Question

3. In a thunderstorm, the wind velocity in meters per second can be described by the function

v(p)=5.7998−p‾‾‾‾‾‾‾√,

where

p

is the air pressure in millibars. What is the air pressure of a thunderstorm in which the wind velocity is 49.3 meters per second? Round your answer to the nearest tenth of a millibar.

Answer 1

p = 923.2 millibar

Explanation:

Given that,

The wind velocity can be described by the function as follows :

`v(p)=5.7√(998-p)`

Where

p is the air pressure in millibar

Put v = 49.3 m/s in the above function.

`49.3=5.7√(998-p)`

Squaring both sides,

`49.3^2=(5.7√(998-p))^2\\\\2430.49=32.49(998-p)\\\\(2430.49)/(32.49)=(998-p)\\\\74.80=(998-p)\\\\p=998-74.80\\\\p=923.2\ mbar`

So, the required pressure is equal to 923.2 millibar.