Water flows at 0.67 m/s through a 3.0 cm diameter hose that terminates in a 0.35 cm diameter nozzle. Assume laminar non-viscous steady-state flow. (a) At what speed does the water pass through the nozzle? 49.22 m/s (b) If the pump at one end of the hose and the nozzle at the other end are at the same height, and if the pressure at the nozzle is 1 atm, what is the pressure at the pump? atm I have found the answer for part A. I have set up an equation for part B: I have to find P1 P2 = 1 atm v1 = .67 m/s v2 = 49.22 m/s let D = density of water = 1000kg/m^3 P1 + 1/2 D * V1^2 = P2 + 1/2 D * V2^2 P1 = P2 + 1/2 D * V2^2 - 1/2 D * V1^2 P1 = 1atm + 1/2(1000kg/m^3) * (2422.6084 m/s) - 1/2(1000kg/m^3) * (.4489m/s) How do I convert this or get answer for P1 in terms of atm?