A 700700700-Newton rightward net force is being applied to a 101010-kg object. What is the magnitude of the rightward accelerati
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Missing details: figure of the problem is attached.
We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,
where:
is the pressure difference between the two ends
is viscosity of the fluid
L is the length of the pipe
is the volumetric flow rate, with
being the section of the tube and
the velocity of the fluid
r is the radius of the pipe.
We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:
is the dynamic water viscosity at
L=4.0 cm=0.04 m
and r=1 mm=0.001 m
Using these data in the formula, we get:
However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.
Answer:
a) Revolutions per minute = 2.33
b) Centripetal acceleration = 11649.44 m/s²
Explanation:
a) Angular velocity is the ratio of linear velocity and radius.
Here linear velocity = 72 m/s
Radius, r = 0.89 x 0. 5 = 0.445 m
Angular velocity
Frequency
Revolutions per minute = 2.33
b) Centripetal acceleration
Here linear velocity = 72 m/s
Radius, r = 0.445 m
Substituting
Centripetal acceleration = 11649.44m/s²