Answer:
66.35m/s
Explanation:
Para resolver el ejercicio es necesario la aplicación de las ecuaciones de continuidad, que expresan que
[tex]A_1V_1 =A_2 V_2[/tex]
From our given data we can lower than:
[tex]R_i = \frac{0.24}{2} = 0.12m[/tex]
[tex]R_f = \frac{0.05}{2} = 0.025m[/tex]
So using the continuity equation we have
[tex]A_1V_1 =A_2 V_2[/tex]
[tex]V_2 = \frac{A_1V_1}{A_2}[/tex]
[tex]V_2 = \frac{(\pi(0.12^2))(2.88)}{(\pi (0.25)^2)}[/tex]
[tex]V_2 = 66.35m/s[/tex]
Therefore the velocity at the exit end is 66.35m/s