# Answer to Question #20528 in Mechanics | Relativity for Sarah

Question #20528

A submarine is totally submerged and moving upward. Its density is 700Kg/m^3. It was at rest 150 meters below the sea level. Find pressure on it after 3 seconds. (density of sea water is 1025 Kg/m^3).

Expert's answer

First we find what distance will the submarine pass in this 3 seconds upward

To do that we need to find acceleration of it.

Here it is

a=F/m = (F_a - mg)/m = (\rho_water * V * g - \rho_submarine * V * g)/\rho_submarine * V = g*(\rho_water/rho_submarine- 1) = g(1025/700 - 1) = 4.55 m/s^2

Now the distance is

s=at^2/ = 20.48 m

hence, the depth is 150-20.5 = 129.5

so, the pressure is \rho_water * g * h = 129.5*9.8*1025 = 1 300 827.5 Pa

To do that we need to find acceleration of it.

Here it is

a=F/m = (F_a - mg)/m = (\rho_water * V * g - \rho_submarine * V * g)/\rho_submarine * V = g*(\rho_water/rho_submarine- 1) = g(1025/700 - 1) = 4.55 m/s^2

Now the distance is

s=at^2/ = 20.48 m

hence, the depth is 150-20.5 = 129.5

so, the pressure is \rho_water * g * h = 129.5*9.8*1025 = 1 300 827.5 Pa

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