Answer:
The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
Explanation:
Given that,
Mass = 2.15 kg
Distance = 0.0895 m
Amplitude = 0.0235 m
We need to calculate the spring constant
Using newton's second law

Where, f = restoring force


Put the value into the formula


We need to calculate the kinetic energy of the mass
Using formula of kinetic energy

Here, 

Here, 


Put the value into the formula


Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
Answer:
The current is not used up. The electrons flow through the entire circuit and this travel is the current. They flow until they are not charged anymore. That is also why the circuit must be closed or else electrons would escape not just light it up for a second then go out.
Explanation:
Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Most well-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodically reverses direction.
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<em>#</em><em>L</em><em>E</em><em>T</em><em>S</em><em> </em><em>STUDY</em></h2>
Fibrous joint functions as a suture to tightly bind bones together so they do not move.
Answer:

Explanation:
The electrostatic attraction between the nucleus and the electron is given by:
(1)
where
k is the Coulomb's constant
Ze is the charge of the nucleus
e is the charge of the electron
r is the distance between the electron and the nucleus
This electrostatic attraction provides the centripetal force that keeps the electron in circular motion, which is given by:
(2)
where
m is the mass of the electron
v is the speed of the electron
Combining the two equations (1) and (2), we find

And solving for v, we find an expression for the speed of the electron:
