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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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vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses
dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}

(This is correct because the horizontal motion has acceleration zero). Then:

v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}

Then, plugging in the given values, we obtain:

k=\frac{(61.2kg)(15.0m/s)^2}{(2.76m)^2}\\\\k=1808N/m

Finally, the effective spring constant of the firing mechanism is 1808N/m.

3 0
3 years ago
. If force (F), work (W) and velocity (v) are taken as fundamental quantities.
alex41 [277]

Answer:

∴ [T]=[WF−1V−1]

Hope this answer is right!!

7 0
3 years ago
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An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.50 m above the ground. The railcar
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When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is direc
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Answer: F

Out of the page.

Explanation:

For an electron with a charge of -e, the magnitude of the force on it is F = BeV

Where

F = force on the electron

e = charge ( electrons )

V = velocity

B = magnetic field

F is the force acting on all the electrons in a wire which gives rise to the F = BIL

Where

I = current

L = length of the wire

The force F is always at the right angle to the particle's velocity and its direction can be found using the left hand rule.

When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.

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2.0y i think this is guess but if it right then thats good
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