How does end A of the rod react when the charged ball approaches it after a great many previous contacts with end A? Assume that
1 answer:
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Answer:
Explanation:
The general equation for position of Simple harmonic motion is given as:
........(1)
where,
x = Position of the wave
A = Amplitude of the wave
ω = Angular velocity
t = time
In this case, the amplitude is just half the range,
thus,
(Given range = 3cm)
A = 1.5 cm
Now, The angular velocity is given as:
Where, T = time period of the wave =0.27s (given)
or
so, at time t = 55 s, the equation (1) becomes as:
on solving the above equation we get,
here the negative sign depicts the position in the opposite direction of +x
Answer:
The energy of this particle in the ground state is E₁=1.5 eV.
Explanation:
The energy of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:
In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:
So we can rewrite the energy in the ground state as:
Finally