Use the conditions provided in the previous problem. atmosphere statically stable at the base of the mountain, where pressure = 1000 hpa, potential temperature lapse rate of 3.1◦c per 50 hpa. atmosphere statically unstable at the top, where pressure = 850 hpa, potential temperature lapse rate of -0.61◦c per 50 hpa. find the change in relative vorticity of the column of air.

change in relative vorticity  0.0590

Explanation:

Given data

pressure = 1000 hPa

temperature lapse rate q1 = 3.1◦C  per 50 hPa

pressure = 850 hPa

temperature lapse rate q2= -0.61◦C per 50 hPa

to find out

change in relative vorticity

solution

we will apply here formula that is

N = (g /  potential temperature ) × (potential vertical temperature) × exp^1/2    1

here we know g = 9.8 m/s

and q1 = potential temperature=3.3 degree celsius

potential vertical temperature gradient = 3.1 - 0.61  / 1000 -850

potential vertical temperature gradient = 0.0166 degree celsius/hpa

so

N = 9.8 / 2.75 × 0.0166 × exp^1/2

N = 0.0590

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explanation:

change tension