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Anarel [89]
3 years ago
8

Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperatu

re of 2.81 K.
Physics
1 answer:
Rus_ich [418]3 years ago
4 0

Complete Question

Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K. Remember that Stefan's Law gives the Power (Watts) and Intensity is Power per unit Area (W/m2).

Answer:

The intensity is I  = 3.535 *10^{-6} \  W/m^2

Explanation:

From the question we are told that

    The temperature is  T = 2.81 \ K

Now  According to Stefan's law

        Power(P) =  \sigma  *  A  * T^4

Where  \sigma is the Stefan Boltzmann constant with value  \sigma  =  5.67*10^{-8} m^2 \cdot kg \cdot s^{-2} K^{-1}

  Now the intensity of the cosmic background radiation emitted according to the unit from the question is mathematically evaluated as

        I  =  \frac{P}{A}

=>      I  =  \frac{\sigma *  A  * T^4}{A}

=>      I  =  \sigma  *  T^4

substituting values

      I  = 5.67 *10^{-8}  *  (2.81)^4

       I  = 3.535 *10^{-6} \  W/m^2

       

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\begin{gathered} E1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_1v_2^2 \\ where: \\ m_1=m_2=0.25kg \\ v_1=7m/s \\ v_2=-7m/s \\ so: \\ E1=\frac{1}{2}0.25(7^2)+\frac{1}{2}0.25(7^2) \\ E1=6.125+6.125 \\ E1=12.25J \end{gathered}

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According to the conservation of energy:

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