A 5 kg projectile is fired at an angle of 25o above the horizontal. Its initial velocity is 200 m/s and just before it hits the
1 answer:
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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
Answer:
3.15m³
Explanation:
To solve this problem, let us first find the mass of the petrol from the given dimension.
Mass = density x volume
Volume of petrol = 4.2m³
Density of petrol = 0.3kgm⁻³
Mass of petrol = 4.2 x 0.3 = 1.26kg
So;
We can now find the volume of the alcohol
Volume of alcohol =
Mass of alcohol = 1.26kg
Density of alcohol = 0.4kgm⁻³
Volume of alcohol = = 3.15m³