Answer:
X₃₁ = 0.58 m and X₃₂ = -1.38 m
Explanation:
For this exercise we use Newton's second law where the force is the Coulomb force
F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
Since all charges are of the same sign, forces are repulsive
F₁₃ = k q₁ q₃ / r₁₃²
F₂₃ = k q₂ q₃ / r₂₃²
Let's find the distances
r₁₃ = x₃- 0
r₂₃ = 2 –x₃
We substitute
k q q / x₃² = k 4q q / (2-x₃)²
q² (2 - x₃)² = 4 q² x₃²
4- 4x₃ + x₃² = 4 x₃²
5x₃² + 4 x₃ - 4 = 0
We solve the quadratic equation
x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5
x₃ = [-4 ± 9.80] 10
X₃₁ = 0.58 m
X₃₂ = -1.38 m
For this two distance it is given that the two forces are equal
Answer:
Force on the proton will be 
Explanation:
We have given speed of proton is 20.3% of speed of light
Speed of light 
So speed of proton 
Magnetic field B = 0.00629 T
Charge on proton 
Angle between velocity and magnetic field 
Force on the proton is equal to 
The force between them <em>decreases</em><em>,</em> as the square of the distance.
For example ...
-- If you move them apart to double the original distance, the force becomes (1/2²) = 1/4 of the original force.
-- If you move them apart to 3 times the original distance, the force becomes (1/3²) = 1/9 of the original force.
-- If you move them apart to 5 times the original distance, the force becomes (1/5²) = 1/25 of the original force.
(Gravity works exactly the same way.)
Answer:
1.21 or 1.2 acceleration
Explanation:
i think it is 50.0/41.0
which would equal 1.21(depending on if you round it)
Answer:
They move at 2.8 m/s and in the opposite direction.
Explanation:
