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3 years ago

Describe the Optical Fiber Network( i need 2 paragraphs, but give me about 5 sentences or like 3)

2 answers:

6 0

Answer: Fiber Optic Network Fiber-optic networks have been used for decades to transmit large volumes of traffic across the country. The economics of fiber networks have only recently allowed for connecting the fiber directly to the home, creating a fiber-to-the-home (FTTH) network.

Explanation:

Tasya [4]3 years ago

6 0

Optical fibers are about the diameter of a strand of human hair and when bundled into a fiber-optic cable, they’re capable of transmitting more data over longer distances and faster than other mediums. It is this technology that provides homes and businesses with fiber-optic internet, phone and TV services. There are several different types of fiber-optic networks but they all begin with optic cables running from the network hub to the curb near your home or straight to your home to provide a fiber-optic internet connection. The fastest type of fiber network is called Fiber to the Home (FTTH) or Fiber to the Premises (FTTP) because it’s a 100% fiber-optic connection with optical fiber cables installed to terminals directly connected to houses, apartment buildings and businesses.

On the other hand, Fiber to the Curb (FTTC) is a partial fiber connection because the optical cables run to the curb near homes and businesses and copper cables carry the signals from the curb the rest of the way. Similarly, Fiber to the Building (FTTB) is when fiber cable goes to a point on a shared property and the other cabling provides the connection to offices or other spaces. When you’re on a FTTH network, you’ll experience significantly faster upload and download speeds, more bandwidth for multiple devices at home and a reliable connection. And that’s exactly what you’ll get with Verizon Fios, the 100% fiber-optic network.

With Fios Gigabit Connection speeds up to a blistering 940/880 Mbps Speeds up to 940 Mbps download and 880 Mbps upload available in select areas. , you’ll enjoy HD streaming, gaming, video conferencing and so much more on up 100 devices at a time –virtually lag-free. No wonder Fios has been rated #1 by industry leaders time and again. Shop for Verizon Fios and see if fiber internet is available in your area. A whole essay for you my boy ☝☝☝

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A plate clutch is used to connect a motor shaft running at 1500rpm to shaft 1. The motor is rated at 4 hp. Using a service facto

vazorg [7]

Answer:

$(M_t)_{rated}=61.11lb-in$

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = $4 \times 0.7457$ =2.9828 KW

service factor(k)= 2.75

now,

$KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }$

$M_t= 18,989.09 \ N-mm= 168.06 lb-in$

torque rating

$(M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in$

4 0

4 years ago

Calculate total hole mobility if the hole mobility due to lattice scattering is 50 cm2 /Vsec and the hole mobility due to ionize

Ad libitum [116K]

Answer:

The total hole mobility is 41.67 cm²/V s

Explanation:

Data given by the exercise:

hole mobility due to lattice scattering = ul = 50 cm²/V s

hole mobility due to ionized impurity = ui = 250 cm²/V s

The total mobility is equal:

$\frac{1}{u} =\frac{1}{ul} +\frac{1}{ui} \\\frac{1}{u}=\frac{1}{50} +\frac{1}{250} \\u=41.67cm^{2} /Vs$

5 0

3 years ago

Read 2 more answers

Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24°C and leaves at 0.8 MPa and 60°C. The mass flow rate o

Leni [432]

Answer:

(a) The power input to the compressor: $\dot{W}=73.07 kJ/s = 73.07 kW$

(b) The volume flow rate of the refrigerant at the compressor inlet: $\dot{v}=0.209 m^{3}/s$

Explanation:

(a)

We need to check the values of enthalpy (as we have an open system) for both states, being the inlet, state 1 and the outlet, state 2. We will know these values by checking the vapor charts of R134a, I used the ones found in Thermodynamics of Cengel, 7th edition.

Then, our values are:

$h_{1}=235.92kJ/kg\\h_{2}=296.81kJkg$

Now we proceed to know the work with the following expression:

$\dot{W}=\dot{m}(h_{2}-h_{1})$

Now we replace values and our result is:

$\dot{W}=73.07 kJ/s = 73.07 kW$

(b)

To know the volume rate at the compressor inlet, we need to know the specific volume in that phase, as we have that is saturated and at -24°C, we can read our table:

$\nu=0.1739m^{3}/kg$

With our specific volume and the mass rate, we can calculate the volume rate:

$\dot{v}=\nu * \dot{m}\\\dot{v}=0.209 m^{3}/s$

6 0

3 years ago

The temperature in a pressure cooker is 130 degree C while the water is boiling. Determine the pressure inside the cooker.

emmasim [6.3K]

Answer:

The pressure inside the cooker is 1.0804 atm.

Explanation:

Boiling occurs when the vapor pressure becomes equal to atmospheric pressure.

For water, At standard conditions (Pressure = 1 atm) boiling occurs at 373.15 K.

So, Standard conditions:

T₁ = 373.15 K

P₁ = 1 atm

Given ,

The water boils at Temperature = 130 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So, the temperature, T₂ = (130 + 273.15) K = 403.15 K

To find pressure inside the cooker (P₂) :

Applying Amontons's Law as:

$\frac {P_1}{T_1}=\frac {P_2}{T_2}$

So,

$P_2=\frac {P_1}{T_1} \times {T_2}$

$P_2=\frac {1 atm}{373.15 K} \times {403.15 K}$

$P_2=1.0804 atm$

Thus, The pressure inside the cooker is 1.0804 atm.

6 0

3 years ago

A 4-kW electric heater runs for 2 hours to raise the room temperature to the desired level. Determine the amount of electric ene

Anna35 [415]

Answer:

Q' = 8 KW.h

Q'=28800 KJ

Explanation:

Given that

Heat Q= 4 KW

time ,t = 2 hours

The amount of energy used in KWh given as

Q ' = Q x t

Q' = 4 x 2 KW.h

Q' = 8 KW.h

We know that

1 h = 60 min = 60 x 60 s = 3600 s

We know that W = 1 J/s

The amount of energy used in KJ given as

Q' = 8 x 3600 = 28800 KJ

Therefore

Q' = 8 KW.h

Q'=28800 KJ

6 0

3 years ago