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:
c
Explanation:
When a satellite is orbiting the earth , a constant force is being applied on it which means it must has acceleration. Also the direction of satellite is always being changed when it is orbitting to there is always change in the velocity vector which means acceleration.
You can view in the attached diagram to understand how the velocity is being changed.
Answer:
A body is said to be moving with uniform speed, if it covers equal distances in equal intervals of time. ...
Answer:
Current: 1.0 Amperes
The minimum current is flowing through path D
Explanation:
We first find the equivalent resistance to the three resistors in parallel ( which is the total resistance of the circuit) via the equation:

with this info, we can estimate the current going through branch A using Ohm's Law, and the information that the power source is 6 V:

where the current comes in units of Amperes since all other the quantities are given in the SI system, and we can round this answer to 1.0 Amp following the request to round it to the tenth.
The current will be the lowest through the branch with the largest resistor due to the fact that less current will flow through the path of more resistance.
Than means that the lowest current will be registered through branch D where the 50
resistor is.
Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)