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

1 answer:

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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dezoksy [38]

You just have to simpy add the vectors together. Draw a picture to help you. Your verticle run is a vector that has a magntiude of 22m along the x axis with an angle of 0, and your climb has is a vector with a magnitude of 4.8 along the y axis and an angle of 90 degreed with respect to the x axis therefore add all of the x components and y components. you will find that the resultant vector has a xcomponent of 22m and a y component of 4.8m so draw a graph the point will be (22,4.8) and then you can simply measure the angle.

7 0

3 years ago

Read 2 more answers

Identify the constants and variables in 2cm +4y

Harman [31]

constant = 2 and 4

variable = y

Explanation:

A constant is usually a symbol with a known value. In the expression, the constants are 2 and 4.

Their values are accurately known.

A variable is place holder and the value is not known. Variables are not static. They can take different numerical values as defined by an equation:

2cm +4y