Explain why air masses do not mix

1 answer:

3 0

Different densities have to have a reason - different pressure and/or humidity etc. If there is a different pressure, there is a mechanical force that preserves the pressure difference: think about the cyclones that have a lower pressure in the center. The cyclones rotate in the right direction and the cyclone may be preserved by the Coriolis force.

If the two air masses differ by humidity, the mixing will almost always lead to precipitation - which includes a phase transition for water etc. It's because the vapor from the more humid air mass gets condensed under the conditions of the other. You get some rain. In general, intense precipitation, thunderstorms, and other visible isolated weather events are linked to weather fronts.

At any rate, a mixing of two air masses is a nontrivial, violent process in general. That's why the boundary is called a "front". In the military jargon, a front is the contested frontier of a conflict. So your idea that the air masses could mix quickly and peacefully - whatever you exactly mean quantitatively - either neglects the inertia of the air, a relatively low diffusion coefficient, a low thermal conductivity, and/or high latent heat of water vapor. A front is something that didn't disappear within minutes so pretty much tautologically, there must be forces that make such a quick disappearance impossible.

You might be interested in

If the forces acting on an system produce a net force, what can we say about these forces

lutik1710 [3]

Net force refers to the (vector) sum of all the forces acting on something. It's a mathematical construction, so it's not a single identifiable force. The forces themselves are real, but the net force is not an actual force.

Hope this helps you out.

8 0

3 years ago

Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw

Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0

3 years ago

A submerged submarine is stationary. The engines are put on maximum power. The submarine moves forward. The engines maintain max

densk [106]

This is EXACTLY the same scenario as the skydiver jumping
out of the airplane, except the whole thing is turned on its side.

==> The skydiver leaves the airplane.
The force of gravity on him (his weight) makes him accelerate down.
But the air resists his downward motion.
The faster he falls, the more UPWARD force the air exerts on him.
The more upward force the air exerts, the less he accelerates down.
When his falling speed is great enough, he stops accelerating, and
falls with a constant speed. He calls that speed his 'terminal velocity'.

==> The submarine turns on its engines, at maximum power.
The force of the engines makes the sub accelerate forward.
But the water resists its forward motion.
The faster it moves, the more BACKWARD force the water exerts on it.
The more backward force the water exerts, the less it accelerates forward.
When the forward speed is great enough, it stops accelerating, and moves
with a constant speed. I don't know if they use the same term in submarines,
but you might say that speed is the 'terminal velocity' in water.

3 0

3 years ago

A 190 lb man riding a 30 lb bicycle rides up a hill 60 ft high in one minute. What is the horsepower developed during this time?

Semmy [17]

The answer is c

Explanation:

4 0

3 years ago

Dmitrij [34]

Take 10m/s^2 for the gravitational acceleration, as we know this is a free fall, we can use the equation: d=1/2*g*t^2

Substitute g=10m/s^2, t=5s, d=125m

7 0

3 years ago