Answer:
D
Explanation:
19.6 newtons
A 2.00-kilogram object weighs 19.6 newtons on Earth.
Observer A is moving inside the train
so here observer A will not be able to see the change in position of train as he is standing in the same reference frame
So here as per observer A the train will remain at rest and its not moving at all
Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body
So here observer B will see the actual motion of train which is moving in forward direction away from the platform
Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction
So the distance between them will decrease at faster rate
Now as per Newton's II law
F = ma
Now if train apply the brakes the net force on it will be opposite to its motion
So we can say
- F = ma

so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate
It is not affected by the gravity because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train
So there is no effect on train motion
I believe the answer is H for when you bounce it, it has stress when it hits the floor and then goes up giving it kinetic
We need to see what forces act on the box:
In the x direction:
Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.
In the y direction:
N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.
From N-Gcosα=0 we get:
N=Gcosα, we will need this for the force of friction.
Now to solve for Fh:
Fh=ma + Ff + Gsinα,
Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²
Fh=ma + μmgcosα+mgsinα
Now we plug in the numbers and get:
Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N
The horizontal force for pulling the body up the ramp needs to be Fh=71 N.