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kodGreya [7K]
2 years ago
8

you suspect a network cable has a break in it somewhere. which tool would be best to use to determine the location of the break

to determine if you are correct
Physics
2 answers:
Elden [556K]2 years ago
8 0

Answer:

The Optical Time Domain reflector

Explanation:

The Optical Time Domain

optical time-domain reflectometer is an optoelectronic instrument used to characterize an optical fiber. An OTDR is the optical equivalent of an electronic time domain reflectometer.

Reflectometer (OTDR) is useful for testing the integrity of fiber optic cables. It can verify splice loss, measure length and find faults. The OTDR is also commonly used to create a "picture" of fiber optic cable when it is newly installed

34kurt2 years ago
3 0

Answer:

Optical Time Domain Reflector

Explanation:

Localized the break, supplies a graphical trace of where the break occurs in order to detect high loss splice point as far as 25 miles.

You might be interested in
How can you tell the difference between chemical sedimentary rocks and intrusive<br> igneous rocks?
Nookie1986 [14]

Answer:

Sedimentary rocks are usually formed under water when grains of broken rocks are glued together while igneous rocks form when melted rock (magma or lava) cools and metamorphic are rocks that once were igneous or sedimentary rocks but have been changed by pressure and temperature.

4 0
3 years ago
Read 2 more answers
What is the energy of a photon whose frequency is 6.0 x 10^20?
omeli [17]

Answer:

3.75 MeV

Explanation:

The energy of the photon can be given in terms of frequency as:

E = h * f

Where h = Planck's constant

The frequency of the photon is 6 * 10^20 Hz.

The energy (in Joules) is:

E = 6.63 x10^(-34) * 6 * 10^(20)

E = 39.78 * 10^(-14) J = 3.978 * 10^(-13) J

We are given that:

1 eV = 1.06 * 10^(-19) Joules

This means that 1 Joule will be:

1 J = 1 / (1.06 * 10^(-19)

1 J = 9.434 * 10^(18) eV

=> 3.978 * 10^(-13) J = 3.978 * 10^(-13) * 9.434 * 10^(18) = 3.75 * 10^(6) eV

This is the same as 3.75 MeV.

The correct answer is not in the options, but the closest to it is option C.

6 0
3 years ago
Can you explain that gravity pulls us to the Earth &amp; can you calculate weight from masses on both on Earth and other planets
schepotkina [342]
I don't actually understand what your question is, but I'll dance around the subject
for a while, and hope that you get something out of it.

-- The effect of gravity is:  There's a <em>pair</em> of forces, <em>in both directions</em>, between
every two masses.

-- The strength of the force depends on the <em>product</em> of the masses, so it doesn't matter whether there's a big one and a small one, or whether they're nearly equal. 
It's the product that counts.  Bigger product ==> stronger force, in direct proportion.

-- The strength of the forces also depends on the distance between the objects' centers.  More distance => weaker force.  Actually, (more distance)² ==> weaker force.

-- The forces are <em>equal in both directions</em>.  Your weight on Earth is exactly equal to
the Earth's weight on you.  You can prove that.  Turn your bathroom scale face down
and stand on it.  Now it's measuring the force that attracts the Earth toward you. 
If you put a little mirror down under the numbers, you'll see that it's the same as
the force that attracts you toward the Earth when the scale is right-side-up.

-- When you (or a ball) are up on the roof and step off, the force of gravity that pulls
you (or the ball) toward the Earth causes you (or the ball) to accelerate (fall) toward the Earth. 
Also, the force that attracts the Earth toward you (or the ball) causes the Earth to accelerate (fall) toward you (or the ball).
The forces are equal.  But since the Earth has more mass than you have, you accelerate toward the Earth faster than the Earth accelerates toward you.

--  This works exactly the same for every pair of masses in the universe.  Gravity
is everywhere.  You can't turn it off, and you can't shield anything from it.

-- Sometimes you'll hear about some mysterious way to "defy gravity".  It's not possible to 'defy' gravity, but since we know that it's there, we can work with it.
If we want to move something in the opposite direction from where gravity is pulling it, all we need to do is provide a force in that direction that's stronger than the force of gravity.
I know that sounds complicated, so here are a few examples of how we do it:
-- use arm-muscle force to pick a book UP off the table
-- use leg-muscle force to move your whole body UP the stairs
-- use buoyant force to LIFT a helium balloon or a hot-air balloon 
-- use the force of air resistance to LIFT an airplane.

-- The weight of 1 kilogram of mass on or near the Earth is 9.8 newtons.  (That's
about 2.205 pounds).  The same kilogram of mass has different weights on other planets. Wherever it is, we only know one of the masses ... the kilogram.  In order
to figure out what it weighs there, we need to know the mass of the planet, and
the distance between the kilogram and the center of the planet.

I hope I told you something that you were actually looking for.
7 0
3 years ago
30 POINTS!!! CAN U AWNSER IT?? :)
solniwko [45]

Answer:

5235.84 kg

Explanation:

There is one theorem - whose proof I will never remember without having to drag calculus in there - that says that the variation of momentum is equal to the force applied times the time the application last.

F\Delta t = m \Delta v As long as the engine isn't ejecting mass - at this point it's a whole new can of worm - we know the force, we know the variation in speed, time to find the mass. But first, let's convert the variation of speed in meters per second. The ship gains 250 kmh, \Delta v = 69.4 m/s;

45 450 \cdot 8 = 69.4 m \rightarrow m = \frac{45450\cdot 8}{69.4} = 5235.84 kg

7 0
2 years ago
a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg
IrinaK [193]

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

M = \frac{q}{p}

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m

Now using thin lens formula:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\

<u>f = 1 m</u>

6 0
3 years ago
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