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.
Answer:
It can only display one record at a time
Explanation:
Form ;
1. This is a document with spaces (also called placeholders or fields ) in which a series of documents with similar content can be written or selected.
2.This is the most popular method of data entry
3.It may contain images in the background.
4.This can be sorted data regardless of its source of information.
Only option C is wrong.
Therefore the answer C is correct.
The first law of Newton’s law state an object in motion will stay in motion and an object at rest will stay at rest unless acted upon with an outside force. Other know as the law of inertia so yes it does
Answer:
the net force applied to an object is directly proportional to the acceleration undergone by that object
Explanation:
This verbal statement can be expressed in equation form as follows:
a = Fnet / m
Answer:
60 days.
Explanation:
Let the original mass (N₀) = 1 g
Amount remaining (N) = 6.25% of its original mass
= 6.25% × 1
= 6.25/100 × 1
= 0.0625 g
Half life (t½) = 15 days
Time (t) =?
Next, we shall determine the rate of decay. This can be obtained as follow:
Decay constant (K) = 0.693/ half life
K = 0.693 / t½
Half life (t½) = 15 days
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 15
K = 0.0462 / day
Finally, we shall determine the time taken for the sample of the isotope to decay to 6.25% of its original mass.
This can be obtained as follow:
Original amount (N₀) = 1 g
Amount remaining (N) = 0.0625 g
Decay constant (K) = 0.0462 / day
Time (t) =?
Log (N₀/N) = Kt/2.3
Log (1/0.0625) = 0.0462 × t / 2.3
Log 16 = 0.0462 × t / 2.3
1.2041 = 0.0462 × t /2.3
Cross multiply
0.0462 × t = 1.2041 × 2.3
Divide both side by 0.0462
t = (1.2041 × 2.3)/0.0462
t = 59.9 ≈ 60 days
Therefore, the time taken for the sample of the isotope to decay to 6.25% of its original mass is 60 days