Answer:
electrical conduction through a gas in an applied electric field
Explanation:
The proton (positive charge) shall be closer to the charge with the lower magnitude which is at the orgin.
The proton also shall be out of the interval between the two charges so that the pull of one charge cancels with the push of the other.
The region at which those conditions happen is to the left of the origin.
In that case the forces over the proton shall be:
k* (2.4 nC) * p / (x^2) - k*(4.8 nC) * p /( 1.3 + x)^2 = 0
where p is the charge of the proton.
You can simplify k and p:
2.4 / x^2 - 4.8 / (1.3 + x)^2 = 0
You can also simplify by 2.4
1/ x^2 - 2 / (1.3 + x)^2 = 0
(1.3+x)^2 - 2x^2 = 0
1.69 + 2.6x + x^2 - 2x^2 = 0
1.69 + 2.6x - x^2 = 0
x^2 -2.6x - 1.69 =0
Solve using the quadratic formula: x = 3.14 (use only the positive value)
That is the proton shall be place 3.14 units to the left of the origin (positive charge)
Answer:
yea it seems pretty simple.... just follow the instruction
Answer:
<em>Refraction is the change in direction of waves that occurs when waves travel from one medium to another. ... Diffraction is the bending of waves around obstacles and openings. The amount of diffraction increases with increasing wavelength.</em>
Answer:
The correct answer is
p = p₁ + p₂
Explanation:
Newton's second law states that force = the change of momentum produced therefore since the collision is inelstic then the change of momentum of each car is p₁ and p₂ and the force of the collition is proportional to p₁ + p₂ that is
F ∝ p₁ + p₂ and since force is directly proportional to p we have
p = p₁ + p₂