Try to use P2-P1 = ρ/2(v1^2-v2^2)
Derived from P1 + 1/2ρv1^2 = P2 + 1/2ρv2^2
Then A1V1 = A2V2 is from continuity and can be expressed V1 = A2V2/A1
You get the equation P2 - P1 = ρ/2((A2V2/A1)^2 - V2^2) via substitution
And then simplify to 2(P2-P1)/ρ = A2V2/A1 - V2^2
Plug in and solve
A venturi meter is a device for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at a speed v2 through a horizontal section of pipe whose cross-sectional area A2 = 0.0600 m2. The gas has a density of ñ = 1.70 kg/m3. The Venturi meter has a cross-sectional area of A1 = 0.0400 m2 and has been substituted for a section of the larger pipe. The pressure difference between the two sections is P2 - P1 = 140 Pa.
(a) Find the speed v2 of the gas in the larger original pipe.
(b) Find the volume flow rate Q of the gas.
anything would help. i have asked everyone and they don't know either.
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