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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.

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N 1800kg car has an<br> of 3.8m/s? What is it<br> on the car?<br> acceleration<br> force acting
nevsk [136]

Answer:

6840 N

Explanation:

The force acting on the car can be found by using Newton's second law:

F = ma

where

F is the net force on the car

m is the mass of the car

a is its acceleration

For the car in this problem,

m = 1800 kg

a=3.8 m/s^2

Substituting,

F=(1800)(3.8)=6840 N

7 0
3 years ago
the force shown in figure 7-15 moves an object from x=0 to x=0.75 m. How much work is done by the force?
Phantasy [73]
Work is force multiplied by the distance the force moves the object
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3 years ago
What is similar about analogical and symbolic representations?
ICE Princess25 [194]
Symbolic representations are mental pictures that have no direct relationship to the actual object you are thinking about. Instead, these mental pictures are connected by themes that are meaningful to you. Anytime you think of words and descriptions for one main concept, you're thinking symbolically. Analogical representations are mental pictures that have a direct relationship to the actual object you are thinking about. Do cows have their ears on the top or side of their heads? Rather than thinking symbolically about a cow and his ears (tiny, smelly, leather), analogical representations of the cow's ears involve thinking of an actual cow in your head.
3 0
3 years ago
Read 2 more answers
5p - 14 = 8 p + 4 find p
Digiron [165]

Answer:

Explanation:

5p - 14 = 8p + 4

5p = 8p + 18 <-- Moving constants to one side; add the same number of +14 to both sides.

-3p = 18. <-- The same thing with the variable itself.

p = -6 <-- Divide both sides by negative 3.

6 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
Marizza181 [45]

Answer:

c) 2.02 x 10^16 nuclei

Explanation:

The isotope decay of an atom follows the equation:

ln[A] = -kt + ln[A]₀

<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>

[A] = Our incognite

k is constant decay:

k = ln 2 / Half-life

k = ln 2 / 4.96 x 10^3 s

k = 1.40x10⁻⁴s⁻¹

t is time = 1.98 x 10^4 s

[A]₀ = 3.21 x 10^17 nuclei

ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]

ln[A] = 37.538

[A] = 2.01x10¹⁶ nuclei remain ≈

<h3>c) 2.02 x 10^16 nuclei</h3>
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