Answer:
(c) 16 m/s²
Explanation:
The position is
.
The velocity is the first time-derivative of <em>r(t).</em>
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The acceleration is the first time-derivative of the velocity.

Since <em>a(t)</em> does not have the variable <em>t</em>, it is constant. Hence, at any time,

Its magnitude is 16 m/s².
Answer:
cooooooooooooooollllllllll
Explanation:
They have to have a positive charge and negative so then thats what u get
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