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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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A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Oiled Steel fry
Alinara [238K]

Answer:

N  = 6.67 N

Explanation:

The frictional or frictional force is a force that arises from the contact of two bodies and opposes movement.

The friction is due to imperfections and roughness, mainly microscopic, that exist on the surfaces of the bodies. Upon contact, these roughnesses engage with each other making movement difficult. To minimize the effect of friction, either the surfaces are polished or lubricated, since the oil fills the imperfections, preventing them from snagging.

As the frictional force depends on the materials and the force exerted on one another, its magnitude is obtained by the following expression:

f = μ*N    Formula (1)

where:  

f is the friction force  (N)

μ is the coefficient of friction

N is the normal force (N)

Data

f = 0.2 N : frictional force between the steel spatula and the Oiled Steel frying pan

μ = 0.03 :coefficient of kinetic friction between the two materials

Calculating of normal force

We replace data in the formula (1)

f = μ*N  

0.2  = 0.03*N  

N  = 0.2 / 0.03

N  = 6.67 N

5 0
3 years ago
two test cars of equal mass moving towards each other collide on a horiontal frictionless surface. Before the collision, car A h
LiRa [457]

Answer:

4 m/s

Explanation:

m1 = m2 = m

u1 = 20 m/s, u2 = - 12 m/s

Let the speed of composite body is v after the collision.

Use the conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

m x 20 - m x 12 = (m + m) x v

20 - 12 = 2 v

8 = 2 v

v = 4 m/s

Thus, the speed of teh composite body is 4 m/s.

4 0
2 years ago
On the Moon the acceleration due to gravity is about one sixth that on Earth. If a golfer on the Moon imparted the same initial
swat32

Explanation:

We know that that the range of the ball on the earth

R_{earth}=\frac{v_o^2sin2\theta}{g_{earth}}

therefore, range of the ball on moon

R_{moon}=\frac{v_o^2sin2\theta}{g_{moon}}

R_{moon}=\frac{v_o^2sin2\theta}{g_{earth}/12}

therefore,

R_{moon}=6R_{earth}

Therefore, the range of ball will be 6 times on the moon than that on earth

6 0
2 years ago
A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the no
padilas [110]

Explanation:

The weight of the car is equal to, W_c=m\times g...........(1)

Where

m is the mass of car

g is the acceleration due to gravity

The normal or vertical component of the force is, F_N=mg\ cos\theta

or

F_N=mg\ cos(13).............(2)

The horizontal component of the force is, F_H=mg\ sin\theta

Taking ratio of equation (1) and (2) as :

\dfrac{F_N}{W_c}=\dfrac{mg\ cos\theta}{mg}

\dfrac{F_N}{W_c}=cos(13)

\dfrac{F_N}{W_c}=0.97

or

\dfrac{F_N}{W_c}=\dfrac{97}{100}

Hence, this is the required solution.

5 0
3 years ago
The turnbuckle is tightened until the tension in the cable AB equals 2.3 kN. Determine the vector expression for the tension T a
brilliants [131]

Answer:

a) 0.83984 i + 0.41992 j - 2.0996 k KN

b) T_ac = 1.972888 KN

Explanation:

Given:

- The tension in cable AB = 2.3 KN

Find:

a) Determine the vector expression for the tension T as a force acting on member AD.

b) Also find the magnitude of the projection of T along the line AC.

Solution:

part a)

- Find unit vector AB:

                            vector (AB) = 2 i + j - 5 k

                             mag (AB) = sqrt (2^2 + 1^2 + 5^2)

                             mag (AB) = sqrt(30)

                             unit (AB) =  ( 1 / sqrt(30) )* ( 2 i + j - 5 k )

- Find Tension vector:

                             vector (T) = unit(AB)* 2.3 KN

                                              = 0.83984 i + 0.41992 j - 2.0996 k

- The projection of T onto AC can be found from the dot product of vector T to unit vector (AC)

- For unit vector (AC)

                               vector (AC) = 2 i - 2 j - 5 k

                               mag (AC) = sqrt (2^2 + 2^2 + 5^2)

                               mag (AC) = sqrt(33)

                               unit (AC) =  ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

- Compute the projection:

                                T_ac = vector T . unit (AC)

T_ac = (0.83984 i + 0.41992 j - 2.0996 k)  . ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

                                T_ac = 0.2923947572 - 0.146973786 - 1.827467232

                                T_ac = 1.972888 KN

4 0
3 years ago
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