$q = \frac{1}{2} n v^{2}$
The dynamic pressure q generated by a fluid moving with velocity v can be found using the formula above, where n is the constant density of the fluid. An aeronautical engineer uses the formula to find the dynamic pressure of a fluid moving with velocity v and the same fluid moving with velocity 1.5v. What is the ratio of the dynamic pressure of the faster fluid to the dynamic pressure of the slower fluid?

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Solution

For the first fluid the pressure $q_{1} = \frac{1}{2} n {v_{1}}^{2}$
For the first fluid the pressure $q_{2} = \frac{1}{2} n {v_{2}}^{2}$
But $v_{2} = 1.5 v_{1}$
Thus, $q_{2} = \frac{1}{2} n {(1.5 v_{1})}^{2}$
$=> q_{2} = \frac{1}{2} \times \frac{9}{4} \times n {(v_{1})}^{2} = \frac{9}{4} \times q_{1}$
Required ratio $= \frac{q_{2}}{q_{1}} = \frac{9}{4}$

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For the first fluid the pressure q1=12nv12For the first fluid the pressure q2=12nv22But v2=1.5v1Thus, q2=12n(1.5v1)2=>q2=12×94×n(v1)2=94×q1Required ratio =q2q1=94