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EleoNora [17]
4 years ago
8

What happens to the molecules when a gas is compressed?

Physics
2 answers:
Daniel [21]4 years ago
5 0

Answer: took the test. Answer is : There is a rise in temperature.

Explanation:

DerKrebs [107]4 years ago
3 0

Answer:

they rise in temperature

Explanation:

when there being compressed theres more pressure causing heat

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A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice
Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of \sqrt{3} v_{e}

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , v_{i}=2 v_{e}. Here the radius is R, so find the escape velocity as follows,

            \frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0

            \frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}

            v_{e}^{2}=\frac{2 G M}{R}

            v_{e}=\sqrt{\frac{2 G M}{R}}

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as,  U_{i}=-\frac{G M m}{R}

R tend to infinity when far away from the planet, so v_{f}=0

Then, kinetic energy at initial would be,

                  k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}

Similarly, kinetic energy at final would be,

                k_{f}=\frac{1}{2} m v_{f}^{2}

Here, v_{f}=\text { final velocity }

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

                 U_{i}+k_{i}=k_{f}+v_{f}

                 \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0

                  \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}

'm' and \frac{1}{2} as common on both sides, so gets cancelled, we get as

                   4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}

We know, v_{e}=\sqrt{\frac{2 G M}{R}}, it can be wriiten as \left(v_{e}\right)^{2}=\frac{2 G M}{R}, we get

                4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}

                v_{f}^{2}=3\left(v_{e}\right)^{2}

Taking squares out, we get,

                v_{f}=\sqrt{3} v_{e}

4 0
3 years ago
you are asked to calculate an object's velocity, in order to do so you must know the object's direction and speed. distance and
Anna11 [10]

You are asked to calculate an object's velocity.


In order to do so, you must know the object's speed and the direction

in which it's moving, or you must have other information that makes it

possible for you to calculate its speed and direction.

6 0
3 years ago
Read 2 more answers
05. The time required to complete one lap around a perfectly circular track having a radius of 1,835 meters is 86
likoan [24]

Answer:

v = 134.06 m/s

Explanation:

Given that,

Radius of a circular track is 1,835 m

Time required to complete one lap around a perfectly circular track is 86 seconds

We need to find the car's velocity. Velocity is equal to,

v=d/t

On circular path,

v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 1835}{86}\\\\v=134.06\ m/s

So, car's velocity is 134.06 m/s.

7 0
4 years ago
Define distance and displacement with illustration​
Deffense [45]

Answer:

<h2><em>Distance</em></h2>

<em>The </em><em>length</em><em> </em><em>of </em><em>the </em><em>actual </em><em>path </em><em>travelled by </em><em>a </em><em>body </em><em>is </em><em>called </em><em>distance </em><em>travelled </em><em>by </em><em>a </em><em>body.It </em><em>is </em><em>a </em><em>scalar </em><em>Quantity.</em><em>I</em><em>t</em><em> </em><em>is </em><em>measured</em><em> </em><em>in </em><em>meter(</em><em>m)</em><em> </em><em>in </em><em>SI </em><em>system.</em>

<h2><em>Displacement</em></h2>

<em>The </em><em>shortest </em><em>distance</em><em> </em><em>from </em><em>initial </em><em>position</em><em> </em><em>to </em><em>the </em><em>final </em><em>position</em><em> </em><em>of </em><em>a </em><em>body </em><em>is </em><em>called </em><em>displacement</em><em> </em><em>of </em><em>the </em><em>body.It </em><em>is </em><em>a </em><em>vector</em><em> </em><em>Quantity.</em><em>I</em><em>t</em><em> </em><em> </em><em>is </em><em>measured</em><em> </em><em>in </em><em>meter(</em><em>m)</em><em> </em><em>in </em><em>SI </em><em>system.</em><em>.</em>

<em>Please </em><em>see </em><em>the </em><em>attached </em><em>picture.</em><em>.</em><em>.</em>

<em>It </em><em>is </em><em>the </em><em>example </em><em>of </em><em>distance </em><em>and </em><em>displacement.</em><em>.</em><em>.</em><em>.</em>

<em>Hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em>

3 0
3 years ago
What is the power involved in lifting a 1.0-kg object 1.0 m in 1.0 s??
MAVERICK [17]
Mass  = 1kg 
Distance  = 1m 
Time = 1s
Force= Mass x Acceleration due to graviy
          =  1 x 9.8 = 9.8
Velocity = Distance  / time
              = 1 / 1 =1m/s
Power = Force x velocity
           = 9.8 x 1 = 9.8 W
5 0
3 years ago
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