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riadik2000 [5.3K]
3 years ago
7

A. How does the speed of light in a vacuum change when observed from two different frames of reference?

Physics
2 answers:
Katyanochek1 [597]3 years ago
5 0

Answer:

a) The velocity of the light is invariant, this means that is always the same independent of the frame in which you are looking.

b) Inertial frames, this means that the frame is "not accelerating" (it can be fixed in place or moving with constant velocity)

C) When you are in the train, you are in the same frame than the clock, so you will no see nothing interesting, now, if you are outside of the train you will see a time dilation, this means that each "tick" of the clock is slightly longer than the one that the observer is holding. This happens because of the invariance of the speed of light, a way of seeing it is the next:

Suppose that the clock works sending a light signal, it travels a distance "d", it rebounds in a surface and comes to a receptor, when the photon hits the receptor, a "tick" of the clock is recorded, which would mean that a second has passed.

If the clock is still, the photon travels the distance d two times at a velocity c, so the "tick" of the clock is each "2d/c" seconds.

Now, in the moving clock, the distance that the photon moves is slightly higher (because now it also moves with the clock) suppose that this distance is D, so now the time is "2D/c", which will be bigger than the one that we had before.

anyanavicka [17]3 years ago
4 0

a) It is absolute, so it does not change.

b) Inertial ones.

c) Inside the train the time will slow down relatively to the outside clock. So if one travel at nearly the speed if light for 2 hours on his clock, for outdoor observers it will look like 3 hours.

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(c) The velocity is segment B is <u>4</u><u>0</u><u>m</u><u>/</u><u>s</u><u>.</u> And in the diagram there is no change in velocity.

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Answer:

(7.90 × 10⁻¹⁵) J

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(4.646 × 10⁻¹³) = (9.11 × 10⁻³¹) × a

a = (4.646 × 10⁻¹³)/(9.11 × 10⁻³¹)

a = (5.10 × 10¹⁷) m/s²

Now, using the equations of motion, we can obtain the velocity with which the electron reaches the positive plate

u = initial velocity of the electron = 0 m/s (since the electron was initially at rest)

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y = distance covered by the electron = 1.7 cm = 0.017 m

v² = u² + 2ay

v² = 0² + 2(5.10 × 10¹⁷)(0.017)

v² = (1.734 × 10¹⁶)

v = 131,677,182.5 m/s = (1.32 × 10⁸) m/s

Kinetic energy with which the electron hits the positive plate = (1/2)(m)(v²) = (1/2)(9.11 × 10⁻³¹)(1.32 × 10⁸)² = (7.90 × 10⁻¹⁵) J

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