\[250 \, \text{Н/см}^2 = 250 \cdot 10^4 \, \text{Па} = 2500000 \, \text{Па}\]
\[h = \frac{P}{\rho g} = \frac{2500000}{1000 \cdot 9.81} \approx 254.84 \, \text{м}\]
\[F = PA = 2500000 \cdot 0.45 = 1125000 \, \text{Н} = 1125 \, \text{кН}\]
Ответ: Глубина 254.84 м, сила 1125 кН