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
= / R
² = mR
ω
²
ω
² = GM
= / R
³
ω
= √(GM
= / R
³)
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:
7.55 km/s
Explanation:
The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

where
is the gravitational constant
is the mass of the telescope
is the mass of the Earth
is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)
v = ? is the orbital velocity of the Hubble telescope
Re-arranging the equation and substituting numbers, we find the orbital velocity:

They will subtract to form a combined wave with a lower amplitude
Since work is the change in kinetic energy, the efficiency of a machine can be stated as the percentage of the output work divided by the input work minus the work lost from to friction and heat. Multiply Eff by 100% to get the efficiency percentage.
Force=mass x acceleration
f= 0.5 x40
f=20N