Answer:
F_total = 29.4 N, directed to the right of particle 2
Explanation:
We must solve this problem in parts, first we calculate each force and then we apply Newton's law to add the forces.
Let's use Coulomb's law to calculate each force
F = 
particles 1 and 2
q₁ = 8.0 10⁻⁶ C, q₂ = 3.5 10⁻⁶ C x₁₂ = 0.10 m
F₁₂ = 9 10⁹ 8.0 3.5 10⁻¹² / 0.1²
F₁₂ = 2.59 10¹ N
Since the two charges are of the same sign, this force is repulsive and is directed towards the positive side of the x axis.
particles 2 and 3
q₂ = 3.6 10⁻⁶ C, q₃ = 2.5 10⁻⁶ C, x₂₃ = 0.15 m
we calculate
F₂₃ = 9 10⁹ 3.5 2.5 10⁻¹²/ 0.15²
F₂₃ = 3.5 N
as the charge is of different sign, the force is attractive, therefore it is directed to the right of the load 2
Now we add the forces as vectors
F_total = ∑ F = F₁₂ + F₂₃
F_total = 25.2 +3.5
F_total = 29.4 N
directed to the right of particle 2
D, all of these are voltage sources. A battery can power small things, generators can power larger things, and outlets can charge, or power things.
Answer:
For example, 1300 with a bar placed over the first 0 would have three significant figures (with the bar indicating that the number is precise to the nearest ten).
Explanation:
hope it helps :)
Choice 'b' is one possible way to state
Newton's second law of motion.
The other choices are meaningless.
The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.
Answer:
D. At x=0, it's acceleration is at a maximum
Explanation:
As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.
Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.
From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.
The same can be said as the box travels backward from point A to -A
