Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.
Hope i helped
Have a good day :)
Initial velocity (Vi) = 25 m/s
acceleration (a) = 
time interval (t) = 5 sec
let us assume that final velocity after 5 sec be Vf
As acceleration is constant, we can apply the the equation of motion with constant acceleration i.e. 
Hence, 
so, the velocity of bicyclist will be 5 m/s after 5 sec
Answer:
A I think
Explanation:
because what is the most frequency a because it has more frequency I think I'm not that sure
Answer:
17.6 N
Explanation:
The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

where
is the change in momentum of the football
is the time elapsed
The change in momentum can be written as

where
m = 0.55 kg is the mass of the football
u = 0 is the initial velocity (the ball starts from rest)
v = 8.0 m/s is the final velocity
Combining the two equations and substituting the values, we find the force exerted on the ball:

Answer:
The answer to the question is
The roller coaster will reach point B with a speed of 14.72 m/s
Explanation:
Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h
Where m = mass
g = acceleration due to gravity = 9.81 m/s²
h = starting height of the roller coaster
we have the given variables
h₁ = 36 m,
h₂ = 13 m,
h₃ = 30 m
v₁ = 1.00 m/s
Total energy at point 1 = 0.5·m·v₁² + m·g·h₁
= 0.5 m×1² + m×9.81×36
=353.66·m
Total energy at point 2 = 0.5·m·v₂² + m·g·h₂
= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m
The total energy at 1 and 2 are not equal due to the frictional force which must be considered
Total energy at point 2 = Total energy at point 1 + work done against friction
Friction work = F×d×cosθ = (
× mg)×60×cos 180 = -117.72m
0.5·m·v₂² + 127.53·m = 353.66·m -117.72m
0.5·m·v₂² = 108.41×m
v₂² = 216.82
v₂ = 14.72 m/s
The roller coaster will reach point B with a speed of 14.72 m/s