Answer:
Just as distance and displacement have distinctly different meanings (despite their similarities), so do speed and velocity. Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance. A fast-moving object has a high speed and covers a relatively large distance in a short amount of time. Contrast this to a slow-moving object that has a low speed; it covers a relatively small amount of distance in the same amount of time. An object with no movement at all has a zero speed.
The total momentum of the system is preserved through the collision.
Note that momentum is
P = m*v
where m = mass
v = velocity.
Initial momentum:
P1 = (30000 kg)*(2 m/s) = 60000 (kg-m)/s for the moving car
P2 = 0 for the starionary car.
Final momentum:
P3 = (30000 + 30000)*v = 60000v (kg-m)/s
Because momentum is preserved,
P3 = P1 + P2
60000v = 60000
v = 1 m/s
The final velocity is 1 m/s.
Answer: 1.0 m/s
Answer:
F, = 12N. F, = 2 N. Block. 4) a 20.0-kg mass moving at 1.00 m/s.
Explanation:
Actual displacement that he required to move
towards North
Displacement that he moved due to snow is
at 47 degree North of East
now in vector component form we can say



now the displacement that is more required to reach the destination is given as



so the magnitude of the displacement is given as


its direction is given as

so it is 5.54 km towards 5.38 degree North of West.