Answer:
Easy search it on g o o g l e
Answer:
B 14.5 m/s to the east
Explanation:
We can solve this problem by using the law of conservation of momentum.
In fact, if the system is isolated, the total momentum of the system must be conserved.
Here the total momentum before the stuntman reaches the skateboard is:

where
M = 72.0 kg is the mass of the stuntman
v = 15.0 m/s is his initial velocity (to the east)
The total momentum after the stuntmen reaches the skateboard is:

where
m = 2.50 kg is the mass of the skateboard
v' is the final velocity of the stuntman and the skateboard
Since momentum must be conserved, we have

And solvign for v',

And since the sign is the same as v, the direction is the same (to the east).
Momentum of a body is calculated by multiplying the mass of a moving body with its velocity. When a body is at rest it has zero momentum since the velocity is also zero.
In this case the momentum of the canoe will be;
38 kg × 2.2 m/s = 83.6 kgm/s
Therefore, the correct answer is 83.6 kg m/s
Answer:
The answer is below
Explanation:
The speed of the boat in still water is perpendicular to the speed of the water flow. Therefore the speed relative to the ground (V), the speed of flow and the speed of the boat in still water form a right angled triangle. Hence the speed relative to the ground is given as:
V² = 56² + 126²
V² = 19012
V = 137.9 m/s