Answer:
No, the distance from the last stop to the school and the time it takes to travel that distance are required.
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
A = 4\pi r^2
A = 4\pi (2\mu m /2)^2 (10^{-6}m/1\mu m)^2 (1mm/10{-3})^2
A = 1.33*!0^{-5}MM^2
Answer:
The planet Jupiter completes one revolution of the sun in 362710000 seconds. Long time, right?
Explanation:
3.154x10^7=3.154x10000000=31540000
11.5x31540000=362710000
Answer:
it can be calculated by measuring the final distance away from a point, and then subtracting the initial distance