## Essential University Physics: Volume 1 (4th Edition) Clone

a) We use the equation for the depth to find: $h = \frac{p-p_0}{\rho g}= \frac{2.02\times 10^5 - 1.01\times 10^5}{1030\times 9.81}= 10 \ m$ b) We can now find the pressure: $= \frac{100}{10} \times 11 \times 10^ 3 = 110 \ MPa$