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antoniya [11.8K]
4 years ago
12

A circuit-switching scenario in whichNcs users, each requiring a bandwidth of 25 Mbps, must share a link of capacity 150 Mbps.

Engineering
1 answer:
Monica [59]4 years ago
3 0

Answer:

0.09

Explanation:

Packet switching involves breaking a message into packets and sending them independently. Since the user only needs to transmit 10 percent of the time, the probability that a given (specific) user is transmitting = 10% = 0.1

The  probability that a user is not transmitting = 100% - 10% = 90% = 0.9

Therefore, the probability that a given (specific) user is transmitting, and the remaining users are not transmitting = 0.1 * 0.9 = 0.09

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Nadusha1986 [10]

Answer:

co-op, apartment, townhome, manor etc

6 0
3 years ago
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What is the heat loss coefficient that has a symbol Uair and is found from (volumetric flow * density * specific heat capacity)
aleksley [76]

Answer:

Uair = 0.0749 KW/k = 74.9 W/k

Explanation:

The natural air change per hour is given by the formula:

Natural Air Change per Hour = ACPH = 60*Volume Flow/Volume

where,

ACPH = 0.4

Volume Flow = ? in ft³/min

Volume = 19456 ft³

Therefore,

0.4 = (60 min)(Volume Flow)/(19456 ft³)

Volume Flow = (0.4)(19456 ft³)/(60 min) = (129.7 ft³/min)(1 min/60 s)

Volume Flow =  (2.16 ft³/s)(0.3048 m/1 ft)³ = 0.061 m³/s

Now, we find heat loss coefficient:

Uair = Volumetric Flow*Density of air*Specific Heat Capacity of air

Uair = (0.061 m³/s)(1.225 kg/m³)(1 KJ/kg.k)

<u>Uair = 0.0749 KW/k = 74.9 W/k</u>

6 0
3 years ago
How is the fuel introduced into the Diesel engine?
Ugo [173]

Answer:

diesel fuel is pumped at high pressure to the injectors which are responsible for entering the fuel into the combustion chamber,

when the piston is at the top the pressure is so high that it explodes the fuel (diesel) that results in a generation of mechanical power

5 0
3 years ago
). A 50 mm diameter cylinder is subjected to an axial compressive load of 80 kN. The cylinder is partially
Delicious77 [7]

Answer:

\frac{e'_z}{e_z} = 0.87142

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

                          εx = - [ σx - ν( σy + σz ) ] / E

                          εy = - [ σy - ν( σx + σz ) ] / E

                          εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

                          εx =  [ ν*σz ] / E

                          εy = [ ν*σz ] / E

                          εz = - [ σz ] / E   .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

                         εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

                         εx' = - [ σx' - ν( σy' + σz ) ] / E =  [ ν*σz ] / 2E

                         εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

                         εx' = - [ σy' - ν( σx' + σz ) ] / E =  [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note:  σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

                        - [ σx' - ν( σx'+ σz ) ] =  [ ν*σz ] / 2

                          σx' ( 1 - v ) = [ ν*σz ] / 2

                          σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

                           εz' = - [ σz - ν( σy' + σx' ) ] / E

                           εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

                          εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E  ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

                            \frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ]  }{-\frac{s_z}{E}}  \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142... Answer

5 0
3 years ago
Is god real??? Because theres no proof
kakasveta [241]

Answer:

thats a good question that i have been trying to figure out. idk but people seem to have their own gods

Explanation:

6 0
3 years ago
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