5. In an AC circuit the applied emf e = E0
1 answer:
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Answer:
v₂ = 70 m / s
Explanation:
For this exercise let's use Bernoulli's equation
where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂
indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute
ρ g y₁ = ½ ρ v₂²
v₂ =
let's calculate
v₂ = √( 2 9.8 250)
v₂ = 70 m / s
Answer:
Explanation:
<u>LC Circuit</u>
It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:
= charge of the capacitor in any time
= initial charge of the capacitor
=angular frequency of the circuit
= current through the circuit in any time
The charge in an LC circuit is given by
The current is the derivative of the charge
We are given
It means that
From eq 1:
From eq 2:
Squaring and adding the last two equations, and knowing that
Operating
Solving for
Now we know the value of , we repeat the procedure of eq 1 and eq 2, but now at the second time
, and solve for
Solving for
Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.
Finally