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
To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.
The tension in the vertical plane will be equivalent to the centripetal force therefore

Here,
m = mass
v = Velocity
r = Radius
The tension in the horizontal plane will be subject to the action of the weight, therefore

Matching both expressions with respect to the tension we will have to


Then we have that,


Rearranging to find the velocity we have that

The value of the angle is 14.5°, the acceleration (g) is 9.8m/s^2 and the radius is



Replacing we have that


Therefore the speed of each seat is 4.492m/s
Answer:
Ok I'm not 100% on this one but, try 3 lifes sorry if u get it wrong D:
Explanation:
The unit of pressure is the Pascal (Pa), equivalent to

.
B) The beam of light will bend three times: small angle of refraction through container wall; larger angle of refraction through water; small angle passing out of a container.
Eliminate