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abruzzese [7]
3 years ago
5

Points A, B, and C lie along a line from left to right, respectively. Point B is at a lower electric potential than point A. Poi

nt C is at a lower electric potential than point B. What would best describes the subsequent motion, if any, of a positively-charged particle released from rest at point B?
Physics
1 answer:
arlik [135]3 years ago
6 0

Answer:

Please see below as the answer is self-explanatory.

Explanation:

  • If the potential at B is lower than A, and the potential at C is lower than B, this means that there is an electric field, directed from A to C.
  • If a positively-charged particle is released at rest at point B, it will be accelerated by the electric field  (which is a force per unit charge, so it produces an acceleration) in the same direction than the field (because it is a positive charge) towards point C.
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3 years ago
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A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve
Troyanec [42]

The velocity of tennis racket after collision is 14.96m/s

<u>Explanation:</u>

Given-

Mass, m = 0.311kg

u1 = 30.3m/s

m2 = 0.057kg

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Since m2 is moving in opposite direction, u2 = -19.2m/s

Velocity of m1 after collision  = ?

Let the velocity of m1 after collision be v

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Therefore,

m1u1 - m2u2 = m1v1 + m2v2

v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2

v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96

Therefore, the velocity of tennis racket after collision is 14.96m/s

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Answer:

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