Question

A loon is a diving bird equally at home "flying" in the air or water. What swimming velocity under water will produce a dynamic pressure equal to that when it flies in the air at 35 mph?

Answer 1

Answer:

The swimming velocity under water is 1.23 mph

Explanation:

dynamic pressure $= \frac{\rho v^2}{2}$

where;

ρ is density of the fluid

v is the velocity of the fluid

$\frac{\rho _wV^2}{2} = \frac{\rho _aV^2}{2}$

ρw is the density of water = 1.94 slug/ft³

ρa is the density of air = 0.00238 slug/ft³

given velocity of air in mph = 35 mph, converting to ft/s is as follows

$35.\frac{miles}{hour}(\frac{5280 .ft}{1.mile})(\frac{1.hour}{60.mins})(\frac{1.min}{60.sec}) = 51.33\frac{ft}{s}$

From this equation $\frac{\rho _wV^2}{2} = \frac{\rho _aV^2}{2}$, we determine the swimming velocity under water.

$\frac{1.94 *V^2}{2} = \frac{0.00238*(51.33)^2}{2}\\\\0.97*V^2 = 3.135\\\\V^2 = \frac{3.135}{0.97} = 3.232\\\\V = \sqrt{3.232} \\\\V = 1.798\frac{ft}{s}$

in mph; $1.798\frac{ft}{s} (\frac{1 .mile}{5280.ft})(\frac{3600.s}{1.hour}) =1.23 mph$

Therefore, the swimming velocity of the loon under water that will produce a dynamic pressure equal to that when it flies in the air is 1.23 mph