Answer: Option (A) is the correct answer.
Explanation:
Laser helps in zooming an image on which it is directed or falls so that it becomes easier for a large population to understand what the teacher or guide is talking about.
For example, when Greg is giving slide show presentation to a large audience then in order to make his slides visualized by everyone in the room he uses laser.
Therefore, we can conclude that a laser best help him with the presentation as he can use the laser to enlarge images on the slides.
Force can be expressed as the product of mass and acceleration. Mathematically, that's F = m(a). Plugging the given into the equation, we have F = (13.5 kg)(9.5 m/s²) = 128.3 kg.m/s² or 128.3 N<span>. </span>
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.
Ions were once atoms with the same number of electrons and protons. Since they have opposite charges atoms are neutral. When they become ions the lose or gain electrons and become unbalanced.These different charges are attracted to each other via electric forces<span>.</span>
The answer is c. +2.0 µC
To calculate this, we will use Coulomb's Law:
F = k*Q1*Q2/r²
where F is force, k is constant, Q is a charge, r is a distance between charges.
k = 9.0 × 10⁹ N*m/C²
It is given:
F = 7.2 N
d = 0.1 m = 10⁻¹ m
Q1 = -4.0 µC = 4 * 1.0 × 10⁻⁶ = 4.0 × 10⁻⁶
Q2 = ?
Thus, let's replace this in the formula for the force:
7.2 = 9.0 × 10⁹ * 4.0 × 10⁻⁶ * Q2/(10⁻¹)²
7.2 = 9 * 4 * 10⁹⁻⁶ * Q2/10⁻¹°²
7.2 = 36 × 10³ * Q2 / 10⁻²
Multiply both sides of the equation by 10⁻²:
7.2 × 10⁻² = 36 × 10³ * Q2
⇒ Q2 = 7.2 × 10⁻² / 36 × 10³ = 7.2/36 × 10⁻²⁻³ = 0.2 × 10⁻⁵ = 2 × 10⁻⁶
Since µC = 1.0 × 10^-6:
Q2 = 2 * 1.0 × 10^-6 = 2 µC
