What is the atomic bomb equivalent to a 7.0 earthquake
1 answer:
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The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force 
which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,
so
The frictional force can be rewritten as
where
, 
. Re-arranging, we can solve this equation to find 
, the coefficient of dynamic friction:
Therefore, since the resultant of the two forces must be zero,
so
The frictional force can be rewritten as
where
Answer:
d = 1.13*10^{-4}m = 0.113mm
Explanation:
To find the minimum diameter, that allow to antiproton circulate in the chamber without touching the walls, you use the following formula for the radius of the trajectory of a charged particle in a constant magnetic field.
(1)
r: radius of the trajectory
m: mass of the antiproton = 9.1*10-31 kg
v: velocity of the antiproton = 4.0*10^4 m/s
B: magnitude of the magnetic field = 4.0mT = 4.0*10^-3 T
q: charge of the antiproton = +1.6*10^{-19}C
You replace the values of the parameters in (1):
Then, the diameter of the chamber must be, at least:
d=2r = 2(5.68*10^-5) = 1.13*10^{-4}m = 0.113mm