The equator has no continental borders.
Answer:
As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.
(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.
(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.
(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.
(iv) At point A, the driver will feel the lightest.
(v)The car can go that much fast without losing contact with the road at A can be determined as follow:
Fn=0 - lose contact with road
Fg= mv²/r
mg=mv²/r
v=sqrt (gr)
Answer:
The answer to the question is as follows
The acceleration due to gravity for low for orbit is 9.231 m/s²
Explanation:
The gravitational force is given as
Where 
= Gravitational force
G = Gravitational constant = 6.67×10⁻¹¹
m₁ = mEarth = mass of Earth = 6×10²⁴ kg
m₂ = The other mass which is acted upon by and = 1 kg
rEarth = The distance between the two masses = 6.40 x 10⁶ m
therefore at a height of 400 km above the erth we have
r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m
and = 
= 9.231 N
Therefore the acceleration due to gravity = /mass
9.231/1 or 9.231 m/s²
Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²
equilibrium i think if not sorry
Heck yeah i'll follow!!!!!!
