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:
See below
Explanation:
Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N
no work is done by this force
Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N
work of friction = 7.55 * 2 m = 15.1 j
Force Downplane = mg sin 53 = 62.61 N
work = 62.61 * 2 = 125.22 j
Net Force downplane = force downplane - force friction = 55.06 N
net Work = force * distance = 55.06 N * 2 M = 110.12 j
Answer:
O- to track how fast a heart beats
Explanation: