Solar Nebula
Our solar system began forming within a concentration of interstellar dust and hydrogen gas called a molecular cloud. The cloud contracted under its own gravity and our proto-Sun formed in the hot dense center. The remainder of the cloud formed a swirling disk called of the solar nebula.
Answer:
The power
Explanation:
We know that the work definition is given by the following expression:
W = F * d
where:
F = force [Newtons] [N]
d = distance [meters] [m]
W = work [Joules]
And the expression that defines the work done by unit of time is called - <u>Power</u>, therefore:
P = W/t
where:
P = power [watts] [w]
W = work [Joules] [J]
t = time [seconds] [s]
C) the moon does not have a strong magnetic field
To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.
Gravitational potential energy can be defined as

As M=m, then

Where,
m = Mass
G =Gravitational Universal Constant
R = Distance /Radius
PART A) As half its initial value is u'=2u, then



Therefore replacing we have that,

Re-arrange to find v,



Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s
PART B) With a final separation distance of 2r, we have that

Therefore




Therefore the velocity when they are about to collide is 
Answer:
h=12.41m
Explanation:
N=392
r=0.6m
w=24 rad/s

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion


moment of inertia

Kinetic energy of the rotation motion

Kinetic energy translational

Total kinetic energy

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height
