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You have to think according to the situation. Consider a tank with high pressure, if we open the tap means water flows with high velocity and eventually after sometime the velocity will start to reduce because of pressure is decreasing accordingly. Velocity and pressure is directly proportional.

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I'll answer the easy question first -- you are looking at it the right way. Now for the other question... it's really impossible to say that "most" flows are locally compressible. Although that's also a lie because every flow is compressible! Incompressibility is an approximation that makes the math easier, but even the slowest flows of air are technically ...

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If you neglect viscosity, Bernoulli's equation (just Navier-Stokes without frictional or stress terms) will get you into the ballpark: $$P_g + \frac{1}{2}\rho_g v_g^2 = P_a$$ Where the $g$ subscripts pertain to the gas and the $a$ subscript to the ambient. The gas density $\rho_g \equiv M / V$ is the ratio of the mass of gas (M) in the tank to the volume ...

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I think it's a good question. Without delving into numerical calculation, I assume the lowest pressure above the wing is in the area where the curvature of flow is greatest. As far as the upwash, I'm not sure, but the leading air may be pulled upward by the reduced pressure above. ADDED: This is essentially what @Floris meant in his comment, referring to ...

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A body moving in a fluid will pass through fluid at a rate of $Av$ where $A$ is the cross-sectional area and $v$ the velocity. This clearly has units of volume per time. If the fluid is brought to rest relative to the body, the momentum transferred per unit time is $A\rho v^2$. But momentum transfer per unit time is force, so we expect that the drag force ...

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Simplified explanation: When you move fast enough, you create a "turbulent wake". It is as though the column of air that you push out of the way ends up traveling at your speed. This requires energy proportional to $v^2$ since the kinetic energy of that column will be proportional to $\frac12\rho v^2$. Thus the work done needed per unit distance is ...

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