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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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Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crys

vredina [299]

Solution :

i. Slip plane (1 1 0)

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

$\cos \phi = \frac{(1,0,0) \cdot (1,1,0)}{1 \times \sqrt2}$

$=\frac{1}{\sqrt2 }$

$\cos \lambda = \frac{(1,0,0) \cdot (1,-1,1)}{1 \times \sqrt3}$

$=\frac{1}{\sqrt3 }$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

ii. Slip plane --- (1 1 0)

Slip direction -- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, -1, 0)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

iii. Slip plane --- (1 0 1)

Slip direction --- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, 1)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

Slip direction ---- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, -1)}{1 \times \sqrt2}=\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, -1, 1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

3 0

2 years ago

Which type of engineering design uses an already existing design

Snezhnost [94]

Answer:

Reverse engineering

Explanation:

Reverse Engineering is the remaking of already made products following the deconstruction and examination of the product to make known the product design, code and architecture features, gain knowledge of the composition and construction in a scientific research approach

Reverse engineering is also known as back engineering and consists of three main stages

1) Recovery implementation

2) Design recovery

3) Recovery analysis.

3 0

3 years ago

Cup-sveg-aph<br><br>I m finding gf<br><br>if anyone interested pls come

Eduardwww [97]

Answer:

rot

Explanation:

5 0

3 years ago

A semiconductor diode has the following parameters: Is = 4.0×10-13 Amps, n = 1.35, and is operated at a temperature of T = 270K.

andrew-mc [135]

Answer:

$V_T=0.02328V\\I=7.77\times 10^{-5} A \\V_D=0.77V$

Explanation:

Let's use Shockley ideal diode equation which relates the current intensity and the potential difference:

$I=I_S (e^{\frac{V_D}{nV_T} } -1)$

Where:

$I=Diode\hspace{3}current\\I_S=Reverse\hspace{3}bias\hspace{3}saturation\hspace{3}current\\V_D=Voltage\hspace{3}across\hspace{3}the\hspace{3}diode\\V_T=Thermal\hspace{3}voltage\\n=Ideality\hspace{3}factor$

Thermal voltage at any temperature it is a known constant defined by:

$V_T=\frac{kT}{q}$

Where:

$k= Boltzmann\hspace{3}constant \approx1.38\times 10^{-23} \\T=Absolute\hspace{3}temperature\\q=Charge\hspace{3}of\hspace{3}an\hspace{3}electron \approx 1.6\times 10^{-19} C$

(a)

Using the data provided:

$V_T=\frac{(1.38\times 10^{-23})*(270) }{1.6\times 10^{-19} }= 0.0232875V$

(b)

Using the data provided and Shockley ideal diode equation

$I=(4\times 10^{-13} )*(e^{\frac{0.6}{1.35*0.0232875} }-1)=7.773505834\times10^{-5}\approx 7.77 \times10^{-5}A\\I\approx0.0777mA$

$I=I_S (e^{\frac{V_D}{nV_T} } -1)\\\\Multiply\hspace{3}both\hspace{3}sides\hspace{3}by\hspace{3}I_S\\\\\ \frac{I}{I_S} = e^{\frac{V_D}{nV_T} } -1\\\\Add\hspace{3}1\hspace{3}both\hspace{3}sides\\\\$

$\frac{I}{I_S} +1 =e^{\frac{V_D}{nV_T} } \\\\Take\hspace{3}the\hspace{3}natural\hspace{3}logarithm\hspace{3}of\hspace{3}both\hspace{3}sides\\\\\frac{V_D}{nV_T} =log(\frac{I}{I_S} +1) \\\\Multiply\hspace{3}both\hspace{3}sides\hspace{3}by\hspace{3}nV_T\\\\V_D=nV_T*log(\frac{I}{I_S} +1)$

Finally, using the data provided:

$V_D=(1.35)(0.0232875)*log(\frac{20\times 10^{-3} }{4\times 10^{-13} }+1)=0.77448729\approx 0.77V$

7 0

3 years ago

Estimating is important in construction industry because

Goshia [24]

Answer:

a( contractors need to know the amount of markup)

because the contractor should have a long term vision for a proper satisfaction to the people.

8 0

3 years ago