All categories

3 years ago

If neutron stars are squeezed harder they collapse into black hole; how would this transition occur?

Physics

1 answer:

antoniya [11.8K]3 years ago

6 0

Answer:

If the neutron star's mass is then increased, neutrons become degenerate, breaking up into their constituent quarks, thus the star becomes a quark star; a further increase in mass results in a black hole.

Explanation:

hope this helps have a good night :)

You might be interested in

What type of friction prevents a pile of rocks from falling apart?

BlackZzzverrR [31]

A because it some type of friction

6 0

3 years ago

What percentage of all traffic deaths occur in cars traveling less than 60 mph?

Alina [70]

This is a bit too broad. Maybe this is something to google up

5 0

3 years ago

A wave created by shaking a rope up and down is called a.

Sidana [21]

A wave created by shaking a rope up and down is called a:
(According to me and certified experts) Transverse wave.

If this helps, then please give Brainliest!

5 0

2 years ago

A 19 nC charge is moved in a uniform electric field. The electric field does 5.3 μJ of work as the charge moves from point A to

Marizza181 [45]

Answer:

The potential difference between points A and B is 278.95 volts.

The potential difference between points B and C is -642.10 volts.

The potential difference between points A and C is -363.15 volts.

Explanation:

Given :

Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C

Work is done to move a charge from point A to B, W₁ = 5.3 μJ

Work done to move a charge from point B to C, W₂ = -12.2 μJ

Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.

The relation between work done and potential difference is:

W = qV

V = W/q ....(1)

Using equation (1), the potential difference between points A and B is:

$V_{1}=\frac{W_{1} }{q}$

Substitute the suitable values in the above equation.

$V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }$

V₁ = 278.95 V

Using equation (1), the potential difference between points B and C is:

$V_{2}=\frac{W_{2} }{q}$

Substitute the suitable values in the above equation.

$V_{2} =\frac{-12.2\times10^{-6} }{19\times10^{-9} }$

V₂ = -642.10 V

The potential difference between points A and C is:

V₃ = V₁ + V₂

V₃ = 278.95 - 642.10

V₃ = -363.15 V

8 0

3 years ago

Calculate the kinetic energy of a 64.g bullet moving at a speed of 411.ms . Round your answer to 2 significant digits.

Mandarinka [93]

Answer:

The kinetic energy of the bullet is 5.4 × 10³ J

Explanation:

Hi there!

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the bullet.

v = speed of the bullet.

Let´s convert the mass unit to kg so that our result is in Joules:

64 g · ( 1 kg / 1000 g) = 0.064 kg

Then, the kinetic energy will be the following:

KE = 1/2 · 0.064 kg · (411 m/s)²

KE = 5.4 × 10³ J

8 0

3 years ago