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3 years ago

HELP ME HOW ARE BLACK HOLES FROMED

Physics

2 answers:

V125BC [204]3 years ago

6 0

Answer:

Stellar black holes form when the center of a very massive star collapses in upon itself.

VladimirAG [237]3 years ago

3 0

Answer:

A black hole can be formed by the death of a massive star. When such a star has exhausted the internal thermonuclear fuels in its core at the end of its life, the core becomes unstable and gravitationally collapses inward upon itself, and the star's outer layers are blown away.

Explanation:

You might be interested in

Calculate the most probable speed of an ozone molecule in the stratosphere

Marysya12 [62]

Answer:

$v_{mp}=305.83 m/s$

Explanation:

The temperature in stratosphere is generally about 270 K

molecular weight of an ozone molecule = 48 gm/mole

now formula for most probable velocity

$v_{mp}= \sqrt{\frac{2RT}{M} }$

plugging the values we get

$v_{mp}= \sqrt{\frac{2\8.314\times270}{48} }$

$v_{mp}=305.83 m/s$

7 0

4 years ago

If the Sun suddenly went dark, we would not know it until its light stopped arriving on Earth. How long would that be, in second

Gre4nikov [31]

Answer: 500 s

Explanation:

Speed $v$ is defined as a relation between the distance $d$ and time $t$ :

$v=\frac{d}{t}$

Where:

$v=3(10)^{8}m/s$ is the speed of light in vacuum

$d=1.5(10)^{11}m$ is the distance between the Earth and Sun

$t$ is the time it takes to the light to travel the distance $d$

Isolating $t$ :

$t=\frac{d}{v}$

$t=\frac{1.5(10)^{11}m}{3(10)^{8}m/s}$

Finally:

$t=500 s$

5 0

3 years ago

One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what

Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

$F_y=ma_y$ (2)

$F_x=ma_x$ (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

$F_y=W+n=0$

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

$F_y=(755)(-9.81)+n=0$

$n=7406.55 N$ (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

$F_x=F+f=ma_x$

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

$1988+f=m(1.36)$

$f=(755)(1.36)-1988=-961.2 N$

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

$|f|=\mu_k n$

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

$\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13$

4 0

3 years ago

Water enters a typical garden hose of diameter 0.016 m with a velocity of 3 m/s. Calculate the exit velocity of water from the g

Vladimir79 [104]

Answer:

v₂ = 306.12 m/s

Explanation:

We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:

A₁v₁ = A₂v₂

where,

A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²

v₁ = entrance velocity = 3 m/s

A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²

v₂ = exit velocity = ?

Therefore,

(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂

v₂ = (0.006 m³/s)/(0.0000196 m²)

5 0

3 years ago

A 0.73-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar

Tasya [4]

Answer:

A) B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West

Explanation:

A) For this exercise we must use the expression of Faraday's law for a moving body

fem = $- \frac{d \phi }{dt}$

fem = $- \frac{d (B l y}{dt}= - B l v$ - d (B l y) / dt = - B lv

B = $- \frac{fem}{l \ v}$

we calculate

B = - 7.9 10⁻⁴ /(0.73 20)

B = 5.4 10⁻⁵ T

B) to determine which side of the bar is positive, we must use the right hand rule

the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.

Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West

4 0

3 years ago