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
The answer is C. in sort of a way. You can't technically see black matter. However, it is holding the galaxies together.
Air speed is how fast you are peddling over the ground, wind speed is how fast the wind is blowing above the ground.Depending on the correlation between the two with direction thrown the pilot knows what sort of resistance he is flying against.
Answer:
The depth of the water at this point is 0.938 m.
Explanation:
Given that,
At one point
Wide= 16.0 m
Deep = 3.8 m
Water flow = 2.8 cm/s
At a second point downstream
Width of canal = 16.5 m
Water flow = 11.0 cm/s
We need to calculate the depth
Using Bernoulli theorem

Put the value into the formula



Hence, The depth of the water at this point is 0.938 m.
Answer:
0.84 cm
Explanation:
u = Object distance = 0.35 cm
v = Image distance = -0.6 cm (near point is considered as image distance and negative due to sign convention)
f = Focal length
From lens equation

Focal length of the lens is 0.84 cm