Temperature and elevation, if it is cold in Idaho and warm on the eastern end of a mountain side in california (or if warm air is going in that direction) then the cold air, being more dense, will go towards california while the cold air in Idaho will become warm. Same goes for the rest of the world
Answer:
Explanation:
Given
mass of rock 
Elevation of Rock 
Distance traveled by rock with time

where, u=initial velocity
t=time
a=acceleration
here initial velocity is zero
when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m



velocity at this time



Answer:
D. Hypnosis can make the subjects talk, but they talk only about their childhoods.
Explanation:
Answer:
1.97 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.8 m/s²

Solving the above equation we get

So, the time the package was in the air is 1.97 seconds
Answer:
It has been learned in this lesson that the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period. ... Once calculated, this area represents the displacement of the object.
Explanation: