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andreev551 [17]
3 years ago
7

Explain to your friend, who is willing to accept that light moves at the same speed in any frame, why clocks on a passing train

are not synchronized. If it helps, assume diat Anna is at tire middle of the train.
Physics
1 answer:
vovikov84 [41]3 years ago
7 0

First we start from the definition of synchronization itself. This is known as the rate of change in the occurrence of an event to occur at the same time. If the light clocks of a train are verified it will be possible to observe that there are different verification times between each of them different. This is because the light of the clocks at each end of the train does not appear instantaneously over them, but travels from one direction to another; the light at the rear must have traveled a distance that will cause the first clock to have changed position.

In this way, if a clock of these is observed at the same time, the light will generate that the first one is out of sync with the last one, which would cause the clocks not to be synchronized.

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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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3 years ago
A ball rolls off a table and it traveling with a horizontal velocity of 2 m/s and 1 point
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Answer:

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4 0
2 years ago
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A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th
solniwko [45]

Answer:

a = R\alpha\sqrt{1 + \alpha^2t^4}

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here we know that

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a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}

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