By what factor is the total pressure greater at a depth of 810 m in water than at the surface where pressure is one atmosphere? (water density = 1.0 × 10³ kg/m³,1 atmosphere pressure = 1.01 × 10⁵ N/m²,and g = 9.8 m/s²) A) 8 B) 79 C) 89 D) 69 E) 7 938

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The total pressure at a depth of 810 m in water is the sum of the atmospheric pressure and the pressure due to the water column. The pressure due to the water column is given by:$$P = ho g h$$where:* P is the pressure (N/m^2)* ρ is the density of the water (kg/m^3)* g is the acceleration due to gravity (m/s^2)* h is the depth of the water (m)Substituting the given values into the equation, we get:$$P = (1.0 	imes 10^3 	ext{ kg/m}^3)(9.8 	ext{ m/s}^2)(810 	ext{ m}) = 7.9 	imes 10^6 	ext{ N/m}^2$$The total pressure at a depth of 810 m in water is therefore:$$P_{total} = P_{atm} + P_{water} = 1.01 	imes 10^5 	ext{ N/m}^2 + 7.9 	imes 10^6 	ext{ N/m}^2 = 8.0 	imes 10^6 	ext{ N/m}^2$$The total pressure at a depth of 810 m in water is therefore 79 times greater than the pressure at the surface.

By What Factor Is the Total Pressure Greater at a Depth