The average speed <em>appears to be</em> (distance) / (time) =
(length of the cable) / (time from when a pulse goes in until it comes out the other end) .
That's 1,200,000 meters/ 0.006 second = 2 x 10^8 = <em>2 hundred million m/sec</em>
That figure is about 66.7% of the speed of light in vacuum.
The reason I went through all of this detail was to point out that this is
NOT necessarily the speed of light in this glass, for two reasons.
1). The path of light through an optical fiber is not straight down the middle. In the original fibers of 20 or 30 years ago, the light bounced back and forth off the inside walls of the fiber, and zig-zagged its way along the length. In current modern fibers, it still zig-zags, but it's a more gentle, up-and-down curved path. In either case, the distance covered by the light inside the fiber is more than the straight length of the cable, and the time it takes it to come out the other end is more than its actual speed inside the glass would have meant if it could have traveled straight through the pipe.
2). This problem talks about an optical fiber that's 1,200km long. There is loss in optical fiber, and you're NOT going to get light all the way through a single piece of it that's something like 745 miles long. It takes electronic repeaters, "boosters", and regenerators every few miles to keep it going, and these devices add "latency" or time delay in the process of going through them. That delay in the electronics shows up as apparent delay through the fiber-optic cable, and it makes the speed through the glass appear to be slower than it actually is.
Answer:
A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm
, this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.
we have a fluctuation of the intensity emitted by the stars. Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.
Explanation:
The radiation of a black body is characterized by its temperature, with Wien's law of displacement we can find the maximum wavelength emitted by each star.
λ T = 2,898 10⁻³
therefore the emission the star of A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm
The emission of the premiere is in the ultraviolet light range, as this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.
The burst with A = 4300K has a bad emission maximum = 6.7395 10⁻⁷ m = 673.95 nm, which corresponds to an emission in the visible in the orange range, giving a blackbody spectrum of this range, but since the emission is formed by two stars, we see that when the two are placed one in front of the other the intensity of the emission must increase significantly and when they are placed next to each other the intensity reaches its minimum, consequently we have a fluctuation of the intensity emitted by the stars.
Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.
More massive attacking objects speed slows down and it transfer its some kinetic energy to less massive target objects whereas less massive target objects starts its motion
Answer:
Explanation:
the unit of work is derived unit because joule is defined the work done by the force aftab 1 newton causing the displacement of one metre something newton metre(n-m) is also used to measuring work
The hint in this problem is in the word "about" or
approximately how many pounds.
So we could find the solution by estimation through rounding
off.
We know that almonds cost about 3.50 per pound and
a bag costs about $7.00.
Dividing 7 by 3.50 per pound, will give us $7 ÷ $3.50 = 2
pounds.
The answer is 2 pounds.
