The minimum speed of the water must be 3.4 m/s
Explanation:
There are two forces acting on the water in the pail when it is at the top of its circular motion:
- The force of gravity, mg, acting downward (where m is the mass of the water and g the acceleration of gravity)
- The normal reaction, N also acting downward
Since the water is in circular motion, the net force must be equal to the centripetal force, so:

Where:

v is the speed of the pail
r = 1.2 m is the radius of the circle
The water starts to spill out when the normal reaction of the pail becomes zero:
N = 0
When this occurs, the equation becomes:

And substitutin the values of g and r, we find the minimum speed that the water must have in order not to spill out:

Learn more about circular motion:
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Answer:0.0704 kg
Explanation:
Given
initial Absolute pressure
=210+101.325=311.325



as the volume remains constant therefore



therefore Gauge pressure is 337.44-101.325=236.117 KPa
Initial mass 

Final mass 

Therefore
=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back