Question

In a thunderstorm, the wind velocity in meters per second can be described by the function v(p) = 5.7 √998 - 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?

Answer 1

Answer:

923.1927 millibars

Step-by-step explanation:

The wind velocity is given by the function:

$v(p)=5.7\sqrt{(998-p}$

#We make p the subject of the formula:

$v(p)=5.7\sqrt{(998-p}\\\\v=5.7\sqrt{(998-p}\\\\\frac{v}{5.7}=\sqrt{(998-p}\\\\\frac{v^2}{5.7^2}=998-p\\\\p=998-\frac{v^2}{5.7^2}$

#Substitute the given values in the function above to solve for p;

$p=998-\frac{v^2}{5.7^2}\\\\=998-\frac{49.3^2}{5.7^2}\\\\=923.1927\ millibars$

Hence, the air pressure is 923.1927 millibars.