Release an electron initially at rest in the presence of an electric field. The electron tends to go to the region of 1. same el
1 answer:
Answer:
The electron tends to go to the region of 4. higher electric potential.
Explanation:
When a charged particle is immersed in an electric field, it experiences a force given by
where
q is the charge of the particle
E is the electric field
The direction of the force depends on the sign of the charge. In particular:
- The force and the electric field have the same direction if the charge is positive
- The force and the electric field have opposite directions if the charge is negative
Therefore, an electron (negative charge) moves in the direction opposite to the electric field lines.
However, electric field lines go from points at higher potential to points at lower potential: so, electrons move from regions at lower potential to regions of higher potential.
Therefore, the correct answer is
The electron tends to go to the region of 4. higher electric potential.
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Answer: = 0.0050 M
= 0.0155 M
Explanation:
Initial moles of = 0.072 mole
Volume of container = 3.9 L
Initial concentration of
The given balanced equilibrium reaction is,
Initial conc. 0.018 M 0
At eqm. conc. (0.018-x) M (2x) M
The expression for equilibrium constant for this reaction will be,
we are given :
Now put all the given values in this expression, we get :
So, the concentrations for the components at equilibrium are:
Hence, concentrations of and
are 0.0050 M ad 0.0155 M respectively.
Answer:
a) It is moving at when reaches the ground.
b) It is moving at when reaches the ground.
Explanation:
Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:
(1)
with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:
(2)
with m the mass and v the velocity.
Using (2) on (1):
(3)
In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):
(4)
Using (4) on (3):
(5)
That's the equation we're going to use on a) and b).
a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:
b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:
Solving for initial velocity (when the boulder left the volcano):