Answer:
66.35m/s
Step-by-step explanation:
Para resolver el ejercicio es necesario la aplicación de las ecuaciones de continuidad, que expresan que
![A_1V_1 =A_2 V_2](https://img.qammunity.org/2020/formulas/physics/college/52zb8i2hixtjnv1jm7ffwlqjyt1sxv8ubr.png)
From our given data we can lower than:
![R_i = (0.24)/(2) = 0.12m](https://img.qammunity.org/2020/formulas/physics/college/42v7d663648zz0pa6xabowtumq1heonl3i.png)
![R_f = (0.05)/(2) = 0.025m](https://img.qammunity.org/2020/formulas/physics/college/3q7v50hf878of2eh4k41ub7v6cny10g29h.png)
So using the continuity equation we have
![A_1V_1 =A_2 V_2](https://img.qammunity.org/2020/formulas/physics/college/52zb8i2hixtjnv1jm7ffwlqjyt1sxv8ubr.png)
![V_2 = (A_1V_1)/(A_2)](https://img.qammunity.org/2020/formulas/physics/college/9zspbpm4d3ojvmpgztxt8a9uvppv45hs18.png)
![V_2 = ((\pi(0.12^2))(2.88))/((\pi (0.25)^2))](https://img.qammunity.org/2020/formulas/physics/college/upmxysodopyjdve3smti3pnim9uwtc9s2p.png)
![V_2 = 66.35m/s](https://img.qammunity.org/2020/formulas/physics/college/t1xhqf70lag0k5vltbznzb5rlplhr43oza.png)
Therefore the velocity at the exit end is 66.35m/s