Answer to Question #337530 in Physics for Vince

Question #337530

water flow at a rate of 0.25 cm³/min in a horizontal pupe of varying cross sectional areas.Calculate the velocity of the water at a section where the rafius of the pupe is 6cm and (b) at a section where the rafius is 9cm

1
Expert's answer
2022-05-05T13:56:45-0400
v=Q/A=Q/(πR2)v=Q/A=Q/(\pi R^2)

(a)

v1=0.25cm3/min/(3.14(6cm)2)=2.2103cm/min=0.13cm/sv_1={\rm 0.25\: cm^3/min/(3.14*(6\: cm)^2)}\\ \rm=2.2*10^{-3}\:cm/min=0.13\: cm/s

(b)

v1=0.25cm3/min/(3.14(9cm)2)=9.8104cm/min=0.059cm/sv_1={\rm 0.25\: cm^3/min/(3.14*(9\: cm)^2)}\\ \rm=9.8*10^{-4}\:cm/min=0.059\: cm/s


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