Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. 
, the amount of charge stored in this capacitor, will stay the same.
The formula 
relates the electric potential across a capacitor to:
- , the charge stored in the capacitor, and
, the capacitance of this capacitor.
While stays the same, moving the two plates apart could affect the potential 
by changing the capacitance of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of . Neither will that change the area of the two plates.
However, as (the distance between the two plates) increases, the value of will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula can be rewritten as:
.
The value of (charge stored in this capacitor) stays the same. As the value of becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
Answer:
Option C
Explanation:
From the question we are told that:
Mass 
Radius 
Time 
Generally the equation for Tension is mathematically given by
Therefore
, toward the center of the circle
Option C
Answer:
<em>The first choice (32m/s) is the closest to the answer</em>
Explanation:
The magnitude of a vector is the distance between the initial and the end point of the vector.
Being Vx and Vy the horizontal and vertical components of the vector V respectively, the magnitude of V is calculated as:
The components of the velocity of the physics student's projectile launcher are Vx=28 m/s and Vy=15 m/s.
Calculate the magnitude of the velocity:
The first choice (32m/s) is the closest to the answer
Answer:
23.8 m
Explanation:
The distance travelled by the zebra can be calculated by using the equation:
where
u is the initial velocity
t is the time
a is the acceleration
For this zebra,
u = 0 since it starts from rest
is the acceleration
Substituting t = 5 s, we find the distance travelled by the zebra:
