a 2.0*10^3 kg car accelerates from rest under the actions of two forces. One is a forward force of 1140 N provided by traction b
1 answer:
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Answer:
Explanation:
From the given information:
Let the first weight be = 80 kg
The weight of the buddy be = 120 kg
The weight of Bubba be = 60 kg
Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:
The length of the boat be = 4 m
∴
We can find the center of mass of the system by using the formula:
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