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3 years ago

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 $\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5.10\times 10^{-3}\times 3.70\times 10^{-6}}}=7279.70\ rad/sec$

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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Let's start with energy conservation law:
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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:
$3v=v_A+2v_B$
Now we have our system of equation that we have to solve:
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\ 3v=v_A+2v_B$
Part A
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.
$3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=\frac{3v-v_A}{2}$
Now we have to plug this in the first equation:
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\$
We will multiply the first equation with 2 and divide by m:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\$
Now we plug in the second equation into first one:
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$a=1\\
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Solutions are:
$v_A=\frac{3v+\sqrt{5v^2+\frac{4E}{m}}}{2},\:v_A=\frac{3v-\sqrt{5v^2+\frac{4E}{m}}}{2}$
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Again we end up with quadratic equation. Coefficients are:
$a=3\\
b=-6v\\
c=6v^2-\frac{3E}{2m}$
Solutions are:
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3 years ago

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sergejj [24]

Answer:

Explanation:

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