Water flows at 12 m/s in a horizontal pipe with a pressure of 3.0 104 n/m2. if the pipe widens to twice its original radius, what is the pressure in the wider section?

Total pressure is constant and PT = P = 1/2*ρ*v^2 So p1 + 1/2*ρ*v1^2 = p2 + 1/2*ρ*v2^2 and from continuity we have ρ*A1*v1 = ρ*A2*v2 So v2 = v1*A1/A2 since r2 = 2r1 then A2 = 4A1 so v2 = v1/4 Now from above p2 = p1 + 1/2*ρ*v1^2 - 1/2*ρ*v2^2 = p1 + 1/2*ρ*v1^2 - 1/2*ρ*v1^2/4^2 So p2 = 3.0x10^4Pa + 1/2*1000kg/m^3*(12m/s)^2 - 1/2*1000kg/m^3*12^2/16 = 9.75x10^4Pa = 9.8x10^4Pa