Answer:
29.4 uN
Explanation:
The electric force between two charges can be calculated using Coulomb's Law. According to this law the force between two point charges is given as:
where k is a proportionality constant known as the Coulomb's law constant. Its value is 
Nm²/C²
r = distance between charges = 70 cm = 0.7 m
q1 = q2 = 4nC = 
C
The negative sign indicates that the charges are negative. In the formula we will only use the magnitude of the charges.
Using these values in the formula, we get:
Therefore, the magnitude of repulsive force between the given charges will be 29.4 uN
Formula from physics to get the answer.
Answer:
CB = 4.45 x 10⁻⁹ F = 4.45 nF
Explanation:
The capacitance of a parallel plate capacitor is given by the following formula:
C = ε₀A/d
where,
C = Capacitance
ε₀ = Permeability of free space
A = Area of plates
d = Distance between plates
FOR CAPACITOR A:
C = CA = 17.8 nF = 17.8 x 10⁻⁹ F
A = A₁
d = d₁
Therefore,
CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1
FOR CAPACITOR B:
C = CB = ?
A = A₁/2
d = 2 d₁
Therefore,
CB = ε₀(A₁/2)/2d₁
CB = (1/4)(ε₀A₁/d₁)
using equation 1:
CB = (1/4)(17.8 X 10⁻⁹ F)
<u>CB = 4.45 x 10⁻⁹ F = 4.45 nF</u>
Answer:
4
Explanation:
a=vf-v
a=20-0÷5
a=4meter per second square
Answer: 3kg: 14.7 6kg: 29.4 9kg: 44.1
Explanation: just did it on Edge
