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jok3333 [9.3K]
3 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]3 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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Bob is pushing a box across the floor at a constant speed of 1.2 m/s, applying a horizontal force whose magnitude is 75 N. Alice
Nadya [2.5K]

Answer:75 N

Explanation:

Given

Bob is pushing the box with a force of 75 N with constant speed. Constant speed indicates that net force on the block is zero i.e. friction force is balancing the applied force.

so friction force f=75 N

If Alice is pushing an identical box with constant speed so net force acting on box must be equal to an applied force.

As the friction force is 75 N so Alice must be applying a force of 75 N  

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A circuit has resistors R 1 = 468 Ω and R 2 = 125 Ω R1=468 Ω and R2=125 Ω , and two batteries V 1 = 12.0 V and V 2 = 3.00 V V1=1
Greeley [361]

Answer:

0.0192A

Explanation:

Since, the reading of the galvanometer is 0 A, the voltage across the resistance R will be:

Step 1

VR = V2

VR = 3.00v

Step 2

Calculating the current through the resistance R as below,

IR = V1 - V2 /R1

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How much positive and negative charges is there in a cup of water?
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Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w
Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

=  (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

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3 years ago
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