A baseball is launched horizontally from a height of 1.8 m. The baseball travels 0.5 m before hitting the ground.
2 answers:
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To solve this problem we will apply the concept of current defined as the electron charge flow by the number of electrons per second. That is,
I = q*N
Here q is Flow of electric charge in one second and N the number of electron flow per second.
A the same time the power is described as the applied voltage for the current.
P = VI
We know the charge of electron, Coulombs, then the current is
And the power in the Beam is
Answer:
Lets take mass of A = m
Mass of B = m
Lets take spring constant = K
Displacement of mass A = x
Displacement of mass B= x'
Given that
x ' > x
a)
We know that time period of spring mass system given as
Time does not depends on the displacement.
So time will be same for both.
b)
We know that acceleration given as
a= ω² .x
ω=natural frequency ( ω² = m K)
Here x ' > x
So Acceleration of mass B is greater than mass A.
c)
Velocity at equilibrium position
V= ω .x
Here x ' > x
So velocity of mass B is greater than mass A.
d)
As we know that at equilibrium point acceleration of the mass is zero.So both the mass have same acceleration and the value of this acceleration will be zero m/s².