A siphon tube is filled with gasoline and closed at each end. One end is inserted into a gasoline tank 0.20 m below the surface of the gasoline. The outlet is placed outside the tank at a distance 0.45 m below the surface of the gasoline. The tube has an inner cross-sectional area of 3.8 × 10-4 m2. The density of gasoline is 680 kg/m3. Ignoring viscous effects, what is the velocity of the gasoline in the tube shortly after the tube is opened? P1+.5*rho*v1^2+rho*g*y1=P2+.5*rho*v2^2+rho*g*y2 1 atm = 1.05E5 So I plugged in 1.05E5+0 (because v at top=0)+680*9.8*-.2 = 1.05E5+.5*680*v2^2+680*9.8*-.44 Completing this calculation gives v2= 2.17m/s which is incorrect. Any suggestions on what I am doing wrong? And to find the flow rate I would use the equation Q=Av and use the provided area given with the v discovered?