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vlabodo [156]
2 years ago
11

For a particular RLC series circuit, the capacitive reactance is 7.95 Ω , 7.95 Ω, the inductive reactance is 47.7 Ω , 47.7 Ω, an

d the maximum voltage across the 69.5 Ω 69.5 Ω resistor is 28.5 V . 28.5 V. What is the maximum voltage across the circuit?
Physics
1 answer:
I am Lyosha [343]2 years ago
7 0

Answer:

51.3165 V

Explanation:

The maximum voltage across the circuit is given as,

V' = IXc + IXl + Vr .................. Equation 1

Where V' = The maximum voltage across the circuit, I = current, Xc = capacitive reactance, Xl = inductive reactance, Vr = voltage cross the resistor.

From Ohm's Law,

V = IR........... Equation 2

I = V/R........... Equation 3

Given: V = 28.5 V, R = 69.5 Ω

Substitute into equation 3

I = 28.5/69.5

I = 0.41 A.

Also Given: Xc = 7.95 Ω, Xi = 47.7 Ω, Vr = 28.5 V.

Substitute into equation 1

V' = 7.95(0.41) + 47.7(0.41) + 28.5

V' = 3.2595+19.557+28.5

V' = 51.3165 V

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also we know that

a_c = \frac{v^2}{r}

11.6 = \frac{2.5^2}{R}

R = 0.54 m

so the coordinates of the center will be

(x, y) = (3.20 , (3.5 + 0.54))

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3 years ago
Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line
Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

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Assume that

vDE to be in direction of - j

vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE     (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s =  21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°

4 0
3 years ago
A(n) 30 kg boy rides a roller coaster. The acceleration of gravity is 9.8 m/s 2 . With what force does he press against the seat
forsale [732]

Answer:

Force, F = 187.42 N

Explanation:

It is given that,

Mass of boy, m = 30 kg

Acceleration due to gravity, a=9.8\ m/s^2

Radius of curvature of the roller coaster, r = 15 m

Speed of the car, v = 7.3 m/s

The force acting on the boy are force of gravity and the centripetal force. The net force acting on him is as follows :

F=mg-\dfrac{mv^2}{r}

F=m(g-\dfrac{v^2}{r})

F=30\times (9.8-\dfrac{(7.3)^2}{15})

F = 187.42 N

So, he press against the seat with a force is 187.42 N. Hence, this is the required solution.

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