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Serga [27]
2 years ago
15

In 1974, the Arecibo Radio telescope in Puerto Rico was used to transmit a signal to M13, a star cluster about 25,000 light-year

s away. How long will it take the message to reach M13, and how far has the message travelled so far (in light-years)?
Physics
1 answer:
Arte-miy333 [17]2 years ago
6 0

Answer:

The Arecibo message is a 1974 interstellar radio message carrying basic information about humanity and Earth sent to globular star cluster M13 in the hope that extraterrestrial intelligence might receive and decipher it. The message was broadcast into space a single time via frequency modulated radio waves at a ceremony to mark the remodeling of the Arecibo radio telescope in Puerto Rico on 16 November 1974.The message was aimed at the current location of M13 some 25,000 light years away because M13 was a large and close collection of stars that was available in the sky at the time and place of the ceremony.The message consisted of 1,679 binary digits, approximately 210 bytes, transmitted at a frequency of 2,380 MHz and modulated by shifting the frequency by 10 Hz, with a power of 450 kW. The "ones" and "zeros" were transmitted by frequency shifting at the rate of 10 bits per second. The total broadcast was less than three minutes.

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Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
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Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

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The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

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            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

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