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jok3333 [9.3K]
4 years ago
5

Within the theory of G relativity what, exactly, is meant by " the speed of light WITHIN A VACUUM" ? & what does that have t

o do with the FIRST part of the theory / equation? How do they relate?
Physics
1 answer:
Ber [7]4 years ago
3 0
The speed of light "within a vacuum" refers to the speed of electromagnetic radiation propagating in empty space, in the complete absence of matter.  This is an important distinction because light travels slower in material media and the theory of relativity is concerned with the speed only in vacuum.  In fact, the theory of relativity and the "speed of light" actually have nothing to do with light at all.  The theory deals primarily with the relation between space and time and weaves them into an overarching structure called spacetime.  So where does the "speed of light" fit into this?  It turns out that in order to talk about space and time as different components of the same thing (spacetime) they must have the same units.  That is, to get space (meters) and time (seconds) into similar units, there has to be a conversion factor.  This turns out to be a velocity.  Note that multiplying time by a velocity gives a unit conversion of
seconds \times  \frac{meters}{seconds} =meters
This is why we can talk about lightyears.  It's not a unit of time, but distance light travels in a year.  We are now free to define distance as a unit of time because we have a way to convert them.  
As it turns out light is not special in that it gets to travel faster than anything else.  Firstly, other things travel that fast too (gravity and information to name two).  But NO events or information can travel faster than this.  Not because they are not allowed to beat light to the finish line---remember my claim that light has nothing to do with it.  It's because this speed (called "c") converts space and time.  A speed greater than c isn't unobtainable---it simply does not exist.  Period.  Just like I can't travel 10 meters without actually moving 10 meters, I cannot travel 10 meters without also "traveling" at least about 33 nanoseconds (about the time it takes light to get 10 meters)  There is simply no way to get there in less time, anymore than there is a way to walk 10 meters by only walking 5.  
We don't see this in our daily life because it is not obvious that space and time are intertwined this way.  This is a result of our lives spent at such slow speeds relative to the things around us.
This is the fundamental part to the Special Theory of Relativity (what you called the "FIRST" part of the theory)  Here is where Einstein laid out the idea of spacetime and the idea that events (information) itself propagates at a fixed speed that, unlike light, does not slow down in any medium.  The idea that what is happening "now" for you is not the same thing as what is "now" for distant observers or observers that are moving relative to you.  It's also where he proposed of a conversion factor between space and time, which turned out to be the speed of light in vacuum.
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The natural force that causes you to lose power as you climb a hill is known as inertia.
Juliette [100K]
Yeah your answer is correct

7 0
3 years ago
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An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e
White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

5 0
3 years ago
A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci
larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 =  e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

t = 2.2 ms

<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

e^(2t/RC) = e

2t/RC = 1

t = RC/2

t = (265 x 12 x 10⁻⁶)/2

t = 1.59 x 10⁻³ s

t = 1.59 ms

Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

Learn more about energy stored in capacitor here: brainly.com/question/14811408

#SPJ1

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2 years ago
Which property of gold allows it to be used this way?
densk [106]

the electric conductivity of gold is very high

3 0
4 years ago
How do you find the total magnification of a microscope?
Alla [95]
In order to find total magnification of a microscope, you need to multiply the power of eyepiece and objective lens.

Hope this helps!
6 0
4 years ago
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