Answer:
q₁ = + 1.25 nC
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Known data
q₃=5 nC
q₂=- 3 nC
d₁₃= 2 cm
d₂₃ = 4 cm
Graphic attached
The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.
For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).
The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).
The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)
Calculation of q1
F₁₃ = F₂₃

We divide by (k * q3) on both sides of the equation



q₁ = + 1.25 nC
Efficiency = (Wanted) energy out ÷ energy in × 100
Energy in = 400J
Wanted Energy out = 240J
Energy cannot be used up, only transferred, so the remaining energy is most likely to be transferred into unwanted energy (loss of energy) such as heat energy.
Efficiency = 240 ÷ 400 × 100
Efficiency = 0.6 × 100
Efficiency = 60%
Answer:

Explanation:
The textbooks say that the maximum range for projectile motion (with no air resistance) is 45 degrees.
Answer:
false. it's positive terminal of an electric cell
If my memory serves me well, if we want to know the velocity that an object is traveling, we must know the <span>direction and speed. Velocity includes two these points listed in the previous sentence which means the answer is D.</span>