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enyata [817]
3 years ago
5

Is a neutron star also a black hole?

Physics
1 answer:
coldgirl [10]3 years ago
8 0

No.  A neutron star is the weird remains of a star that blew its outer layers off
in a nova event, and then had enough mass left so that gravity crushed its
electrons into its protons, and then what was left of it shrank down to a sphere
of unimaginably dense neutron soup.  But it didn't have enough mass to go
any farther than that.

A black hole is the remains of a star that had enough mass to go even farther
than that.  No force in the universe was able to stop it from contracting, so it
kept contracting until its mass occupied no volume ... zero.  It became even
more weird, and is composed of a substance that we don't know anything about
and can't describe, and occupies zero volume.

Contrary to popular fairy tales, a black hole doesn't reach out and "suck things in".
It's just so small (zero) that things can get very close to it.  You know that gravity
gets stronger as you get closer to an object, so if the object has no size at all, you
can get really really close to it, and THAT's where the gravity gets really strong.
You may weigh, let's say, 100 pounds on the Earth.  But you're like 4,000 miles
from the center of the Earth.  What if all of the earth's mass was crammed into
the size of a bean.  Then you could get 1 inch from it, and at that distance from
the mass of the Earth, you would weigh 25,344,000,000 pounds. 
But Earth's mass is not enough to make a black hole.  That takes a minimum
of about 3 times the mass of the sun, which is right about 1 million times the
Earth's mass.   THEN you can get a lightweight black hole.
Do you see how it works now ?

I know.  It all seems too fantastic to be true. 
It sure does.

You might be interested in
Drag the correct labels to the images. Each label can be used more than once.
coldgirl [10]

Answer:

plato answer.

Explanation:

4 0
3 years ago
A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a
Stella [2.4K]

Answer:

y = 67.6 feet,   y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

         tan θ = y / L

For the small triangle. Projector up to the person

         tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

         y₀ / (L -d) = y / L

         y = y₀  L / (L-d)

The distance from the screen (d), we look for it with kinematics

         v = d / t

        d = v t

we replace

         y = y₀ L / (L - v t)

         y = 5.2 22 / (22 - 3 t)

         y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

           y = 114.4 / (22-13)

           y = 67.6 feet

7 0
3 years ago
The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi
melomori [17]

Complete Question:

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experiences the smallest drop in temperature, and which one experiences the largest drop? Sample A: 4.0 kg of water [c = 4186 J/(kg·C°)] Sample B: 2.0 kg of oil [c = 2700 J/(kg·C°)] Sample C: 9.0 kg of dirt [c = 1050 J/(kg·C°)]

Answer:

A. Smallest B. Largest.

Explanation:

Assuming no heat exchange except for the heat removed from any sample (which we know is the same for the three ones), and that the process is done using only conduction, we can use the equation that relates the heat lost or gained by one object, with the mass of the object and the consequent change in temperature, as follows:

Q = c*m*ΔT, where c, is a proportionality constant called specific heat, which is different for each material.

As we know that the heat removed is the same for the three samples, we can equate the right sides of the equation for each sample, as follows:

cw*mw*ΔTw = co*mo*ΔTo = cd*md*ΔTd

Replacing by the givens, we have:

4.0 kg. 4,186 J/kgºC*ΔT(ºC) = 2.0 kg*2,700 J/kgºC*ΔT(ºC) =9.0kg*1,050J/kgºC*ΔT(ºC)

As the three expressions must be equal each other, it's clear that the unknown term (the drop in temperature) must compensate the product of the mass times the specific heat.

This product is the following for the three samples:

Water: 4.0 kg*4,186 J/kgºC = 16,744 J/ºC

Oil : 2.0 kg*2,700 J/kgºC    = 5,400 J/ºC

Dirt: 9.0 * 1,050 J/kgºC        = 9,450 J/ºC

Clearly, we see that in order to keep the heat exchange equations equal each other, the water must suffer the smallest drop in temperature, and the oil must experience the largest one.

So, the sample A experiencies the smallest drop in temperature, and sample B does the largest one.

5 0
2 years ago
Select the correct answer from each drop-down menu.
bagirrra123 [75]

Answer:

(1) An object that’s negatively charged has more electrons than protons.

(2) An object that’s positively charged has fewer electrons than protons.

(3) An object that’s not charged has the same number of electrons than protons.

Explanation :

Objects have three subatomic particles that are Electrons, protons, and neutrons.

Protons and neutrons are found in the nucleus and electrons rotate or move outside the nucleus. Naturally, protons are positively charged, neutrons have no charge, and electrons are negatively charged.

Therefore, an object that is negatively charged has more electrons than protons.  An object that is not charged has the same number of electrons than protons. An object that is positively charged has fewer electrons than protons.

8 0
3 years ago
A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i
tresset_1 [31]

Answer:

a) Acceleration of the car is given as

a_{car} = -21 m/s^2

b) Acceleration of the truck is given as

a_{truck} = 10.15 m/s^2

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

m_1v_1 + m_2v_2 = (m_1 + m_2)v

now we have

1400 (6.32) + 2900(0) = (1400 + 2900) v

v = 2.06 m/s

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

a_{car} = \frac{v_f - v_i}{t}

a_{car} = \frac{2.06 - 6.32}{0.203}

a_{car} = -21 m/s^2

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

a_{truck} = \frac{v_f - v_i}{t}

a_{truck} = \frac{2.06 - 0}{0.203}

a_{truck} = 10.15 m/s^2

5 0
3 years ago
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