The electric field is always perpendicular to the surface outside of a conductor. TRUE
<span> If an electron were placed on an electric field line, it would move in a direction perpendicular to the field. FALSE, it would move in an anti-parallel direction because its charge is negative </span>
<span>Electric field lines originate on positive charge and terminate on negative charge. TRUE ; but they can also go to infinity </span>
It is possible for two electric field lines to cross each other.
<span> Usually FALSE; though technically possible at special points where field is zero. </span>
If an electron and a positron were in the presence of a very strong electric field, they would move away from each other.
<span> TRUE; one is positive, and one is negative. If the field is strong enough, the action of the field will overcome the mutual attraction between them </span>
It is not possible for the electric field to ever be zero. FALSE: it IS possible, inside a conductor for instance
If a proton were placed on an electric field line, it would move in a direction anti-parallel to the field.
<span> FALSE: being positive, it would move in the SAME direction as the field</span>ic
The correct answers are <span>starting friction and </span>static friction
Friction slows down all forces, but starting friction slows down or stops completely the start of motion.
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;
where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²
Therefore, the mutual force between the two point charges is 319.64 N
Answer:
they help us allocate a particular place in case one needs to allocate or find a place or something
We will use the formula / equation to determined the time.
Distance = ½ * (vi + vf) * t
48100 = ½ * (26.3 + 41.9) * t
t = 48100 ÷ 34.1 = 1410.557185 seconds
We will use the formula / equation to determined the acceleration.
vf = vi + a * t
41.9 = 26.3 + a * 1410.557185
a = (41.9 – 26.3) ÷ 1410.557185 = 0.011059459 m/s^2
We will use the formula / equation to determined the acceleration.
vf^2 = vi^2 + 2 * a * d
41.9^1 = 26.3^2 + 2 * a * 48100
a = (41.9^2 – 26.3^2) ÷ 96200 = 0. 011059459 m/s^2
Since both answers are the same, I believe the acceleration is correct.
