Answer:
The required angular speed the neutron star is 10992.32 rad/s
Explanation:
Given the data in the question;
mass of the sun M
= 1.99 × 10³⁰ kg
Mass of the neutron star
M
= 2( M )
M = 2( 1.99 × 10³⁰ kg )
M = ( 3.98 × 10³⁰ kg )
Radius of neutron star R = 13.0 km = 13 × 10³ m
Now, let mass of a small object on the neutron star be m
angular speed be ω.
During rotational motion, the gravitational force on the object supplies the necessary centripetal force.
GmM = / R² = mRω²
ω² = GM = / R³
ω = √(GM = / R³)
we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²
we substitute
ω = √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)
ω = √( 2.65466 × 10²⁰ / 2.197 × 10¹²
ω = √ 120831133.3636777
ω = 10992.32 rad/s
Therefore, The required angular speed the neutron star is 10992.32 rad/s
Answer:
The best option is for the following option m = 15 [g] and V = 5 [cm³]
Explanation:
We have that the density of a body is defined as the ratio of mass to volume.
where:
Ro = density = 3 [g/cm³]
Now we must determine the densities with each of the given values.
<u>For m = 7 [g] and V = 2.3 [cm³]</u>
![Ro=7/2.3\\Ro=3.04 [g/cm^{3} ]](https://tex.z-dn.net/?f=Ro%3D7%2F2.3%5C%5CRo%3D3.04%20%5Bg%2Fcm%5E%7B3%7D%20%5D)
<u>For m = 10 [g] and V = 7 [cm³]</u>
<u />![Ro=10/7\\Ro=1.42[g/cm^{3} ]\\](https://tex.z-dn.net/?f=Ro%3D10%2F7%5C%5CRo%3D1.42%5Bg%2Fcm%5E%7B3%7D%20%5D%5C%5C)
<u />
<u>For m = 15 [g] and V = 5 [cm³]</u>
<u />![Ro=15/5\\Ro=3[g/cm^{3} ]\\](https://tex.z-dn.net/?f=Ro%3D15%2F5%5C%5CRo%3D3%5Bg%2Fcm%5E%7B3%7D%20%5D%5C%5C)
<u />
<u>For m = 21 [g] and V = 8 [cm³]</u>
<u />![Ro=21/8\\Ro=2.625[g/cm^{3} ]\\](https://tex.z-dn.net/?f=Ro%3D21%2F8%5C%5CRo%3D2.625%5Bg%2Fcm%5E%7B3%7D%20%5D%5C%5C)
<u />
Answer:
Sck my p3nis
Explanation:
if you do so, then your mom will have coronavirus.
High temperature gives the hydrogen atoms enough energy to overcome the electrical repulsion between the protons. Fusion requires temperatures of about 100 million Kelvin (approximately six times hotter than the sun's core).
2.c
3.b
1.a
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