Answer:
Energy Emitted, [tex]E_{J} = 0.1044 J[/tex]
Given:
Power of the jet emitted by black hole, [tex]P_{J} = 1044 W[/tex]
Rotational period of black hole, [tex]\Delta T = 0.1 ms = 0.1\times 10^{-3}[/tex]
Solution:
Energy emitted during one rotation of the black hole is given by:
[tex]P_{J} = \frac{Energy emitted, E_{J}}{\Delta T}[/tex]
Therefore,
[tex]E_{J} = P_{J}\times \Delta T[/tex]
Now, putting the given values in the above formula:
[tex]E_{J} = 1044\times 0.1\times 10^{-3} = 0.1044 J[/tex]
