Answer:
<em>Electrical power</em>
Explanation:
<em>Electrical power:</em> Electric power (P) is defined as the energy (W) used or produced or transformed per seconds.
Power = Energy produced/transformed/time taken.
P = W/t................................ Equation 1
Where t = time taken
But,
W = VIt................................ Equation 2
Substituting equation 2 into equation 1
P = VIt/t
P = VI............................ Equation 3
or
P = V²/R or P = I²R.
The unit of power is J/s or Volt-ampere is called watt (W).
Large amount of electric power are required in the industry
1 kW = 1000 W
1 MW = 1000000 W
1 h.p = 3/4 kW.
The correct answer is A continúe moving with constant velocity
Answer:
x₂ = 0.1715 d
1) false
2) True
3) True
4) false
5) True
Explanation:
The field electrifies a vector quantity, so we can add the creative field by these two charges
E₂-E₁ = 0
k q₂ / r₂² - k q₁ / r1₁²= 0
q₂ / r₂² = q₁ / r₁²
suppose the sum of the fields is zero at a place x to the right of zero
r₂ = d + x
r₁ = d -x
we substitute
q₂ / (d + x)² = q₁ / (d-x)²
we solve the equation
q₂ / q₁ (d-x)² = (d + x) ²
let's replace the value of the charges
q₂ / q₁ = + 2q / + q = 2
2 (d²- 2xd + x²) = d² + 2xd + x²
x² -6xd + d² = 0
we solve the quadratic equation
x = [6d ± √ (36d² - 4 d²)] / 2
x = [6d ± 5,657 d] / 2
x₁ = 5.8285 d
x₂ = 0.1715 d
with the total field value zero it is between the two loads the correct solution is x₂ = 0.1715 d
this value remains on the positive part of the x axis, that is, near charge 1
now let's examine the different proposed outcomes
1) false
2) True
3) True
4) false
5) True
