## College Physics (4th Edition)

(a) The pressure increase at a depth of 35.0 m below the surface is $343~kPa$ (b) The pressure decrease in going 35 m above the surface is $412~Pa$
(a) The pressure increase is the gauge pressure at a depth of 35.0 m below the surface: $P_g = \rho~gh$ $P_g = (10^3~kg/m^3)(9.80~m/s^2)(35.0~m)$ $P_g = 343\times 10^3~N/m^2$ $P_g = 343~kPa$ The pressure increase at a depth of 35.0 m below the surface is $343~kPa$ (b) To find the pressure decrease in going 35 m above the surface, we can find the gauge pressure for a height difference of 35 m: $P_g = \rho~gh$ $P_g = (1.20~kg/m^3)(9.80~m/s^2)(35~m)$ $P_g = 412~N/m^2$ $P_g = 412~Pa$ The pressure decrease in going 35 m above the surface is $412~Pa$.