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mixer [17]
2 years ago
10

Convert the following as instructed:

Physics
1 answer:
Elis [28]2 years ago
7 0
3.4m
6700g
172800s
1 day
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A tow rope pulls a 1450 kg truck, giving it an acceleration 1.25 m/s^2. What force does the rope exert?
bagirrra123 [75]

force=mass × acceleration

mass=force ÷ acceleration

acceleration=force ÷ mass

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1) Calculate the potential energy of a 5.00 kg object sitting on a 3.00 meter high ledge.
maria [59]

Answer:

15kg

Explanation:

5 0
3 years ago
A railroad car of mass 2.52 104 kg is moving with a speed of 3.86 m/s. It collides and couples with three other coupled railroad
aleksley [76]

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

         p_{f} = (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

         Em_{f} = 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

6 0
3 years ago
The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr
Makovka662 [10]

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.

8 0
3 years ago
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