Complete Question
A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted,
A :
a force repels the glass out of the capacitor.
B :
a force attracts the glass into the capacitor.
C :
no force acts on the glass.
D :
a net charge appears on the glass.
E :
the glass makes the plates repel each other.
Answer:
The correct option is B
Explanation:
Generally when the glass dielectric is slowly inserted between the plated,
The positive plate of the capacitor will induce a negative charge on the glass while the negative plate of the capacitor will induce a positive charge on glass which a electric field that posses an electric force that will attract the glass
Answer:
Dx = -0.5
Dy = -0.25
Explanation:
Two vectors are given in rectangular components form as follows:
A = i + 6j
B = 3i - 7j
It is also given that:
A - B - 4D = 0
so, we solve this to find D vector:
(i + 6j) - (3i - 7j) - 4D = 0
- 2i - j = 4D
D = - (2/4)i - (1/4)j
D = - (1/2)i - (1/4)j
<u>D = - 0.5i - 0.25j</u>
Therefore,
<u>Dx = -0.5</u>
<u>Dy = -0.25</u>
Answer:
Approximately 
.
Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.
Explanation:
Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant 
near the surface of the earth.
For an object that is accelerating constantly,
,
where
is the initial velocity of the object,
is the final velocity of the object.
is its acceleration, and
is its displacement.
In this case, is the same as the change in the ball's height: 
. By assumption, this ball was dropped with no initial velocity. As a result, 
. Since the ball is accelerating due to gravity, 
.
.
In this case, would be the velocity of the ball just before it hits the ground. Solve for
.
.
Answer:
the brightest found are Blue - White with
Explanation:
The energy emission of objects increases with their temperature, specifically Wien described the process in an expression
T = 2,898 10⁻³
With this expression we can find the temperature of the stars by the color they emit.
Specifically the Sun has a color of 550 nm which corresponds to 5400K
bright stars have a BLUE color corresponding to 7500K
the brightest found are Blue - White with a temperature of 20000K
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