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Natalija [7]
3 years ago
15

If I turn on a light in a spaceship traveling 1C BACKWARDS, what happens to the photons? Speed -0-, or 1C in opposite direction?

Physics
1 answer:
Advocard [28]3 years ago
5 0

If the spaceship's Physicist happens to be hanging out of one side
of the ship, and he measures the speed of the photons as they pass
him and leave the ship, he'll see them passing him at 'c' ... the speed
of light.

When those photons pass somebody who happens to be in their
path, and he decides to measure their speed, he'll see them move
past him at 'c' ... the speed of light.
 
It doesn't matter whether the observer who measures them is
moving, or at what speed.

And it doesn't matter what source the photons come from, or
whether the source is moving, or at what speed.

And it doesn't matter what the photons' wavelength/frequency is ...
anything from radio to gamma rays.

The photons pass everybody at 'c' ... the speed of light.

Yes, I hear you.  That can't be true. It's crazy.
Maybe it's crazy, but it's true.

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. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 1.00 × 104 kg. The thrust of its engin
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Answer:

a) The module's acceleration in a vertical takeoff from the Moon will be 1.377 \frac{m}{s^2}

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Explanation:

a) During a vertical takeoff, the sum of the forces in the vertical axis will be equal to mass times the module's acceleration. In this this case, the thrust of the module's engines and the total module's weight are the only vertical forces. (In the Moon, the module's weight will be equal to its mass times the Moon's gravity acceleration)

T-(m*g)=m*a

Where:

T= thrust =3 *10^{4} N

m= module's mass =1 *10^{4} N

g= moon's gravity acceleration =1.623 \frac{m}{s^2}

a= module's acceleration during takeoff

Then, we can find the acceleration like this:

a=\frac{T}{m} -g=\frac{3*{10}^4 N}{1*{10}^4 kg}-1.623\frac{m}{s^2}

a=1.377 \frac{m}{s^2}

The module's acceleration in a vertical takeoff from the Moon will be 1.377 \frac{m}{s^2}

b) To takeoff, the module's engines must generate a thrust bigger than the module's weight, which will be its mass times the Earth's gravity acceleration.

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Then we can say that a thrust of 3*10^{4} N won't be able to lift off the module from the Earth because it's smaller than the module's weight (9.8 *10^{4} N).

8 0
4 years ago
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<h3>What is Gravitational interaction?</h3>

This is defined as the interaction between a particle or body resulting from their mass. This type of interaction is usually weak and occurs in all distances possible.

It is not gravitational interaction, because the rod attracts paper only against the gravitational force of the earth and there is no attraction between both bodies under a different condition.

This is therefore the reason why it is not possible to explain the interaction of the rod and pieces of paper as a gravitational interaction.

Read more about Gravitational interaction here brainly.com/question/25624188

#SPJ1

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