Determine the amount of energy being radiated through hawking radiation.
Answers
Answered by
0
The theoretical energy outflow PHH from a nonrotating, non-charged Schwarzschild black hole (i.e., not a very realistic object) follows the Stefan–Boltzmann–Schwarzschild–Hawking “black hole radiation power law”:
PHH= ℏℏc66/(15360πG22M22),
where ℏℏ = reduced Planck constant, c = lightspeed, and G = gravitational constant(it is not clear how the “fudge factor” 15 630 is derived).
For a black hole of 5M⨀⨀(five solar masses, ≈ 103030kg), this yields ≈ 3.6×10–30–30 watt (J s–1–1).
However, this will not happen until the CMBhas “cooled off” to below the BH “
PHH= ℏℏc66/(15360πG22M22),
where ℏℏ = reduced Planck constant, c = lightspeed, and G = gravitational constant(it is not clear how the “fudge factor” 15 630 is derived).
For a black hole of 5M⨀⨀(five solar masses, ≈ 103030kg), this yields ≈ 3.6×10–30–30 watt (J s–1–1).
However, this will not happen until the CMBhas “cooled off” to below the BH “
Answered by
0
The friction between the mat and the ground increases in the second case which reduces the net force applied on it. This is because frictional force is directly proportional to the 'weight' of the body that moves with respect to a surface. So in the second case, the weight of the body (mat+Person) is higher and would have more friction that would oppose the motion.
Similar questions
Math,
7 months ago
English,
7 months ago
English,
7 months ago
Computer Science,
1 year ago
Math,
1 year ago