Ask question

Ask question

All categories

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

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.

You might be interested in

About 60 percent of all known galaxies are classified as _____.

masha68 [24]

The answer that fills in the blank is the galaxies which are six percent is classified as the elliptical planet. The 60 percent are considered to be elliptical planets because they move in a way of the elliptical path that considers them to be more of an elliptical planet.

8 0

3 years ago

Read 2 more answers

A reaction taking place in a container with a piston-cylinder assembly absorbs 4850. J of heat and has a change in the internal

Maslowich

Answer:

$V_f = 16.3 Ltr$

Explanation:

As per first law of thermodynamics we know that

heat absorbed = change in internal energy + Work done

so we have

$Heat = 4850 J$

$\Delta U = 1618.5 J$

so work done is given as

$4850 = 1618.5 + W$

$W = 3231.5 J$

now we know that at constant pressure the work done is given as

$W = P\Delta V$

$3231.5 = P(V_f - V_i)$

$3231.5 = 5.5 \times 1.01 \times 10^5(V_f - 10.5 \times 10^{-3})$

$V_f = 16.3 \times 10^{-3} m^3$

or we have

$V_f = 16.3 Ltr$

6 0

3 years ago

A roller-coaster car has a mass of 1040 kg when fully loaded with passengers. As the car passes over the top of a circular hill

rusak2 [61]

Answer:

a.6373.5 N

b.-3837.6 N

Explanation:

Mass of roller coaster=m=1040 kg

Radius=r=24 ,

a. $v=9.4m/s$

Normal force= $F_N$

According to question

$mg-F_N=\frac{mv^2}{r}$

Where $g=9.81 m/s^2$

Substitute the values

$1040\times 9.81-F_N=\frac{1040\times (9.4)^2}{24}$

$10202.4-F_N=3828.9$

$F_N=10202.4-3828.9=6373.5 N$

b. $v=18m/s$

$g=9.81 m/s^2$

$1040\times 9.81-F_N=\frac{1040\times (18)^2}{24}$

$10202.4-F_N=14040$

$F_N=10202.4-14040=-3837.6 N$

8 0

4 years ago

A solenoid having an inductance of 6.95 μh is connected in series with a 1.24 kω resistor. (a) if a 12.0 v battery is connected

Serggg [28]

In electrical circuit, this arrangement is called a R-L series circuit. It is a circuit containing elements of an inductor (L) and a resistor (R). Inductance is expressed in units of Henry while resistance is expressed in units of ohms. The relationship between these values is called the impedance, denoted as Z. Its equation is

Z = √(R^2 + L^2)
Z = √((1.24×10^3 ohms)^2 + (6.95×10^-6 H)^2)
Z = 1,240 ohms

The unit for impedance is also ohms. Since the circuit is in series, the voltage across the inductor and the resistor are additive which is equal to 12 V. Knowing the impedance and the voltage, we can determine the maximum current.
I = V/Z=12/1,240 = 9.68 mA
But since we only want to reach 73.6% of its value, I = 9.68*0.736 = 7.12 mA. Then, the equation for R-L circuits is
$I= \frac{V( 1- e^{-t/τ} )}{R}$ , where τ = L/R = 6.95×10^-6/1.24×10^3 = 5.6 x 10^-9
Then,
$7.12x 10^{-3} = \frac{12( 1- e^{-t/5.6x 10^{-9} } )}{1240}$

t = 7.45 nanoseconds
Part B.) If t = 1.00τ, then t/τ = 1. Therefore,
$I= \frac{12( 1- e^{-1 } )}{1240}$

I = 6.12 mA

3 0

3 years ago

An electric motor is used to run an elevator. The total mass of the elevator car and passenger is 1600 kg. The elevator moves up

Masja [62]

Answer:

Approximately $2.7 \times 10^{4}\; {\rm W}$ , assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$

Explanation:

The weight of the elevator is:

$\begin{aligned}& \text{weight} \\ =\; & m\, g \\ =\; & 1600\; {\rm kg} \times 9.81\; {\rm m \cdot s^{-2}} \\ \approx\; & 15700\; {\rm N}\end{aligned}$ .

Since the speed of the elevator is constant, the acceleration of this elevator would be $0$ .

By Newton's Second Law of Motion, the net force on the elevator (proportional to acceleration) would also be $0\!$ . All external forces on the elevator need to be balanced in every direction.

The only two vertical forces on the elevator are:

the weight of the elevator (downward gravitational pull from the earth,) and
the upward pull from the motor.

These two forces need to balance one another. Since the weight of the elevator is approximately $15700\; {\rm N}$ , the upward pull of the motor would be $15700\; {\rm N}\!$ . in magnitude.

The direction of this upward pull is the same as the direction of the motion of this elevator. Thus, the work that the motor did on the elevator would be positive:

$\begin{aligned}& \text{work} \\ =\; & F\, s \\ \approx\; & 15700\; {\rm N} \times 15.0\; {\rm m} \\ \approx\; & 2.35 \times 10^{5}\; {\rm J}\end{aligned}$ .

Since the velocity of the elevator is constant, instantaneous power output of the motor would be equal to the average power of the motor:

$\begin{aligned}& \text{power} \\ =\; & \frac{\text{work}}{t} \\ \approx\; & \frac{2.35 \times 10^{5}\; {\rm J}}{8.82\; {\rm s}} \\ \approx\; & 2.7 \times 10^{4}\end{aligned}$ .

4 0

3 years ago