Answer:
The ratio of the orbital time periods of A and B is 
Solution:
As per the question:
The orbit of the two satellites is circular
Also,
Orbital speed of A is 2 times the orbital speed of B
(1)
Now, we know that the orbital speed of a satellite for circular orbits is given by:
where
R = Radius of the orbit
Now,
For satellite A:
Using eqn (1):
(2)
For satellite B:
(3)
Now, comparing eqn (2) and eqn (3):
Force, F = ma
Where m = mass in kg, a = acceleration in m/s², Force, F is in N.
F = ma
2000 = m*2.2
2.2m = 2000
m = 2000/2.2
m ≈ 909.09
Mass is ≈ 909.09 kg.
Answer:
the force of the friction is A-0.52
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:
m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,
m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂' --> equation 1
The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is
(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2
Solving equations 1 and 2 simultaneously, v₁' = -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
