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3 years ago

What does the principle of the constancy of lightspeed tell us about the speed of light?

Physics

1 answer:

Thepotemich [5.8K]3 years ago

7 0

Answer:

The speed of light is not relative, it is always c no matter how fast or in what way the observer is moving

Explanation:

The speed of light. In a vacuum, it is by definition a universal constant of value 299,792,458 m / s (usually close to 3 · 108 m / s), or what is the same 9.46 · 1015 m / year; The second figure is used to define the interval called the light year. It is symbolized by the letter c, from the Latin celéritās (in Spanish celerity or rapidity), and is also known as the Einstein constant. [Citation needed] The value of the speed of light in a vacuum was officially included in the System International Units as a constant on October 21, 1983, thus passing the meter to be a unit derived from this constant. The speed through a medium other than the "vacuum" depends on its electrical permittivity, its magnetic permeability, and other electromagnetic characteristics. In material media, this speed is lower than "c" and is encoded in the index of refraction. In more subtle vacuum modifications, such as curved spaces, Casimir effect, thermal populations or presence of external fields, the speed of light depends on the energy density of that vacuum .

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If planet A is three times as far from planet C, then the period of its orbit will be __ times as long

liubo4ka [24]

I may be wrong, but I think you're trying to say that Planet-A is
<em>3 times as far from the sun</em> as Planet-C is.

If that's the real question, then the answer is that the period of Orbit-A
is about<em> 5.2</em> times as long as the period of Orbit-C .

Orbital period ≈ (proportional to) (the orbital distance) ^ 3/2 power.

This was empirically demonstrated about 350 years ago by Johannes
and his brilliant Kepple, and derived about 100 years later by Newton
from his formula for the forces of gravity.

6 0

3 years ago

b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul

Nat2105 [25]

Answer:

b) Betelgeuse would be $\approx 1.43 \cdot 10^{6}$ times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is $\approx 1.37 \cdot 10^{-5} }$ the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

$B = \displaystyle{\frac{L}{4\pi d^2}}$ (1)

where $B$ is the brightness, $L$ is the star luminosity and $d$ , the distance from the star to the point where the brightness is calculated (measured). Thus:

b) $B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}}$ and $B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}}$ where $L_{Sun}$ is the Sun luminosity ( $3.9 x 10^{26} W$ ) but we don't need to know this value for solving the problem. $ly$ is light years.

Finding the ratio between the two brightness we get:

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }$

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is $1.581 \cdot 10^{-5}\ ly$ . Then

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }$

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0

3 years ago

How to calculate moments with 3 separate weights of different amounts at different points?

Rina8888 [55]

I don't completely understand your drawing, although I can see that you certainly
did put a lot of effort into making it. But calculating the moment is easy, and we
can get along without the drawing.

Each separate weight has a 'moment'.
The moment of each weight is:

(the weight of it) x (its distance from the pivot/fulcrum) .

That's all there is to a 'moment'.

The lever (or the see-saw) is balanced when (the sum of all the moments
on one side) is equal to (the sum of the moments on the other side).

That's why when you're on the see-saw with a little kid, the little kid has to sit
farther away from the pivot than you do. The kid has less weight than you do,
so he needs more distance in order for his moment to be equal to yours.

6 0

3 years ago

Contact force that acts to resist sliding between two touching surfaces

Black_prince [1.1K]

I think that the answer is friction

3 0

3 years ago

A child slides down a hill on a toboggan with an acceleration of 1.8 m/s^2. If she starts at rest, how far has she traveled in :

DENIUS [597]

Explanation:

It is given that,

The acceleration of the toboggan, $a=1.8\ m/s^2$