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Elenna [48]
1 year ago
9

When one member of a binary star system is a black hole, and astronomers detect flickering x-rays coming from the system, where

are these x-rays usually coming from?
Physics
1 answer:
puteri [66]1 year ago
8 0

When one member of a binary star system is a black hole, and astronomers detect flickering x-rays coming from the system, these x-rays usually coming from a disk of material around the black hole (material that has been pulled from the companion star and is falling toward the black hole).

A binary black hole (BBH) is a system consisting of two black holes in close orbit around each other. Like black holes themselves, binary black holes are often divided into stellar binary black holes, formed either as remnants of high-mass binary star systems or by dynamic processes and mutual capture; and binary supermassive black holes, believed to be a result of galactic mergers.

For many years, proving the existence of binary black holes was made difficult because of the nature of black holes themselves and the limited means of detection available.

However, in the event that a pair of black holes were to merge, an immense amount of energy should be given off as gravitational waves, with distinctive waveforms that can be calculated using general relativity.

Therefore, during the late 20th and early 21st century, binary black holes became of great interest scientifically as a potential source of such waves and a means by which gravitational waves could be proven to exist. Binary black hole mergers would be one of the strongest known sources of gravitational waves in the universe, and thus offer a good chance of directly detecting such waves.

As the orbiting black holes give off these waves, the orbit decays, and the orbital period decreases. This stage is called binary black hole inspiral. The black holes will merge once they are close enough. Once merged, the single hole settles down to a stable form, via a stage called ringdown, where any distortion in the shape is dissipated as more gravitational waves. In the final fraction of a second the black holes can reach extremely high velocity, and the gravitational wave amplitude reaches its peak.

Learn more about binary black hole here : brainly.com/question/16199119

#SPJ4

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Answer:

it has no acceleration

Explanation:

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3 years ago
How much force acts on one surface of a rectangular bedroom wall that is 2.20 m by 3.20 m when atmosspheric pressure is?
irakobra [83]
<span>We know that pressure is the force applied into a surface, in our case the wall of the room, so then first we will calculate the surface of this wall: S = 2.2 * 3.2 = 7.04 m2 Then we also know the atmospheric pressure in normal conditions is 1 atm. That is the same 1 atm = 101325 Pascals or 101325 N/m2 Now we need to use the formula : P = F/S where P is pressure, F is force and S is surface to calculate the force: F = P * S = 101325 * 7.04 = 713,328 Newtons Conclusion: the force acts on the wall due the air inside the room is 713,328 N</span>
3 0
3 years ago
There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is
Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

4 0
3 years ago
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Ratling [72]

Answer:

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7 0
3 years ago
I need help on 4 and 5 please
Ivan

Answer:

c

Explanation:

6 0
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