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Using the kinetic energy 1/2mv^2 formula
5*10^5 is the answer
Explanation:
- Newton's first law of motion:
"An object at rest (or in uniform motion) remains at rest (or in uniform motion) unless acted upon an unbalanced force
In this situation, we can apply Newton's first law to the keys of the keyboard that are not hit by the fingers of the man. In fact, as no force act on the keys, they remain at rest.
- Newton's second law of motion:
"The acceleration experienced by an object is proportional to the net force exerted on the object; mathematically:

where F is the net force, m is the mass of the object, and a its acceleration"
In this case, we can apply Newton's second law to the keys of the keyboard that are hit by the man: in fact, as they are hit, they experience a downward force, and therefore they experience a downward acceleration.
"Newton's third law of motion:
"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"
Here We can apply Newton's third law to the pair of objects finger-key: in fact, as the finger apply a force on the key (action force), then the key exerts a force back on the finger (reaction force), equal and opposite.
Answer:
s
Explanation:
From the question we are told that
The outer ring with a radius of 30 m
inner Gravity Approximately 9.80 m/s'
Outer Gravity Approximately 5.35 m/s.
Generally the equation for centripetal force is given mathematically as
Centripetal acceleration enables Rotation therefore?

Considering the outer ring,




Therefore solving for Period T
Generally the equation for solving Period T is mathematically given as


s
Well, to be honest, there is no official way to find out which volcano erupted recently as scientists don't have that type of equipment ready or in use at all; also there is no official way [...] because there are always eruptions occurring.
If you want the most known, "Calbuco" is your answer. It erupted in 2014 and it is found in Chile.