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

Engineering

2 answers:

Elena L [17]3 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]3 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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A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maxi

saw5 [17]

Answer:

W = 11,416.6879 N

L ≈ 64.417 cm

Explanation:

The maximum shear stress, $\tau_{max}$ , is given by the following formula;

$\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )$

$t_w$ = 1 cm = 0.01

h = 29 cm = 0.29 m

$h_w$ = 25 cm = 0.25 m

b = 15 cm = 0.15 m

$I_c$ = The centroidal moment of inertia

$I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )$

$I_c$ = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴

Substituting the known values gives;

$I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right ) = 1.2257083\bar 3 \times 10^{-4}$

$I_c$ = 1.2257083 $\bar 3$ × 10⁻⁴ m⁴

From which we have;

$4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )$