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mamaluj [8]
3 years ago
9

Galaxy B moves away from galaxy A at 0.577 times the speed of light. Galaxy C moves away from galaxy B in the same direction at

0.731 times the speed of light. How fast does galaxy C recede from galaxy A?
Physics
1 answer:
Pepsi [2]3 years ago
7 0

Answer:

The value is  p = 0.7556 c

Explanation:

From the question we are told that

   The speed at which galaxy B moves away from galaxy A is  v =  0.577c

Here c is the speed of light with value  c = 3.0 *10^{8} \  m/s

     The speed at which galaxy C moves away from galaxy B is  u  =  0.731 c

Generally from the equation of  relative speed we have that  

     u   =  \frac{p - v}{ 1 - \frac{ p * v}{c^2} }

Here p is the velocity at which galaxy C recede from galaxy A so

     0.731c   =  \frac{p - 0.577c }{ 1 - \frac{ p * 0.577c}{c^2} }

=>   0.731c  [1 - \frac{ p * 0.577}{c}]  = p - 0.577c

=>   0.731c  -  0.4218 p = p - 0.577c

=>   0.731c  + 0.577c = p  + 0.4218 p

=>   1.308 c  = 1.731 p

=>    p = 0.7556 c

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i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

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q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

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\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

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Solving for i_2

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