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4 years ago

In which state of matter has the least kinetic energy

Physics

2 answers:

Savatey [412]4 years ago

8 0

Solids have the least kinetic energy

steposvetlana [31]4 years ago

8 0

Answer:

Solid state

Explanation:

Kinetic energy is the energy associated with the motion, in a solid particles are close together so they don't move much, inside a solid the electrons of the atoms vibrate constantly but they are fixed in their position because of how the structure of the atoms behaves in this state. In conclusion, the solid-state has some kinetic energy associated with the vibration of electrons but this energy is much smaller than in liquids or gases.

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A car (mass 1200 kg, speed 100 km/h) and a truck (mass 2800 kg, speed 50 km/h) are moving in the same direction along a highway.

Sloan [31]

Answer:

Speed of the wreck after the collision is 65 km/h

Explanation:

When a car hits truck and sticks together, the collision would be totally inelastic. Since the both the vehicles locked together, they have the same final velocity.

Mass of car $m_{1}=1200 kg$

Mass of truck $m_{2}=2800 kg$

Initial speed of the car $u_{1}=100 km /h$

Initial speed of the truck $u_{2}=50 km /h$

The final velocity of the wreck will be

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

since final speed are same, $v_{1}=v_{2}=v$

$m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v$

$1200\times 100 +2800 \times 50 =(1200+2800)v\\v=\frac{260000}{4000} \\v=65 km/h$

5 0

3 years ago

Why do the earth have many states

IrinaVladis [17]

Landforms pushing up mountains and hills. water and wind can wear down land and create valleys and canyons

6 0

3 years ago

What advantage do weather do weather satellites have over ground based weather stations

docker41 [41]

Answer:

Satellites can gather weather data from much higher altitudes than land-based instruments can.

3 0

3 years ago

Two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a certain re

anzhelika [568]

Complete Question

Two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.

A) Which of the following statements about the motion of the electrons in the given reference frame will be true at the instant the two speeds reach their separations?

A) Electrons A is moving faster than electron B.

B) Electron B is moving faster than electron A.

C) Both electrons are moving at the same (nonzero) speed in the opposite direction

.D) Both electrons are moving at the same (nonzero) speed in the same direction.

E) Both electrons are momentarily stationary.

2) What is the minimum separation $r_{min}$ that the electrons reach?

Answer:

The correct option is E

$r_{min} = \frac{kq^2}{4 mv^2}$

Explanation:

From the question we are told that

The mass of each electron is m

The charge of each electron is q

The speed of electron A is v

The speed of electron B is 3v

Generally at their point of separation the repulsion force is equal to the force that is propelling the electrons due to this the electrons are momentarily stationary

Generally the total initial kinetic energy of both electron is mathematically represented as

$K_{inT} = K_A + K_B$

=> $K_{inT} = \frac{1}{2}m (v)^2 + \frac{1}{2} m (3v)^2$

=> $K_{inT} = \frac{1}{2} (mv^2 + 9v^2m)$

=> $K_{inT} = 5mv^2$

Generally the total final kinetic energy of both electron is mathematically represented as

$K_{fT} = \frac{1}{2} *m * v^2 + \frac{1}{2} *m * v^2$

Here v is the velocity due to the repulsion force

$K_{fT} = mv^2$

Generally the final potential energy of the both electrons is

$P_f = \frac{ k * q^2}{r_{min}}$

Here k is the coulombs constant

So according to energy conservation law

$K_{inT} = K_{fT} + P_f$

=> $5mv^2 = mv^2 + \frac{ k * q^2}{r_{min}}$

=> $r_{min} = \frac{kq^2}{4 mv^2}$

3 0

3 years ago

At the top of a roller coaster 20 m above the ground the 5 kg car is moving at 2 m/s how fast is it moving at the bottoms of the

krek1111 [17]

Answer:

VB = 19.9 [m/s]

Explanation:

In order to solve these problems, we must use the principle of energy conservation. Which tells us that the energy between two points after an instant of time must be equal. This is we have two points A & B, in point A is the highest point while Point B is the lowest

Now we need to identify the types of energy at each point.

<u>For point A</u>

At this point we have two energies, the kinetic energy since the roller coaster moves at a speed of 2 [m/s], in the same way there is potential energy since the roller coaster is 20 [m] above ground level.

<u>For point B</u>

At Point B we only have kinetic energy, since it is located at zero meters with respect to the ground. In this way we can determine the velocity at this point.

And the energy is expressed by means of the following expression:

$E_{A}=E_{B}\\\frac{1}{2} *m*v_{A}^{2} +m*g*h_{A}=\frac{1}{2} *m*v_{B}^{2}$

where:

m = mass = 5 [kg]

VA = velocity of the roller coaster in point A = 2[m/s]

hA = elevation of the roller coaster = 20 [m]

vB = velocity of the roller coaster in point B [m/s]

Now replacing:

$\frac{1}{2}*5*(2)^{2} +(5*9.81*20)=\frac{1}{2}*(5)*v_{B}^{2}\\991=2.5*v_{B}^{2}\\v_{B}x^{2} =991/2.5\\v_{B}=\sqrt{396.4} \\v_{B}=19.9[m/s]$

4 0

3 years ago