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kodGreya [7K]
3 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]3 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

34kurt3 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.

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2.A 5 Ohm resistor is connected to a 9 Volt battery. How many Joules of thermal energy are produced in 7 minutes?
IceJOKER [234]

Answer:

Explanation:

2 )

power of an electric device = V² / R where V is volts and R is resistance

putting given data

power = 9²/ 5

= 16.2 J/s

energy produced in 7 minutes

= 16.2 x 7 x 60

= 6804J .

3 ) Power of an electrical device

= V² / R

= V X I where I  is current

= 4.5 x .5

= 2.25 W or J/s

4 )

energy used in 3 minutes with power of 2.25 W

= 2.25 x 3 x 60

= 405 J .

7 )

power of a electrical device

= V x I

IR x I  where R is resistance .

= I²R

putting given data

power = .005² x 50

= 1.25 x 10⁻³ W .

8 )

Energy used up by a 60 W bulb in 2.5 hours

= 60 x 2.5 x 60 x 60

= 5.4 x 10⁵ J .

7 0
3 years ago
An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper
dimulka [17.4K]

Answer:

The power output of this engine is  P =  17.5 W

The  the maximum (Carnot) efficiency is  \eta_c  = 0.7424

The  actual efficiency of this engine is  \eta _a  = 0.46

Explanation:

From the question we are told that

    The temperature of the hot reservoir is  T_h = 1250 \ K

      The temperature of the cold reservoir  is  T_c  =  322 \ K

     The energy absorbed from the hot reservoir is E_h  = 1.37 *10^{5} \ J

       The energy exhausts into  cold reservoir is  E_c  = 7.4 *10^{4} J

The power output is mathematically represented as

      P  =  \frac{W}{t}

Where t is the time taken which we will assume to be 1 hour =  3600 s  

W is the workdone which is mathematically represented as

      W =  E_h  -E_c

substituting values

       W = 63000 J

So

    P =  \frac{63000}{3600}

    P =  17.5 W

The Carnot efficiency is mathematically represented as

          \eta_c  =  1 - \frac{T_c}{T_h}

         \eta_c  =  1 - \frac{322}{1250}

         \eta_c  = 0.7424

The actual efficiency is mathematically represented as

        \eta _a  =   \frac{W}{E_h}

substituting values

         \eta _a  =  \frac{63000}{1.37*10^{5}}

         \eta _a  = 0.46

     

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Only your weight would change. Mass is a measurement of how much matter something is made up of.
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