Problem 1. The wind near the surface is westerly, with a velocity of 6 m/s. The

Question

Problem 1. The wind near the surface is westerly, with a velocity of 6 m/s. The wind at 6 km height in the atmosphere is southerly with a velocity of 14 m/s. Calculate the magnitude of the wind shear between these two levels. Estimate the direction of the wind shear vector as a compass direction (e.g., ENE for a shear vector pointing eastnorth east). Suppose the CAPE is 1500 J/kg. What is the value of the bulk Richardson number, R? Based on this, what type of thunderstorm would you expect at this location? ( distance above the earth is not requrired )

Problem 2. For the case in problem 1, which side of a mature thunderstorm cell would a new cell be more likely to form? Explain in two or three sentences why this is the case.

answer for problem B

Explanation / Answer

Answer 1) The wind shear r between the two levels is 14-6m/s= 8m/s.

The wind shear vector between this two levels is 8m/s SW

Richardson number = CAPE/SHEAR= 1500/8=189.99=190

Supercell storm is going to develop based on the Richardson number.

Answer 2) In case of a Supercell thunderstorms a new cell forms bove the previous. This is because we define a Supercell thunderstorms with a deep rotating updraft. This storm is very harmful to people and is referred to as rotating thunderstorms.

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