Question

If the density of water is 1,00 g/cm° and atmospheric pressure is 1,013 x 105 Pa, the depth in water (in m) where the absolute pressure is four times atmospheric

pressure, is equal to:

Answer 1

$$p=p_a+g\rho h$$

$$4p_a=p_a+g\rho h$$

$$h=\frac{3p_a}{g\rho}$$

$$h=\frac{3*1.013*10^5}{9.8*1000}=31\:\rm m$$