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:
Explanation:
a )
momentum of baseball before collision
mass x velocity
= .145 x 30.5
= 4.4225 kg m /s
momentum of brick after collision
= 5.75 x 1.1
= 6.325 kg m/s
Applying conservation of momentum
4.4225 + 0 = .145 x v + 6.325 , v is velocity of baseball after collision.
v = - 13.12 m / s
b )
kinetic energy of baseball before collision = 1/2 mv²
= .5 x .145 x 30.5²
= 67.44 J
Total kinetic energy before collision = 67.44 J
c )
kinetic energy of baseball after collision = 1/2 x .145 x 13.12²
= 12.48 J .
kinetic energy of brick after collision
= .5 x 5.75 x 1.1²
= 3.48 J
Total kinetic energy after collision
= 15.96 J
The correct answer to this qustion is velocity and time
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