Answer:
Maximum speed of water that leaves the machine is 20.5 m/s
Explanation:
As we know that the angular frequency of the tub is
f = 652 rpm
so we have
[tex]f = 10.87 rounds/sec[/tex]
now we have
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 68.3 rad/s[/tex]
Now tangential speed is given as
[tex]v = R\omega[/tex]
so we will have
[tex]v = (0.299)(68.3)[/tex]
[tex]v = 20.5 m/s[/tex]