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DanielleElmas [232]
4 years ago
11

In an oscillating LC circuit, the total stored energy is U and the maximum current in the inductor is I. When the current in the

inductor is I/2, the energy stored in the capacitor is
Physics
1 answer:
Deffense [45]4 years ago
5 0

Answer:

The definition of that same given problem is outlined in the following section on the clarification.

Explanation:

The Q seems to be endless (hardly any R on the circuit). So energy equations to describe and forth through the inducer as well as the condenser.  

Presently take a gander at the energy stored in your condensers while charging is Q.

⇒  U =\frac{Qmax^2}{C}

So conclude C doesn't change substantially as well as,

When,

⇒  Q=\frac{Qmax}{2}

⇒  Q^2=\frac{Qmax^2}{4}

And therefore only half of the population power generation remains in the condenser that tends to leave this same inductor energy at 3/4 U.

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a ladybug sits at the outer edge of a merry-go-round and a gentleman bug sits halfway between her and the axis of rotation. The
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What current flows through a 2.56-cm-diameter rod of pure silicon that is 20.0 cm long, when 1.00 ✕ 103 V is applied to it? (Suc
vfiekz [6]

Answer:

Current, I = 0.0011 A

Explanation:

It is given that,

Diameter of rod, d = 2.56 cm

Radius of rod, r = 1.28 cm = 0.0128 m

The resistivity of the pure silicon, \rho=2300\ \Omega-m

Length of rod, l = 20 cm = 0.2 m

Voltage, V=1\times 10^3\ V

The resistivity of the rod is given by :

R=\rho\dfrac{L}{A}

R=2300\ \Omega-m\dfrac{0.2\ m}{\pi (0.0128\ m)^2}

R = 893692.30 ohms

Current flowing in the rod is calculated using Ohm's law as :

V = I R

I=\dfrac{V}{R}

I=\dfrac{10^3\ V}{893692.30\ \Omega}

I = 0.0011 A

So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.

6 0
3 years ago
HELP ASAP PLEASE!!!
Montano1993 [528]

Answer:

The answer is A, B, C, D

Explanation:

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7 0
2 years ago
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g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p
kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r = \frac{mv}{qB}

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0
4 years ago
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