Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s
Explanation:
We can solve this problem by using Bernoulli's equation:
(1)
where
is the pressure in the pipe
is the pressure in the constricted section
is the density of the oil
is the velocity of the oil in the pipe
is the velocity of the oil in the constricted section
Also, according to the continuity equation,

where
is the cross-sectional area of the pipe, with
is the radius
is the cross-sectional area of the constricted section, with
is the radius
So the equation becomes

So we can write

Substituting into eq.(1),

And solving the equation for
:

And the velocity in the constricted section is

Learn more about flow rate:
brainly.com/question/9805263
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