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Elodia [21]
3 years ago
5

Radio Frequency IDentification (RFID) tags and readers are a category of low-end wireless devices that people may not recognize

as forming a computer network. Which of the statements below are true about RFID?
i. RFID technology takes many forms, used in smartcards, implants for pets, passports, library books and more
ii. with the EPC (Electronic Product Code) RFID tags are small, inexpensive devices that have a unique identifier and a small amount of memory that can be read and written by an RFID reader
iii. the tags are the intelligence in the system, analogous to base stations and access points in WiFi networks
iv. in the EPC Gen 2 Physical Layer the data is transmitted in the same way as in all other wireless situations we have seen so far
Engineering
1 answer:
Fittoniya [83]3 years ago
8 0

Answer:

See explaination.

Explanation:

Radio Frequency Identification (RFID) tags and readers uses the electromagnetic waves to identify and track the attached objects.

A tag is attached to the object which is to be identified or tracked, and reader is used to read the response and send the acknowledgement. Therefore, RFID tags and readers are used in many industries, passports, transportations and pet identification etc.

i.

RFID technology is used in smartcards, implants for pets, passports and library books to identify and track the persons, objects and pets etc.

Hence, we can say that option (i) is true.

ii.

Electronic Product Code (EPC) is a small code stored in the RFID tag. The code stored in the memory is 96 bits which are used to identify the organization which manages the data, unique number to identify the product and a number to identify the particular tag and etc.

EPC is a unique identification number, it can read and be written by the RFID reader. It is used in supply chains instead of a bar code even though expansive.

Hence, option (ii) is true.

iii.

Tags are used to identify and track the objects. It doesn’t belong to base stations and access points as a Wi-Fi networks.

Therefore, option (iii) is false.

iv.

The EPC Generation 2 RFID tag is used to improve the security by enabling the authentication features. It is not the similar way of data transmission in the other wireless situations.

Therefore, option (iv) is false.

Finally, the options (i) and (ii) are TRUE, while (iii) and (iv) are FALSE.

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sergejj [24]

Answer:

T₂ =93.77  °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure  ,P₂= 367+1 = 368  kPa

Lets take  temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

{T_2}=T_1\times \dfrac{P_2}{P_1}

Now by putting the values in the above equation we get

{T_2}=300\times \dfrac{368}{301}\ K

The temperature in  °C

T₂ = 366.77 - 273  °C

T₂ =93.77  °C

8 0
3 years ago
A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s
Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})

where

'H' is the initial depth of the layer

C_c is the Compression index

e_o is the inital void ratio

\bar{\sigma_o} is the initial effective stress at the depth

\Delta \bar{\sigma_o} is the change in the effective stress at the given depth

Applying the given values we get

\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04

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3 years ago
Una empresa realizó en el ejercicio de compras al contado por valor
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Q.14
Kay [80]

Answer:

A. optical isolation

Explanation:

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6 0
3 years ago
Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adia
harina [27]

Answer:

Power = 371.28 kW

Explanation:

Initial pressure, P1 = 5 bar

Final pressure, P2 = 1 bar

Initial temperature, T1 = 320°C

Final temperature, T2 = 160°C

Volume flow rate, V = 0.65m³/s

From steam tables at state 1,

h1 = 3105.6 kJ/kg, s1 = 7.5308 kJ/kgK

v1 = 0.5416 m³/kg

Mass flow rate, m = V/v1

m = 1.2 kg/s

From steam tables, at state 2

h2 = 2796.2 kJ/kg, s2 = 7.6597 kJ/kgK

Power developed, P = m(h1 - h2)

P = 1.2(3105.6-2796.2)

P = 371.28 kW

8 0
3 years ago
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