Answer:
The answer to the question is
At an altitude of 1413 km or 1.222·R above the north pole the weight of an object reduced to 67% of its earth-surface value
Explanation:
We make use of the gravitational formula as follows
F = G
where
m₁ = mass of the object
m₂ = mass of the earth
d = distance between the two objects and
G = gravitational constant
if at the altitude the weight is reduced to 67 % of its weight on earth then
with all other variables remaining constant, we have
67% F = G
=0.67× G×
cancelleing like ternss from both sides we have
1/R₂² =0.67/R₁² or r₁²/R₂² =0.67 and R₁/R₂ = 0.8185
or R₂ = R₁/0.8185 or 1.222·R = (6.37×10⁶ m)/0.8185 = 7.783×10⁶ m
Hence at an altitude = 1.413×10⁶ m the weight of the object will reduce to 67% its earth-surface value
Answer:
7.92 m/s
Explanation:
= Mass of raindrop = 
= Mass of mosquito
= Initial Velocity of raindrop = 8.1 m/s
= Initial Velocity of mosquito = 0 m/s
= Velocity of center of mass
For elastic collision

Hence, the velocity of the attached mosquito, falling immediately afterward is 7.92 m/s
Answer:
Distance moved = 42 meters
Mag and Dir = 10 meters northward
Explanation:
Distance is total distance traveled which = 22+16+4 = 42 meters
Displacement is the amount of distance moved from the start point to the end point straight line so our equation for the magnitude is 22-16+4 = 10 meters. Then the direction is northward from the start point. This answer would also be a vector because it has a magnitude and direction.