force=mass × acceleration
mass=force ÷ acceleration
acceleration=force ÷ mass
Answer:
a) v = 2.4125 m / s , b) Em_{f} / Em₀ = 0.89
Explanation:
a) This is an inelastic crash problem, the system is made up of the four carriages, so the forces during the crash are internal and the moment is conserved
Initial
p₀ = m v₁ + 3 m v₂
Final
= (4 m) v
p₀ =p_{f}
m (v₁ + 3 v₂) = 4 m v
v = (v₁ +3 v₂) / 4
Let's calculate
v = (3.86 + 3 1.93) / 4
v = 2.4125 m / s
b) the initial mechanical energy is
Em₀ = K₁ + 3 K₂
Em₀ = ½ m v₁² + ½ 3m v₂²
The final mechanical energy
= K
Em_{f} = ½ 4 m v²
The fraction of energy lost is
Em_{f} / Em₀ = ½ 4m v² / ½ m (v₁² +3 v₂²)
Em_{f} / Em₀ = 4 v₂ / (v₁² + 3 v₂²)
Em_{f} / Em₀ = 4 2.4125² / (3.86² + 3 1.93²)
Em_{f} / em₀ = 23.28 / 26.07
Em_{f} / Em₀ = 0.89
Answer:
The tube should be held vertically, perpendicular to the ground.
Explanation:
As the power lines of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its potential.
And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.