91Ó°ÊÓ

1. The depth at which the special–op soldier must dive in seawater,$d=20m$
2. Height of parachute,$h=7600m$
3. Average density of air,$p=0.87kg/m^3$

We can find the pressure at depth d and height h using the formula for pressure in terms of. The difference between these two pressures gives the change in pressure on a special–op soldier.

Formula:

Pressure at a height hof a substance, $p=pgh$

We can find the gauge pressure at the given height and depth.

Calculation for gauge pressure at depthusing equation (i):

$$\begin{gathered} p_{1=}1024×9.8×20\left(âˆ\mu Densityofseawater=1024kg/m3\right) \\ =2.01×{10}^{5}pa \end{gathered}$$

Calculation for gauge pressure at heighth =7600m using equation (i):

$$\begin{gathered} p_{2=}0.87×9.8×7600 \\ =6.48×{10}^{4}pa \end{gathered}$$

Thus, the change in the pressure can be given by:

$$\begin{gathered} ∆p=p_{1-}{p}_2 \\ =(2.01×{10}^5)-(6.48×{10}^4) \\ =1.362×{10}^5 \\ ~1.4×{10}^{5}pa \end{gathered}$$

Therefore, the change in pressure on a special –op soldier is$1.4×{10}^{5}pa$