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jeka57 [31]
4 years ago
5

Consider five wireless stations A,B,C,D,E. Station

Engineering
2 answers:
Elena L [17]4 years ago
6 0

Solution to 1. When A is sending to B, then communication is established within A,C,E.

Solution to 2. When B is sending to A, A can communicate with other stations and in this case ,there is a possibility of communication with C,D,E as B is sending to A.

Solution to 3. When B is sending to C, communication is established between A,B and D.

Sladkaya [172]4 years ago
5 0

Answer and Explanation:

Communication is possible when sender and reciever both communicates in well manner and communication is only possible when there is a sender available to communicate.

1) Sending data from A to B, there is no possible communication between A and other stations because while A is communicating with B the packet's of station A will interfere with other packets by other stations. Thus while transmission from A to B there is no other communication possible between any other station.

2) In this case also there is no communication possible as B can communicate with other stations except D. If E and C try to send some data to D at the same time then this would interfere B's transmission to A. Thus while transmission from B to A there is no communication possible between any other station.

3) Station B can communicate with E, A and C but not with D. while sending data from B to C the communication will be interfered by the station B's transmission to A. But station E can safely transmit data to D while communication between station B to C. Thus there is one communication possible while sending data from B to C which is E to D.

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an electric circuit includes a voltage source and two resistances (50 and 75) in parallel. determine the voltage source required
ASHA 777 [7]

Answer:

The voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

Explanation:

Given;

Resistance, R₁ = 50Ω

Resistance, R₂ = 75Ω

Total resistance, R = (R₁R₂)/(R₁ + R₂)

Total resistance, R = (50 x 75)/(125)

Total resistance, R = 30 Ω

According to ohms law, sum of current in a parallel circuit is given as

I = I₁ + I₂

I = \frac{V}{R_1} + \frac{V}{R_2}

Voltage across each resistor is the same

1.6 = \frac{V}{R_2}  

V = 1.6 x R₂

V = 1.6 x 75

V = 120 V

Therefore, the voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

This voltage is also the same for 50 ohms resistance but the current will be 2.4 A.

3 0
3 years ago
Read 2 more answers
An aggregate blend is composed of 65% coarse aggregate by weight (Sp. Cr. 2.635), 36% fine aggregate (Sp. Gr. 2.710), and 5% fil
den301095 [7]

Answer:

Explanation:

From the given information:

Addition of all the materials = 65+ 36+ 5 +6 = 112  which is higher than 100 percentage; SO we need to find;

The actual percentage of each material which can be determined as follows:

Percentage of the coarse aggregate will be = 65 × 112/100

= 72.80%

Percentage of the Fine aggregate will be = 36 × 112/100

= 40.32%

Percentage of the filler  will be = 5 × 112/100

= 5.6%

Percentage of the   asphalt binder will be = 6 × 112/100

= 6.72 %

So; the theoretical specific gravity (Gt) of the mixture can be calculated as follows:

Gt = 100/( 72.80/2.635 + 40.32/2.710 + 5.6/2.748 + 6.72/1.088)

Gt = 100/( 27.628 + 14.878 + 2.039 + 6.177)

Gt = 100/ (50.722)

Gt =1.972

Also;Given that the bulk density = 143.9 lb/ft³

LIke-wsie ; as we know that unit weight of water is =62.43lb/cu.ft

Hence, the bulk specific gravity of the mix (Gm) = 143.9/62.43

=2.305

The percentage of air  void  = (Gt -Gm )× 100/ Gt

= (1.972 - 2.305) ×  100/ 1.972

= -16.89%

The percentage of the asphalt binder is =(6.72/1.088*100)/(72.80+40.32+5.6+6.72)/2.305)

= 617.647/54.42

= 11.35%

Thus; the percentage voids in mineral aggregate =  -16.89% + 11.35%

the percentage voids in mineral aggregate = -5.45%

The percent voids filled with asphalt. = 100 × 11.35/-5.45

The percent voids filled with asphalt = - 208.26 %

7 0
3 years ago
Create a class called Candle to represent a candle. It should have four private instance variables: An int for the height, an in
Papessa [141]

Answer:

Explanation:

5

5 0
4 years ago
A circular pipe of 25-mm outside diameter is placed in an airstream at 25 °C and 1-atm pressure. The air moves in cross flow ove
Scrat [10]

Answer: a) Fd = 3.24 N/m

b) Q = 520 w/m

Explanation: please find the attached files for the solution

7 0
3 years ago
A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest
sveticcg [70]

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

y = ax^{2} +bx+c\\dy/dx=2ax+b\\

At highest point of the curve dy/dx=o

2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275

-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\

Station PVT

Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)

3 0
3 years ago
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