Which front is a head on collision or neither warm nor cold air moves

1 answer:

3 0

Stationary wave fronts are formed when two air masses of different densities and temperature range push each other but unable to move the front in any direction.

<u>Explanation:</u>

Weather Fronts

When collision happens between air masses of two with varied densities and the temperature, they form a front. These air masses also differ in humidity. When a cold front and warm front generated and push each other equally, the net force turns to be zero and there is no movement in any of the front.

By this, both of the fronts form a stationary row which separates both the fronts. Both kind of winds flow perpendicular to each other rather than parallel and this what holds the stationary front in place.

If the winds change the directions, the front may also move in the direction of the wind which is dominating the the other one. The weather seems to be often cloudy with this front. Hence, in the state of the formation of a stationary front, neither warm nor cold air moves parallel.

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In which city did the wright brothers take off on their very first flight

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If two particles have equal kinetic energies, are their momenta necessarily equal? explain.

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Answer:

No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

$K.E=\frac{1}{2}mv^{2}$

Similarly the momentum is given by $m\times v$

For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective kinetic energies is given by

$K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}$

$K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}$

Similarly For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective momenta are given by

$p_{1}=m_{1}\times v_{1}$

$p_{2}=m_{2}\times v_{2}$

Now since it is given that the two kinetic energies are equal thus we have

$\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)$

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

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The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres

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Answer:0.0704 kg

Explanation:

Given

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