Answer:

Explanation:

The law of gravitation

Universal gravitational constant [S.I. units]

Mass of Earth [S.I. units]

Mass of a man in a spacecraft [S.I. units]

Earth radius [km]
Distance between man and the earth's surface
![h=261 \mathrm{~km} \quad[\mathrm{~km}]](https://tex.z-dn.net/?f=h%3D261%20%5Cmathrm%7B~km%7D%20%5Cquad%5B%5Cmathrm%7B~km%7D%5D)
ESULT 

I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
A mercury filled balloon would fall faster then water. Mercury is heavier.
The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
you need to be able to have long enough to reach and have it far away from things that are going to cause accidents