Velocity and kinetic energy relationship

Velocity and Kinetic Energy

velocity and kinetic energy relationship

Kinetic energy is frame dependent, just as velocity is. Momentum is proportional to velocity and is frame dependent too, just as velocity is. Now, looking at the. Velocity and Kinetic Energy. As the roller coaster train begins its descent from the lift hill, its velocity increases. This causes the train to gain kinetic energy, which. Originally Answered: How do I describe the relationship between velocity and kinetic energy in this graph? I'm not going to give you the answer outright, as you .

velocity and kinetic energy relationship

It is assumed that while a molecule is exiting, there are no collisions on that molecule. Effusion of gas molecules from an evacuated container.

This is where Graham's law of effusion comes in. It tells us the rate at which the molecules of a certain gas exit the container, or effuse. Thomas Graham, a Scottish chemist, discovered that lightweight gases diffuse at a much faster rate than heavy gases. Graham's law of effusion shows the relationship between effusion rates and molar mass.

According to Graham's law, the molecular speed is directly proportional to the rate of effusion. You can imagine that molecules that are moving around faster will effuse more quickly, and similarity molecules with smaller velocities effuse slower. Because this is true, we can substitute the rates of effusion into the equation below. This yields Graham's law of effusion.

It is important to note that when solving problems for effusion, the gases must contain equal moles of atoms. You can still solve the equation if they are not in equal amounts, but you must account for this. For example, if gas A and gas B both diffuse in the same amount of time, but gas A contains 2 moles and gas B contains 1 mole, then the rate of effusion for gas A is twice as much.

Since both gases are diatomic at room temperature, the molar mass of hydrogen is about 2. When you open a bottle of perfume, it can very quickly be smelled on the other side of the room.

This is because as the scent particles drift out of the bottle, gas molecules in the air collide with the particles and gradually distribute them throughout the air. Diffusion of a gas is the process where particles of one gas are spread throughout another gas by molecular motion. Diffusion of gas molecules into a less populated region.

In reality the perfume would be composed of many different types of molecules: Root Mean Square RMS Speed We know how to determine the average kinetic energy of a gas, but how does this relate to the average speed of the particles?

We know that in a gas individual molecules have different speeds. Collisions between these molecules can change individual molecular speeds, but this does not affect the overall average speed of the system. As we have seen demonstrated through effusion, lighter gas molecules will generally move faster than heavier gases at the same temperature.

But how do you determine the average speed, or velocity, of individual gas molecules at a certain temperature?

Kinetic Energy

One approach would be to mathematically average the speed of the particles. Let's say we have a gas that contains only three molecules. The mean speed is the average of the molecular speeds of the particles of a gas: The RMS speed of our gas is: There is another way of calculate average speed, as defined by the kinetic theory of gases.

velocity and kinetic energy relationship

Obviously real gas particles do occupy space and attract each other. These properties become apparent at low temperatures or high pressures. But in buddy B frame, the Earth is actually moving backward. The friction force of A acceleration increases the momentum of A, and this increase has to be stolen somewhere total momentum does not change.

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So it has to come from Earth, i. The missing part of the energy has been used to accelerate the Earth motion with respect to B. High speed is easy in space is it?

Taking the same problem to space, as much as make sense you will see whymay clarify some issues. But it is really quite different. The only way to accelerate is to exchange momentum with another mass. Other than direct use of the natural force fields such as gravitythe usual way to do that is by exhausting a reaction mass at speed from a rocket engine. You throw the mass one way, and you are pushed the other way with the inverse momentum variation.

Since you carry the mass, the trick is often to increase the exhausted momentum by using high speed with small massesso that you do not end up massless too quickly unless you are a top model.

You may have a screen dial telling you how much energy you spent. But that is of limited use because it will not tell you what you spent it on, as you are accelerating both your craft and the exhausted reaction mass.

Kinetic Energy

The repartition of energy is controled by momentum preservation so that we have in the initial inertial frame as is well known: We can now address the first question: The answer is almost yes in classical mechanicsprovided all other things on board are kept equal. What is true is that the craft will always gain the same amount of velocity with respect to its previous inertial frame, provided it spends its energy in the same way.

The sharing ratio remains open with the data provided here, and so does the velocity increase. Since all inertial frames have constant velocity with respect to each other, and since velocities add vectorially in classical mechanic, I guess all that remains true in all inertial frames.

Relation between momentum and kinetic energy

But as I said, this is true if all other things are kept equal on board. We assumed that somehow the ship mass is kept constant mass is maintained by transfer from another craft: The other thing that must be kept unchanged is the ratio of energy repartition between the ship acceleration and the exhausted mass acceleration, as measured in the inertial frame of the ship just before acceleration.

Of course, this inertial frame changes with each acceleration. For example, the cyclist could encounter a hill just high enough to coast up, so that the bicycle comes to a complete halt at the top.

Kinetic Theory of Gases

The kinetic energy has now largely been converted to gravitational potential energy that can be released by freewheeling down the other side of the hill.

Since the bicycle lost some of its energy to friction, it never regains all of its speed without additional pedaling. The energy is not destroyed; it has only been converted to another form by friction. Alternatively, the cyclist could connect a dynamo to one of the wheels and generate some electrical energy on the descent.

The bicycle would be traveling slower at the bottom of the hill than without the generator because some of the energy has been diverted into electrical energy. Another possibility would be for the cyclist to apply the brakes, in which case the kinetic energy would be dissipated through friction as heat.

Like any physical quantity that is a function of velocity, the kinetic energy of an object depends on the relationship between the object and the observer's frame of reference. Thus, the kinetic energy of an object is not invariant.

Spacecraft use chemical energy to launch and gain considerable kinetic energy to reach orbital velocity. In an entirely circular orbit, this kinetic energy remains constant because there is almost no friction in near-earth space.

However, it becomes apparent at re-entry when some of the kinetic energy is converted to heat.