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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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In a 30.0-s interval, 500 hailstones strike a glass window with an area of 0.600m2 at an angle of 45.0°to the window surface. Ea
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or one hailstone we have;
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Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in
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Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

v^{2} = \frac{dr}{dt}^{2}  +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2} - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

\frac{r}{z}=tan(45)

or:

z = r cot(45) - (2)

and:

\frac{dz}{dt} = \frac{dr}{dt} cot(45)

replacing (2) in (1) we obtain:

v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}  - (3)

Now the kinetic energy is given as:

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And the potential energy is given by:

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