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Maksim231197 [3]
3 years ago
10

A capacitance C and an inductance L are operated at the same angular frequency.

Physics
1 answer:
PIT_PIT [208]3 years ago
6 0

Answer:

(B) 7279.70 rad/sec (C) X_L=37.1265\ ohm\ and \ X_C=37.1265\ ohm

Explanation:

We have given L=5.10mH and C=3.70\mu F

(B) The angular frequency is denoted as \omega whose value is given by \omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5.10\times 10^{-3}\times 3.70\times 10^{-6}}}=7279.70\ rad/sec

(C) The inductive reactance X_L=\omega L=7279.70\times 5.10\times 10^{-3}=37.1265\ ohm

Capacitive reactance  X_C=\frac{1}{\omega C}=\frac{1}{7279.70\times 3.70\times 10^{-6}}=37.1265\ ohm

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We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
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\frac{mv^2}{2}+\frac{2mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+\frac{2mv_{B}^2}{2}
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This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
mv+2mv=mv_A+2mv_B\\ 3mv=mv_A+2mv_B\\
Because both carts have the same mass we can cancel those out:
3v=v_A+2v_B
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\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\ 3v=v_A+2v_B
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We need to solve our system for v_a. We will solve second equation for v_b and then plug that in the first equation.
3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=\frac{3v-v_A}{2}
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\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\
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3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\
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Coefficients are:
a=1\\
b=-6v\\
c=v^2-\frac{E}{m}
Solutions are:
v_A=\frac{3v+\sqrt{5v^2+\frac{4E}{m}}}{2},\:v_A=\frac{3v-\sqrt{5v^2+\frac{4E}{m}}}{2}
Part B
We do the same thing here, but we must express v_a from momentum equation:
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c=6v^2-\frac{3E}{2m}
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A.

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