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3 years ago

In remote areas, your gps devices may lose reception. It’s a good idea to have a

2 answers:

4 0

Answer: MAP!!! it's always a good Idea to have a map nearby, cause you'll never know when you'll need it or like you said, lose reception

Serhud [2]3 years ago

3 0

Answer:

You will need a map, having a map is always very important when electronic gadgets don't work, but you can also take a compass with you.

Explanation:

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assume a five layer network model. There are 700 bytes of application data. There is a 20 bye header at the transport layer, a 2

amm1812

Answer: The overhead percentage is 7.7%.

Explanation:

We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.

We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.

So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:

OH % = (58 / 758) * 100 = 7.7 %

4 0

3 years ago

What was a campaign belief in the 1980 presidential election? Carter called for a stronger national defense. Carter promised to

andreyandreev [35.5K]

Answer:

Explanation:

On edge 2021

6 0

3 years ago

Read 2 more answers

A cooling system load is 96,000 BTUh sensible. How much chilled air is required to satisfy the load if the system is designed fo

Natalija [7]

Answer:

For $20^{\circ}$ - 5.556 lb/s

For $15^{\circ}$ - 7.4047 lb/s

Solution:

As per the question:

System Load = 96000 Btuh

Temperature, T = $20^{\circ}$

Temperature rise, T' = $15^{\circ}$

Now,

The system load is taken to be at constant pressure, then:

Specific heat of air, $C_{p} = 0.24 btu/lb ^{\circ}F$

Now, for a rise of $20^{\circ}$ in temeprature:

$\dot{m}C_{p}\Delta T = 96000$

$\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s$

Now, for $15^{\circ}$ :

$\dot{m}C_{p}\Delta T = 96000$

$\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 15} = 26666.667 lb/h = \frac{26666.667}{3600} = 7.4074 lb/s$

4 0

3 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration

Jobisdone [24]

Answer:

Time =t2=58.4 h

Explanation:

Since temperature is the same hence using condition

x^2/Dt=constant

where t is the time as temperature so D also remains constant

hence

x^2/t=constant

2.3^2/11=5.3^2/t2

time=t^2=58.4 h

4 0

3 years ago

6. During some actual expansion and compression processes in piston–cylinder devices, the gases have been

Katyanochek1 [597]

During some actual expansion and compression processes in piston-cylinder devices, the gases have been are the P1= P2.

<h3>What is the pressure?</h3>

Pressure is something that has the pressure that is physical and that causes the pressure is piston-cylinder devices.

During a few real enlargements and compression procedures in piston-cylinder devices, the gases were located to meet the connection PV n = C, wherein n and C are constants.