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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 $P_1$?

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

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    $\begingroup$ 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. $\endgroup$ – JMac Mar 5 at 19:27
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    $\begingroup$ 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. $\endgroup$ – Gert Mar 5 at 22:25
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    $\begingroup$ 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. $\endgroup$ – Andrew Steane Mar 5 at 22:42
  • $\begingroup$ 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... $\endgroup$ – Aaron Stevens Mar 5 at 22:42
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Because the pipe on the right is smaller, but the same amount of mass flows per second through both sides, the flow on the right side is faster.

For the velocity of a bit of fluid to change as it moves, there must be a net force on it.

That net force come from the difference in pressure.

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