Bernoulli Experiment- why is the pressure on the left greater? [closed]

Is the Pressure $$P_2$$ caused by a reaction force to the fluid-fluid collision, or is it the result of the collisions of the water molecules with the walls at the neck of this experiment. The part which confuses me is that $$P_2$$ needs to be less than $$P_1$$but $$P_2$$ is exerted over a longer distance and the pressure exerted by fluids is calculated by $$hdg$$. Is there a reason why $$P_2$$ is less than $$P_1$$?

closed as unclear what you're asking by Aaron Stevens, ZeroTheHero, GiorgioP, Jon Custer, Feynmans Out for Grumpy CatMar 7 at 0:15

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• hdg is only for hydrostatic fluids, FYI. Also, what do you mean "P2 is exerted over a longer distance"? We don't know anything about the distances over which those pressures are exerted, only some idea about the general area of the pipe cross sections. – JMac Mar 5 at 19:27
• Is there a reason why P2 is less than P1? At steady state flow, bear in mind $v_1 A_1=v_2A_2$, then derive pressures with Bernoulli. – Gert Mar 5 at 22:25
• The reason is conservation of energy. I feel that your question does not show much evidence of having tried to find out for yourself; this is a basic type of problem in fluid flow. – Andrew Steane Mar 5 at 22:42
• Pressure doesn't depend on the "distance over which the force is applied." Also $\Delta P=\rho g\Delta h$ only applies to the difference in pressure between different heights of a static fluid near Earth's surface... – Aaron Stevens Mar 5 at 22:42