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LenKa [72]
3 years ago
14

Suppose your network is connected to another network via a router. Which OSI model layer provides the information necessary to d

irect data between the two networks ?
Physics
1 answer:
RUDIKE [14]3 years ago
8 0

Answer: it is called the "NETWORK LAYER"

Explanation: OSI model means Open System Interconnection model. It have different model layers.

The network layer is the OSI model layer that helps in forwarding packets from network to network, using a network router. It's main function is to connect different network together. It also determine the best virtual circuits(paths) for the data to travel.

Because the question specify that the network is connected to a network router, that means the OSI model layer is network layer.

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Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
3 years ago
An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and
svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s

velocity at the exit=20m/s

for entry

V=\frac{5}{1} =5m/s

2.

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha  =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81  =p2\\P2=-37.5kPa

the pressure at exit is -37.5kPa

7 0
3 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

f = f_0 (\frac{v_0}{v_0-v})

Here,

f_0 = Frequency of Source

v_s = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v  = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

f = f_0 (\frac{v_0}{v_0-v})

300 = f_0 (\frac{343}{343-v})

(300*343) - 300v = 343f_0

Now the second expression will be,

f' = f_0 (\frac{v_0}{v_0-v/2})

290 = (343)(\frac{v_0}{343-v/2})

290*343-145v = 343f_0

Dividing the two expression we have,

\frac{(300*343) - 300v}{290*343-145v} = 1

Solving for v, we have,

v = 22.12m/s

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0
3 years ago
Kwanzaa farts apple juice
8_murik_8 [283]

Answer:

well

Explanation:

7 0
2 years ago
What happens to centripetal acceleration as the radius of curvature decreases and the speed is constant, and why
lutik1710 [3]
It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature.



Hopr it helps :)
4 0
3 years ago
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