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4 years ago

What is tcp in network?

1 answer:

5 0

Transmission Control Protocol (TCP) – a connection-oriented communications protocol that facilitates the exchange of messages between computing devices in a network. It is the most common protocol in networks that use the Internet Protocol (IP); together they are sometimes referred to as TCP/IP.

pls mark me as the brainliest

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EJERCICIO 6

Ket [755]

Answer:

mnfokfnfi3or

Explanation:

can you translate it into english.....

8 0

3 years ago

An escalator in a shopping center is designed to move 50 people, 75 kg each, at a constant speed of 0.6 m/s at 450 slope. Determ

REY [17]

Answer:

Power required to drive the escalator shall be equal to the rate at which the energies of the persons is increased.

$Energy=n\times mass\times g\times h\\\\\therefore Power=n\times mass\times g\times \frac{dh}{dt}\\\\Power=50\times 75\times 9.81\times 0.6sin(45)\\\\Power=15.607kW$

As we infer from the above equation If the velocity of the escalator is doubled then the Power required will also be doubled and become 31.215kW

3 0

4 years ago

Q1) Determine the force in each member of the

Sever21 [200]

Answer:

CD = DE = DF = 0
BC = CE = 15 N tension
FA = 15 N compression
CF = 15√2 N compression
BF = 25 N tension
BG = 55/2 N tension
AB = (25√5)/2 N compression

Explanation:

The only vertical force that can be applied at joint D is that of link CD. Since joint D is stationary, there must be no vertical force. Hence the force in link CD must be zero, as must the force in link DE.

At joint E, the only horizontal force is that applied by link EF, so it, too, must be zero.

Then link CE has 15 N tension.

The downward force in CE must be balanced by an upward force in CF. Of that force, only 1/√2 of it will be vertical, so the force in CF is a compression of 15√2 N.

In order for the horizontal forces at C to be balanced the 15 N horizontal compression in CF must be balanced by a 15 N tension in BC.

At joint F, the 15 N horizontal compression in CF must be balanced by a 15 N compression in FA. CF contributes a downward force of 15 N at joint F. Together with the external load of 10 N, the total downward force at F is 25 N. Then the tension in BF must be 25 N to balance that.

At joint B, the 25 N downward vertical force in BF must be balanced by the vertical component of the compressive force in AB. That component is 2/√5 of the total force in AB, which must be a compression of 25√5/2 N.

The <em>horizontal</em> forces at joint B include the 15 N tension in BC and the 25/2 N compression in AB. These are balanced by a (25/2+15) N = 55/2 N tension in BG.

In summary, the link forces are ...

(25√5)/2 N compression in AB
15 N tension in BC
25 N tension in BF
0 N in CD, DE, and EF
15 N tension in CE
15√2 compression in CF
15 N compression in FA

_____

Note that the forces at the pins of G and A are in accordance with those that give a net torque about those point of 0, serving as a check on the above calculations.

8 0

3 years ago

Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and

Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

⇒ P = 4745.11 lb = 4.745 kips

The pic shown can help to understand the question.

5 0

4 years ago

This is used to keep all cleaning supplies and chemicals safe

bulgar [2K]

A storage room

common sense lol

6 0

3 years ago