years ago

Q. Why do molecules of an object moving faster when dropped to the ground?

Physics

2 answers:

bija089 [108]2 years ago

6 0

Answer:

the cause of variation gravity

Levart [38]2 years ago

3 0

Answer:

Heavier Objects Don't Hit the Ground First

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Double the amount of people because the ramp went from 10-meter to 20-meter, so times by 2. 4 times 2 is 8.
The answer is 8 people.
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1 year ago

On a playground, there is a merry‑go‑round. In order to get it moving, Bonnie applies a force of 31 N31 N . The merry‑go‑round m

nika2105 [10]

Answer:

The magnitude of the torque is 263.5 N.

Explanation:

Given that,

Applied force = 31 N

Distance from the axis = 8.5 m

She applies her force perpendicularly to a line drawn from the axis of rotation

So, The angle is 90°

We need to calculate the torque

Using formula of torque

$\tau=Fd\sin\theta$

Where, F = force

d = distance

Put the value into the formula

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2 years ago

A test car travels in a straight line along the x-axis. The graph in the figure shows the car’s position x as a function of time

svetlana [45]

If it is s-t graph , point is c
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2 years ago

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How could you make a sound wave sound louder

kykrilka [37]

If you clap your hands, the shock causes the air around your hands to begin vibrating. When air particles vibrate, they bump into other particles near them. Then these particles begin to vibrate and bump into even more air particles. When the air particles begin vibrating the air inside your ear, you hear a sound.

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2 years ago

CHEGG You stretch a spring with spring constant k = 1.2x104 N/m to extend 6.0 cm away from its equilibrium position. How much do

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Answer:

The elastic potential energy of the spring change during this process is 21.6 J.

Explanation:

Given that,

Spring constant of the spring, $k=1.2\times 10^4\ N/m$

It extends 6 cm away from its equilibrium position.

We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :

$E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 1.2\times 10^4\times (0.06)^2\\\\E=21.6\ J$

So, the elastic potential energy of the spring change during this process is 21.6 J.

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2 years ago