Answer:
Velocity of truck will be 20.287 m /sec
Explanation:
We have given mass of the truck m = 4000 kg
Radius of the turn r = 70 m
Coefficient of friction
Centripetal force is given
And frictional force is equal to
For body to be move these two forces must be equal
So
Answer:
a) the magnitude of the force is
F= Q() and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E =
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) = , where p = q × s
E(r) = where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q()
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²