in a tornado the wind velocity in meters per second can be described by the function v(p)=12.3 sqrt 1134-3p where 'p' is th air pressure in millibars. What is the air pressure of a tornado in which the wind velocity is 49.3 meters per second?
We are given with the function v(p) = 12.3 √(1134 - 3p) We are asked to find the air pressure when the velocity is 49.3 m/s Substituting this to v(p) 49.3 = 12.3 √(1134 - 3p) 49.3/12.3 = √(1134 - 3p) 4.01 = √(1134 - 3p) 4.01² = √(1134 - 3p)² 16.07 = 1134 - 3p 3p = 1134 - 16.07 3p = 1117.92 p = 1117.92/3 p = 372.64 millibars
