Answer:
a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards
Explanation:
The forces on the athlete are
a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,
therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards
b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles
c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.
The final velocity of the train at the end of the given distance is 7.81 m/s.
The given parameters;
- initial velocity of the train, u = 6.4 m/s
- acceleration of the train, a = 0.1 m/s²
- distance traveled, s = 100 m
The final velocity of the train at the end of the given distance is calculated using the following kinematic equation;
v² = u² + 2as
v² = (6.4)² + (2 x 0.1 x 100)
v² = 60.96
v = √60.96
v = 7.81 m/s
Thus, the final velocity of the train at the end of the given distance is 7.81 m/s.
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Answer:
A reference frame is that frame to which the qualities of an object are related:
For instance - an object may described by - mass, speed, acceleration, size, etc,
It is important to remember that Newton's Laws of motion do not hold in accelerated reference frames -
Einstein's laws of special relativity are only true in frames that move with contant speed to one another
M1*V1 + M2*V2 = M1*V + M2*V.
1400*25 + 1800*20[180+40]=1400*V+1800*V.
Divide both sides by 100:
14*25 + 18*20[220o] = 14V + 18V.
350 + 360[220o] = 32V.
350 - 276-231i = 32V.
74 - 231i = 32V.
242.6[-72.2o] = 32V.
V = 7.6m/s[-72.2o]=7.6m/s[72o] S. of E.
Answer:
Metals, nonmetals and metalloids.
Explanation:
