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3 years ago

Which field in a Transmission Control Protocol (TCP) header is chosen from ephemeral ports?

Engineering

1 answer:

zalisa [80]3 years ago

8 0

Answer:

<u>Source Port</u>

Explanation:

When a connection is establish for an application, it assigns a port that is not in use as the source port. Such ports are known as ephemera ports, which are not related to specific protocol and are only allocated for the period of the connection. All data received on this port (until the connection is disconnected) is sent on to that application.

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When block C is in position xC = 0.8 m, its speed is 1.5 m/s to the right. Find the velocity of block A at this instant. Note th

Amanda [17]

Answer:

The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

$\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant$

Differentiating the above equation:

$\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\$

$\to V_A= \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\$

$= \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}$

3 0

3 years ago

Steam enters a steady-flow adiabatic nozzle with a low inlet velocity (assume ~0 m/s) as a saturated vapor at 6 MPa and expands

Sergio [31]

Yea bro I don’t really know

7 0

3 years ago

Base course aggregate has a target dry density of 119.7 Ib/cu ft in place. It will be laid down and compacted in a rectangular s

djyliett [7]

Answer:

total weight of aggregate = 5627528 lbs = 2814 tons

Explanation:

we get here volume of space to be filled with aggregate that is

volume = 2000 × 48 × 0.5

volume = 48000 ft³

now filling space with aggregate of the density that is

density = 0.95 × 119.7

density = 113.72 lb/ft³

and dry weight of this aggregate is

dry weight = 48000 × 113.72

dry weight = 5458320 lbs

we consider here percent moisture is by weigh

so weight of moisture in aggregate will be

weight of moisture = 0.031 × 5458320

weight of moisture = 169208 lbs

so here total weight of aggregate is

total weight of aggregate = 5458320 + 169208

total weight of aggregate = 5627528 lbs = 2814 tons

3 0

3 years ago

At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge

tangare [24]

Answer:

e= 50 J/kg

Explanation:

Given that

Speed ,v= 10 m/s

Diameter of the turbine = 90 m

Density of the air ,ρ = 1.25 kg/m³

We know that mechanical energy given as

$E=\dfrac{1}{2}mv^2\ J$

That is why mechanical energy per unit mass will be

$e=\dfrac{1}{2}v^2\ J/kg$

Now by putting the values in the above equation we get

$e=\dfrac{1}{2}\times 10^2\ J/kg$

e= 50 J/kg

That why the mechanical energy unit mass will be 50 J/kg.

5 0

3 years ago

To do a research, a researcher needs to be clear about what he/she is doing, why he/she is doing it and the associated implicati

Mashutka [201]

Answer:

True

Explanation:

To carry out a research in order to that it should yield the desirable result, a researcher must be very clear about his objective for the research.

The researcher must be very much clear about all the do's and don'ts and all the where, why, what sort of thing, i.e., the researcher must be clear about his/her intentions of carrying out the research, and what research is being carried out by him/her.

The one doing the research must be clear about the implications of the research.

Unless all these things are not clear at the researcher's end the research will not give the desired results.

4 0

3 years ago