Answer:
A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
Explanation:
We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

Where:
v: is the tangential speed of the disk
t: is the time = 30 s
The tangential speed can be found as follows:

Where:
ω: is the angular speed = 100 rpm
r: is the radius = 50 cm = 0.50 m
Now, the distance traveled by the disk is:

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
I hope it helps you!
S=56, u=0, v=33, a=?, t=3.4
v=u+at
33=3.4 a
a = 9.7m/s^2
Answer:

Explanation:
From the second law of Newton movement laws, we have:
, and we know that a is the acceleration, which definition is:
, so:

The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):

(This is the indefinite integral), the definite one is:

Answer:
r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s
Increasing its velocity will add to the kinetic energy more as the formula for kinetic energy is 0.5*m*v^2. (The speed will be squared making it greater)