Answer
given,
mass of satellite = 545 Kg
R = 6.4 x 10⁶ m
H = 2 x 6.4 x 10⁶ m
Mass of earth = 5.972 x 10²⁴ Kg
height above earth is equal to earth's mean radius
a) satellite's orbital velocity
centripetal force acting on satellite = 
gravitational force = 
equating both the above equation
v = 5578.5 m/s
b) 
T = 14416.92 s
T = 4 hr
c) gravitational force acting
F = 5202 N
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Answer:
The value of the spring constant of this spring is 1000 N/m
Explanation:
Given;
equilibrium length of the spring, L = 10.0 cm
new length of the spring, L₀ = 14 cm
applied force on the spring, F = 40 N
extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm
From Hook's law
Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.
F ∝ e
F = ke
where;
k is the spring constant
k = F / e
k = 40 / 0.04
k = 1000 N/m
Therefore, the value of the spring constant of this spring is 1000 N/m
So for a minute lets ignore the 880 km/h. If it took 4 hours and she flew at 600 km/h 600*4=2400. Now lets Look at the 880 bit. If it took 4 hours and she where to fly at 880 it would've been 880*4=3520. Lets do 2400-600=1800, now we've got the 600 kmh bit done. Now lets see if you fly 880 km/h for one hour then you add 1800+880=2680.
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