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.
One double bond consists 4 electrons, so 2 double bonds means 8 electrons
Answer:
The second law of a vibrating string states that for a transverse vibration in a stretched string, the frequency is directly proportional to the square root of the string's tension, when the vibrating string's mass per unit length and the vibrating length are kept constant
The law can be expressed mathematically as follows;

The second law of the vibrating string can be verified directly, however, the third law of the vibrating string states that frequency is inversely proportional to the square root of the mass per unit length cannot be directly verified due to the lack of continuous variation in both the frequency, 'f', and the mass, 'm', simultaneously
Therefore, the law is verified indirectly, by rearranging the above equation as follows;

From which it can be shown that the following relation holds with the limits of error in the experiment
m₁·l₁² = m₂·l₂² = m₃·l₃² = m₄·l₄² = m₅·l₅²
Explanation: