Waves transfer energy but not matter
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.
Answer:
A. Kindly find attached free body diagram for your reference (smiles I guess I will make a terrible artist)
B. The collision is inelastic because both the husband and the wife moved together with same velocity as he grabs her on the waist
C. The general equation for conservation of momentum in terms of m 1, v 1, m 2, v 2, and final velocity vf
Say mass of husband is m1
Mass of the wife is m2
Velocity of the husband is v1
Velocity of the wife is v2
According to the conservation of momentum principle momentum before impact m1v1+m2v2 =momentum after impact Common velocity after impact (m1+m2)vf
The momentum equation is
m1v1+m2v2= (m1+m2)vf
D. To solve for vf we need to make it subject of formula
vf= {(m1v1) +(m2v2)}/(m1+m2)
E. Substituting our given data
vf=
{(1570*58)+(2550*54)}/(1570+2558)
vf=91060+137700/4120
vf=228760/4120
vf=55.52m/s
Their speed after collision is 55.52m/s
Answer:
the velocity of the car is 0.875 m/s
Explanation:
therefore the V of car is 0.875 m
Using kinematic equation, v^2 - u^2 = 2as. 5^2 - 3^2 = 2a x 16. a = 0.5m/s^2. So particle will deaccelerate at 0.5m/s^2. ( v = final velocity, u= initial velocity, a= acceleration, s= displacement.)
