Let <em>a</em> denote the airplane's velocity in the air, <em>g</em> its velocity on the ground, and <em>w</em> the velocity of the wind. (Note that these are vectors.) Then
<em>a</em> = <em>g</em> + <em>w</em>
and we're given
<em>a</em> = (325 m/s) <em>j</em>
<em>w</em> = (55.0 m/s) <em>i</em>
Then
<em>g</em> = - (55.0 m/s) <em>i</em> + (325 m/s) <em>j</em>
The ground speed is the magnitude of this vector:
||<em>g</em>|| = √[ (-55.0 m/s)² + (325 m/s)² ] ≈ 330. m/s
which is faster than the air speed, which is ||<em>a</em>|| = 325 m/s.
Answer:
B) Because the Space Station is constantly in free-fall around the Earth.
Explanation:
Anything that is falling experiences an upward force on them. For example when a person is going down in a lift they will experience something that is pushing them upwards. This happens due to the fact that the total acceleration the body is feeling is less than the acceleration due to graviity.
The force on a body which is falling is

Where,
m = Mass of object
g = acceleration due to gravity
a = acceleration the object is experiencing.
a = g. So, the force becomes zero and the object experiences weightlessness.
Hence, the astronauts in the space station experience weightlessness due to fact that the Space Station is constantly in free-fall around the Earth.
Answer:
D. The objects particles move faster
The initial is where you are starting and the final postion is where the object ends up