Answer:
Explanation:
1. FALL PROTECTION-GENERAL REQUIREMENTS (29 CFR 1926.501) 6,010 VIOLATIONS
2. HAZARD COMMUNICATION (29 CFR 1910.1200). 3,671
3. SCAFFOLDING (29 CFR 1926.451). 2,813
Answer:
Explanation:
As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron
So here we have to find the work done by electric field on moving electron
So we have
now the distance moved by the electron is given as
so we have
now we have to convert it into keV units
so we have
<span>Radius = 4.6 m
Time for one complete rotation t = 5.5 s.
Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
Force of the cat Fc = 6m, m being the mass.
Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116
So coefficient of static friction = 0.6116</span>
This may helpv^2=u^2+2as. v=0 at top of flight. a=acceleration of gravity(vo^2)/2a=s.
Answer:
C
Explanation:
Because this same question was on my test last week and I got it correct