1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.
2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as
Where,
a = Acceleration
A = Amplitude
= Angular velocity
From a reference system in which the downward acceleration is negative due to the force of gravity we will have to
From the definition of frequency and angular velocity we have to
Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz
Answer:
The magnitude of the tension in he string is equal to the magnitude of the weight of the object.
Explanation:
According to the Newton's 1st law, An object will remain at rest or in uniform motion in a straight line unless acted upon by an unbalanced force.
In here, the elevator is moving with a constant speed. So the object must have the equal constant speed. Which means, it has a uniform motion. According to Newton's 1st law, the total unbalanced force on the object must be zero . As we know, there are only two forces are on the object and they are,
The tension in string(T) , The weight of the object(W) .
∴ F = 0
T - W = 0
So to balanced those forces, the magnitude of the tension in the string must be equal to the magnitude of the weight of the object.
The last row also know as noble gasses row also know as 18 row also know as 8A
