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vitfil [10]
4 years ago
14

In a certain oscillating LC circuit, the total energy is converted from electrical energy in the capacitor to magnetic energy in

the inductor in 1.40 μs. What are (a) the period of oscillation in microseconds and (b) the frequency of oscillation? (c) How long after the magnetic energy is a maximum will it be a maximum again?
Physics
1 answer:
aleksandrvk [35]4 years ago
5 0

Answer:

A = 5.6μs

B = 178.57kHz

C = 2.8μs

Explanation:

A. It takes ¼ of the period of the circuit before the total energy is converted from electrical energy in the capacitor to magnetic energy in the inductor.

t = T/4

T = 4*t

T = 4 * 1.4 = 5.6μs

B. f = 1/T

Frequency is the inverse of period

f = 1 / 5.6*10⁻⁶

f = 178571.4286Hz

f = 178.57kHz

C. time taken for maximum energy to occur is T/2

t = 5.6 / 2 = 2.8μs

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4 0
3 years ago
A cannon of mass 5.71 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7
zheka24 [161]

Answer:

541.14 m/s

Explanation:

We are given that

Mass of cannon=m_1=5.71\times 10^3 kg

Mass of shell,m_2=73.5 kg

Initial velocity of shell,v=547 m/s

We have to find the velocity of shell fired from this loose cannon.

According to law of conservation of momentum

m_1v_1+m_2v_2=m_1u_1+m_2u_2

Initial momentum of system=0

m_1v_1=-m_2v_2

v_1=-\frac{m_2v_2}{m_1}

When the cannon is bolted to the ground then only shell moves and kinetic energy of system equals to kinetic energy of shell

Kinetic energy of shell,K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 73.5(547)^2}=1.09\times 10^7 J

K.E of shell=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2

K.E of shell=\frac{1}{2}m_1(-\frac{m_2v_2}{m_1})^2+\frac{1}{2}m_2v^2_2

K.E of shell=\frac{1}{2}(\frac{m^2_2v^2_2}{m_1}+\frac{1}{2}m_2v^2_2)

2K.E of shell=m_2v^2_2(\frac{m_2}{m_1}+1)

Velocity of shell fired from this loose cannon,v_2=\sqrt{\frac{2k.E}{m_2(\frac{m_2}{m_1}+1)}

v_2=\sqrt{\frac{2\times 1.09\times 10^7}{73.5(\frac{73.5}{5.71\times 10^3}+1)}

v_2=541.14m/s

Hence, the velocity of shell fired from this loose cannon would be 541.14 m/s

5 0
3 years ago
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