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\/(\\pi R^2)"

(a)

"v_1={\\rm 0.25\\: cm^3\/min\/(3.14*(6\\: cm)^2)}\\\\\n\\rm=2.2*10^{-3}\\:cm\/min=0.13\\: cm\/s"

(b)

"v_1={\\rm 0.25\\: cm^3\/min\/(3.14*(9\\: cm)^2)}\\\\\n\\rm=9.8*10^{-4}\\:cm\/min=0.059\\: cm\/s"


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