91Ó°ÊÓ

1. Depth of the surface, d = 100m
2. Density of the ocean water, $\mathrm{p}=1024kg/m^3$
3. Dimension of the pop-out hatch are$1.2m×0.60m$
4. Internal pressure,$P_i=1.00atm$

Using the formula of pressure, we can find the pressure at the given depth. Pressure is force per unit area. Using this, we can find the force required that must be applied to open the hatch.

Formulae:

Pressure applied on a fluid surface,$P_{gauge}=\mathrm{p}gd$ (i)

Pressure applied by force F on area A,

$p=\frac{F}{A}$ (ii)

Comparing equations (i) & (ii), we get

$$\begin{gathered} P_{gauge}=\frac{F}{A} \\ F=pgd×A \\ =\left(1024\frac{kg}{m^3}\right)\left(9.8\frac{m}{s^3}\right)\left(100m\right)(1.2×0.60m)(âˆ\mu p=1024\frac{kg}{m^3}\&A=1.2m×0.60m) \\ =7.2×{10}^{5}N \end{gathered}$$

Therefore, the force which must be applied to a pop-out hatch, which is by to push it out at that depth is$7.2×{10}^{5}N$