Answer:
50% of it .
Explanation:
50% of it is illuminated by the Sun.
 
        
             
        
        
        
Answer:
3054.4 km/h 
Explanation:
Using the conservation of momentum
momentum before separation = 5M × 2980 Km/h where M represent the mass of the module while 4 M represent the mass of the motor 
initial momentum = 14900 M km/h
let v be the new speed of the motor so that the 
new momentum = 4Mv and the new momentum of the module  = M ( v + 94 km/h ) 
total momentum = 4Mv + Mv + 93 M = 5 Mv + 93M
initial momentum = final momentum 
14900 M km/h = 5 Mv + 93M
14900 km/h = 5v + 93 
14900 - 93 = 5v 
v = 2961.4 km/h
the speed of the module = 2961.4 + 93 = 3054.4 km/h 
 
        
             
        
        
        
821
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Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
   = (260 N)*(0.33 m)
   = 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
  = (85.8 N-m)/(0.81 rad/s²)
  = 105.93 kg-m²
Answer: 105.93 kg-m²