Answer to Question #337530 in Physics for Vince

Question

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

Answer 1

$$v=Q/A=Q/(\pi R^2)$$

(a)

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

(b)

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