Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
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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
Answer:
Hindi ko po alam
Explanation:
Kaya sa iba ka nalang mag tanong kac Bobo ako jaan
