We know that:
W=Fs
200J=150N*s
s=200J/150N
s=1,33m
3,89,988 cm/min is the linear velocity
Given,
Diameter of CD = 12 cm
So, Radius of CD = 6 cm
CD is spinning at 10350 rev/min
Firstly , convert rev/min into rad/min
1 rev = 2π radians
10350 rev/min = 10350 × 2π
= 64998 rad/min
Formula used,
where,
is the Linear velocity
is the radius
is the angular velocity
= 6 cm × 64998rad/min
= 3,89,988 cm/min
Thus, linear velocity for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev/min is 389988 cm/min.
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Answer:

Explanation:
Given that,
The mass of a Hubble Space Telescope, 
It orbits the Earth at an altitude of 
We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

Where
is the mass of Earth
Put all the values,

So, the potential energy of the telescope is
.
Explanation:
what is the question? could you pls provide it
Answer:
52 rad
Explanation:
Using
Ф = ω't +1/2αt²................... Equation 1
Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.
Since the object states from rest, ω' = 0 rad/s.
Therefore,
Ф = 1/2αt²................ Equation 2
make α the subject of the equation
α = 2Ф/t².................. Equation 3
Given: Ф = 13 rad, t = 2.5 s
Substitute into equation 3
α = 2(13)/2.5²
α = 26/2.5
α = 4.16 rad/s².
using equation 2,
Ф = 1/2αt²
Given: t = 5 s, α = 4.16 rad/s²
Substitute into equation 2
Ф = 1/2(4.16)(5²)
Ф = 52 rad.