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3 years ago

Is a neutron star also a black hole?

1 answer:

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

Why do high and low tides happen 1 hour later each day? just answer please and it is a written response

8_murik_8 [283]

With each passing day, the high tides occur about an hour later. The moon rises about an hour later each day, too (actually, 54 minutes later). Since the moon pulls up the tides, these two delays are connected. As the earth rotates through one day, the moon moves in its orbit.

6 0

3 years ago

A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.3 = constant.

Galina-37 [17]

Answer:

Change in specific internal Energy $=250\ \rm Btu/lb$

Explanation:

Given:

Mass of the gas, m=0.4 lb
Initial pressure and volume are $p_1=160\ \rm lbf/in^2\ and\ v_1=1\ \rm ft^3\\$

Final pressure and temperature are $p_1=480\ \rm lbf/in^2$
Heat transfer from the gas is 2.1 Btu

Since the process is isotropic we have

$p_1v_1^{1.3}=p_2v_2^{1.3}\\160\times 1^{1.3}=480\times v_2^{1.3}\\v_2=0.43\ \rm ft^3\\$

So the final volume of the gas is calculated.

Work in any isotropic is given by w

$w=\dfrac{p_1v_1-P_2v_2}{n-1}\\\\w=\dfrac{160\times1-480\times0.43}{1.3-1}\\w=-154.67\ \rm Btu\\$

According to the first law of thermodynamics we have

$Q=\Delta U+w\\-2.1=\Delta U-154.67\\\Delta U=152.56\ \rm Btu\\$

So the Specific Internal Change is given by

$\Delta u=\dfrac{\Delta U}{m}\\\Delta u=\dfrac{152.56}{0.4}\\\Delta u=250\ \rm Btu/lb$

So the specific Change in Internal energy is calculated.

6 0

3 years ago

by scale drawing, find the resultant of vectors 70N inclined at100N and direction of the resultant 100N

Thepotemich [5.8K]

Answer: 70n=

100n=

Explanation:

4 0

3 years ago

A block of ice with mass 5.50 kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal

Yuki888 [10]

Answer:

A) 3.13 m/s

B) 5.34 N

C) W = 26.9 J

Explanation:

We are told that the position as a function of time is given by;

x(t) = αt² + βt³

Where;

α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³

Thus;

x(t) = 0.21t² + 0.0204t³

A) Velocity is gotten from the derivative of the displacement.

Thus;

v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)

v(t) = 0.42t + 0.0612t²

v(4.5) = 0.42(4.5) + 0.0612(4.5)²

v(4.5) = 3.1293 m/s ≈ 3.13 m/s

B) acceleration is gotten from the derivative of the velocity

a(t) = v'(t) = 0.42 + 2(0.0612t)

a(4.5) = 0.42 + 2(0.0612 × 4.5)

a(4.5) = 0.9708 m/s²

Force = ma = 5.5 × 0.9708

F = 5.3394 N ≈ 5.34 N

C) Since no friction, work done is kinetic energy.

Thus;

W = ½mv²

W = ½ × 5.5 × 3.1293²

W = 26.9 J

6 0

3 years ago

Please help me guys never mind the calculations

vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} \\ \frac{1}{R} = \frac{3 + 2}{6} = \frac{5}{6} \\ R = \frac{6}{5} = 1.2 \: \: ohm$

5.2) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{8} + \frac{1}{10} \\ \frac{1}{R} = \frac{5 + 4}{40} = \frac{9}{40} \\ R = \frac{40}{9} = 4.4 \: \: ohm$

I hope I helped you^_^

7 0

3 years ago