Ok, I think this is right but I am not sure:
Q = ϵ
0AE
A= π π
r^2
=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
(3x10^6 N/C) =3.3x10^-8 C = 33nC N = Q/e = (3.3x10^-8 C)/(1.60x10^-19 C/electron) = 2.1x10^11 electrons
no, work is = force * distance or displacement
U=10 m/s
v=30 m/s
t=6 sec
therefore, a=(v-u)/t
=(30-10)/6
=(10/3) ms^-2
now, displacement=ut+0.5*a*t^2
=60+ 0.5*(10/3)*36
=120 m
And you can solve it in another way:
v^2=u^2+2as
or, s=(v^2-u^2)/2a
=(900-100)/6.6666666.......
=120 m
Answer:

Explanation:
The relationship between the linear distance covered by an object and its angular displacement is given by the following formula:
s = rθ
where,
s = distance traveled on road = ?
r radius of tires = diameter/2 = 2.2 m/2 = 1.1 m
θ = angular displacement = (60 rev)(2π rad/1 rev) = 377 rad
Therefore,

Answer:
As much I know the gravity on moon is 1.62m/s२.