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shutvik [7]
3 years ago
9

1. How is high-speed internet typically transmitted around the world? over telephone wires b. through fiber optic cables c. beam

ed wirelessly to homes d via satellites​
Engineering
2 answers:
PilotLPTM [1.2K]3 years ago
8 0

Answer:

High speed internet is typically transmitted around the world;

b. Through fiber optic cables

Explanation:

High speed internet is typically transmitted through fiber optic cables such as the Submarine Cables that makes connection through two cables that connects USA to Europe over the Atlantic Ocean, and the Middle East to Europe Terrestrial Cable Systems (MEETS) which is a cable network of 1,400 km and is being built to have an initial capacity of 200 Gbps

Other example of fiber optic cable assisted internet broadband transmission are the Diverse Route for European and Asian Markets (DREAM) terrestrial cable that connects Germany, Kazakhstan, Russia, and China, TEA NEXT that provides connection between borders on the East and West of Russia and Euro-Russia-Mongolia-China (ERMC) terrestrial cable

kicyunya [14]3 years ago
8 0

Answer:

B through fiber cables

Explanation:

hope that helps :)

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It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 610 MPa. Calc
julia-pushkina [17]

Answer:

volume fraction of fibers is 0.4

Explanation:

Given that for the aligned carbon fiber-epoxy matrix composite:

Diameter (D) = 0.029 mm

Length (L) = 2.3 mm

Tensile strength (\sigma_{cd}) = 610 MPa

fracture strength (\sigma_f) = 5300 MPa

matrix stress (\sigma_m) =  17.3 MPa

fiber-matrix bond strength (\tau_c) = 19 MPa

The critical length is given as:

L_C=\sigma_f(\frac{D}{2\tau_c} )=5300*10^6(\frac{0.029*10^{-3}}{19*10^6} )=8.1*10^{-3}=8.1mm

Since the critical length is greater than the length, the aligned carbon fiber-epoxy matrix composite can be produced.

The longitudinal strength is given by:

\sigma_{cd}=\frac{L*\tau_c}{D} .V_f+\sigma_m(1-V_f)

making Vf the subject of the formula:

V_f=\frac{\sigma_{cd}-\sigma_m}{\frac{L*\tau_c}{D} -\sigma_m}

Vf is the volume fraction of fibers.

Therefore:

V_f=\frac{\sigma_{cd}-\sigma_m}{\frac{L*\tau_c}{D} -\sigma_m}=\frac{610*10^6-17.3*10^6}{\frac{2.3*10^{-3}*19*10^6}{0.029*10^{-3}}-17.3*10^6} } =0.4

volume fraction of fibers is 0.4

8 0
3 years ago
Consider the following hypothetical scenario for Jordan Lake, NC. In a given year, the average watershed inflow to the lake is 9
dybincka [34]

Answer:

The lake can withdraw a maximum of 1.464\times 10^{10} cubic feet per year to provide water supply for the Triangle area.

Explanation:

The maximum amount of water that can be withdrawn from the lake is represented by the following formula:

V = V_{in}+V_{p}-V_{e}-V_{out} (Eq. 1)

Where:

V - Available amount of water for water supply in the Triangle area, measured in cubic feet per year.

V_{in} - Inflow amount of water, measured in cubic feet per year.

V_{out} - Amount of water released for the benefit of fish and downstream water users, measured in cubic feet per year.

V_{p} - Amount of water due to precipitation, measured in cubic feet per year.

V_{e} - Amount of evaporated water, measured in cubic feet per year.

Then, we can expand this expression as follows:

V = f_{in}\cdot \Delta t+h_{p}\cdot A_{l}-h_{e}\cdot A_{l}-f_{out}\cdot \Delta t

V = (f_{in}-f_{out})\cdot \Delta t +(h_{p}-h_{e})\cdot A_{l} (Eq. 2)

Where:

f_{in} - Average watershed inflow, measured in cubic feet per second.

f_{out} - Average flow to be released, measured in cubic feet per second.

\Delta t - Yearly time, measured in seconds per year.

h_{p} - Change in lake height due to precipitation, measured in feet per year.

h_{e} - Change in lake height due to evaporation, measured in feet per year.

A_{l} - Surface area of the lake, measured in square feet.

If we know that f_{in} = 900\,\frac{ft^{3}}{s}, f_{out} = 300\,\frac{ft^{3}}{s}, \Delta t = 31,536,000\,\frac{second}{yr}, h_{p} = 32\,\frac{in}{yr}, h_{e} = 55\,\frac{in}{yr} and A_{l} = 47,000\,acres, the available amount of water for supply purposes in the Triangle area is:

V = \left(900\,\frac{ft^{2}}{s}-300\,\frac{ft^{3}}{s} \right)\cdot \left(31,536,000\,\frac{s}{yr} \right) +\left(32\,\frac{in}{yr}-55\,\frac{in}{yr} \right)\cdot \left(\frac{1}{12}\,\frac{ft}{in}\right)\cdot (47000\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)V = 1.464\times 10^{10}\,\frac{ft^{3}}{yr}

The lake can withdraw a maximum of 1.464\times 10^{10} cubic feet per year to provide water supply for the Triangle area.

5 0
3 years ago
a ________________ is the flat, low-lying area adjacent to a stream channel that receives overbank flow
Umnica [9.8K]

Answer:

flood plain

Explanation:

it should be called a flood plain which makes sense because it’s not a deposition.

5 0
2 years ago
An office worker claims that a cup of cold coffee on his table warmed up to 80°C by picking up energy from the surrounding air,
kherson [118]

Answer:

The claim is false and violate the zeroth law of thermodynamics.

Explanation:

Zeroth law of thermodynamics refers to thermal equilibrium among  elements. It states that  elements which different temperatures will reach the same temperature at the endgame if they are close enough to interact each other. This temperaure is called <em>equilibrium temperature and it is always a intermediate value between the element with highest temperature and the element with the lowest one. So there is no way </em> a cup of cold coffee on a table can warm up to 80°C picking up energy from the surrounding air at 25°C because the cup can only reach a temperature closer to the surrounding air temperature which will be the equilimbrium temperature for that case.

4 0
3 years ago
Read 2 more answers
A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici
Pavlova-9 [17]

Answer:

8.24μm

Explanation:

The theory of brittle fracture was used to solve this problem.

And if you follow through with the attachment made a the subject of the formula

Such that,

a = 2x(69x10⁹)x0.3/pi(40x10⁶)²

= 4.14x10¹⁰/5.024x10¹⁵

= 8.24x10^-06

= 8.24μm

This is the the maximum length of the surface flaw

4 0
3 years ago
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