Answer:
Asexual production they can be eukaryotes or prokaryotes
Explanation:
Answer:
What is observed that connects radio emissions in the galactic nucleus with the emissions in the halo or radio lobes? There is a jet of matter coming out of the nucleus, which often points toward the lobes.
Answer:
a). Determine the magnitude of the gravitational force exerted on each by the earth.
Rock: 
Pebble: 
(b)Calculate the magnitude of the acceleration of each object when released.
Rock: 
Pebble: 
Explanation:
The universal law of gravitation is defined as:
(1)
Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.
<em>Case for the rock </em>
<em>:</em>
m1 will be equal to the mass of the Earth
and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth
.

Newton's second law can be used to know the acceleration.

(2)

<em>Case for the pebble </em>
<em>:</em>


Fx = Fcos21.9
Fx = 2.3N * cos21.9
Fx = 2.13N