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
I believe the answer is the second one.
Class 1 lever
Explanation:
In a class 1 lever, the fulcrum is placed between the effort and the load. This lever systems is the most common.
- The effort is the force input and the load is the force output
- The fulcrum is a hinge between the load and effort.
- Movement of the effort and load are in opposite directions.
- There are other classes of lever like the class 2 and 3.
- They all have different load, fulcrum and effort configurations
learn more:
Load related problems brainly.com/question/9202964
Torque brainly.com/question/5352966
#learnwithBrainly
Hope this answer helps, cause Idk, I might be wrong, but I still, I used the correct formulas, so I might be correct
Answer:
Explanation:
Total Resistance, Rt = (0.35 + 0.002 + 5200 + 0.5 + 0.008 + 1100) = 6300 ohms
a) Capacity of transformer, Pt = 150KVA = 150,000 W
Input Voltage, Vp = 2.8 KV = 2800 V
Current, Ip = 150000/2800 = 53.57 A
Input impedance, Zp = Vp/Ip = 2800/5357 = 52.27 ohms
b) i) Input current = 53.57 A
ii) Voltage, V = Ip * Rt = 53.57 X 6300 = 337.5 KV
iii) Power, P = I² * Rt = (53.57)² X 6300 = 18.08 MW
iv) Power factor = 0.83