Answer:
1.11 m/s
Explanation:
The motion of the boat is an example of accelerated motion, since the velocity is not constant. However, we don't need to find the acceleration, because we are only interested in the average velocity of the boat, which is given by:

where d is the total distance covered and t the time taken. In this problem, the boat covered a distance of d = 20 m and it takes t = 18 s, therefore the average velocity is

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:
Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.
Explanation:
The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.
When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.
The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.
The velocity of the orbit is given by the relation,

Where
V - velocity of the orbit at a height h from the surface
R - Radius of the second object
G - Gravitational constant
h - height from the surface
The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.
Supposing the runner is condensed to a point and moves upward at 2.2 m/s.
It takes a time = 2.2/g = 2.2/9.8 = 0.22 seconds to increase to max height.
Now looking at this condition in opposite - that is the runner is at max height and drops back to earth in 0.22 s (symmetry of this kind of motion).
From what height does any object take 0.22 s to fall to earth (supposing there is no air friction)?
d = 1/2gt²= (0.5)(9.8)(0.22)²= 0.24 m
Wires or silver and copper