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Bogdan [553]
3 years ago
7

If the core of a supernova contains about one solar mass, the core will become a

Physics
2 answers:
irina [24]3 years ago
3 0
A solar mass of above 1.4 will become a neutron star according to the Chandrasekhar limit, therefore if there is 1 solar mass it will collapse to form a white dwarf.a solar mass of above 1.4 will become a neutron star according to the Chandrasekhar limit, therefore if there is 1 solar mass it will collapse to form a white dwarf.

Hope this helps!
grin007 [14]3 years ago
3 0

Answer:

It will become a neutron star

Explanation:

For the solar mass of anything above 1.4, the mass will automatically become a neutron star. This is according to the Chandrasekhar limit. This limit determines that 1 solar mass will collapse to form a white dwarf of a solar mass above 1.4 .

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How many milligrams in 1 gram. ??? HELP !!!
blagie [28]
1,000 milligrams = 1 gram
2,000 milligrams = 2 grams
3,000 milligrams = 3 grams
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3 years ago
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A wind turbine is rotating counterclockwise at 0.626 rev/s and slows to a stop in 12.9 s. Its blades are 17.9 m in length. What
iogann1982 [59]

Answer:

276.5 m/s^2

Explanation:

The initial angular velocity of the turbine is

\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s

The length of the blade is

r = 17.9 m

So the centripetal acceleration is given by

a=\omega^2 r

At the instant t = 0,

\omega=3.93 rad/s

So the centripetal acceleration of the tip of the blades is

a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2

5 0
3 years ago
True or false, the older rocks tend to be closer to the tear in the ocean floor
Natali5045456 [20]

Answer:

false!!

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4 0
3 years ago
What impact can a very large volcanic eruption have on earths climate
Vaselesa [24]

Answer:

  1. It could instantly kill all life on Earth
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8 0
4 years ago
To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where
BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

\int E.dA = \frac{q}{\epcilon_0}

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

q = \int \rho dV

q = \rho * \pi r^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r}{2 \epcilon_0}

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

q = \int \rho dV

q = \rho * \pi r_0^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r_0^2}{2 \epcilon_0 r}


7 0
3 years ago
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