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

this could be newton#'s law of cooling. the cooling curve is exponential

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this