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LenaWriter [7]
3 years ago
15

A train traveling at 6.4 m/s accelerates at 0.10 m/s 2 over a distance of 100 m. How large is the train’s final velocity?

Physics
1 answer:
klasskru [66]3 years ago
4 0

The final velocity of the train at the end of the given distance is 7.81 m/s.

The given parameters;

  • initial velocity of the train, u = 6.4 m/s
  • acceleration of the train, a = 0.1 m/s²
  • distance traveled, s = 100 m

The final velocity of the train at the end of the given distance is calculated using the following kinematic equation;

v² = u² + 2as

v² = (6.4)² + (2 x 0.1 x 100)

v² = 60.96

v = √60.96

v = 7.81 m/s

Thus, the final velocity of the train at the end of the given distance is 7.81 m/s.

Learn more here:brainly.com/question/21180604

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vf = 10 m/s. A ball with mass of 4kg and a impulse given of 28N.s with a  intial velocity of 3m/s would have a final velocity of 10 m/s.

The key to solve this problem is using the equation I = F.Δt = m.Δv, Δv = vf - vi.

The impulse given to the ball with mass 4Kg is 28 N.s. If the ball were already moving at 3 m/s, to calculate its final velocity:

I = m(vf - vi) -------> I = m.vf - m.vi ------> vf = (I + m.vi)/m ------> vf = I/m + vi

Where I 28 N.s, m = 4 Kg, and vi = 3 m/s

vf = (28N.s/4kg) + 3m/s = 7m/s + 3m/s

vf = 10 m/s.

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Explanation:

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Due to the process mentioned above , the negatively charged ions are accumulated at the positive electrode and the positively charged ions are accumulated at the negative electrode . This accumulation prevents further attraction of ions at oppositely charged electrodes because the incoming ions face repulsion from already accumulated ions at electrodes.  Further , it creates an emf acting in opposite direction . It reduces the current through the solution. Hence conductivity of solution is reduced.

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Answer:

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Explanation:

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