A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati
2 answers:
Explanation:
The given data is as follows.
Initial velocity; u = 0, Final velocity; v = 19.6 m/s
time; t = 2 seconds
As the relation between initial velocity, final velocity and acceleration is as follows.
v = u + at
Hence, putting the given values into the above formula as follows.
v = u + at
19.6 m/s = 0 +
a = 9.8
Thus, we can conclude that acceleration of the dropped ball is 9.8 .
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Answer:
Explanation:
Consider two particles are initially at rest.
Therefore,
the kinetic energy of the particles is zero.
That initial K.E. = 0
The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are
now, since both the masses have mass m
therefore,
= m/2
The final K.E. of the particles is
Distance between two particles is d and the gravitational potential energy between them is given by
By law of conservation of energy we have
Now plugging the values we get
This the required relation between G,m and d
Answer:
The fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is
Explanation:
From the question we are told that
The time constant
The potential across the capacitor can be mathematically represented as
Where is the voltage of the capacitor when it is fully charged
So at
Generally energy stored in a capacitor is mathematically represented as
In this equation the energy stored is directly proportional to the the square of the potential across the capacitor
Now since capacitance is constant at
The energy stored can be evaluated at as
Hence the fraction of the energy stored in an initially uncharged capacitor is