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sladkih [1.3K]
3 years ago
12

) Assuming different AM regulations; the receiver is using mixer with subtracting format. The frequency selectivity ratio is app

roximately 10 and the AM range is from 750 kHz to 2600 kHz. The intermediate frequeney is most nearly:
Engineering
1 answer:
Zarrin [17]3 years ago
8 0

Answer:

F=710KHZ

Explanation:

From the question we are told that:

Frequency selectivity ratio R=10

AM range 750 kHz to 2600 kHz

Therefore Bandwidth is

 B=2100-680

 B=1420KHZ

Generally the equation for The intermediate frequency is mathematically given by

Intermediate frequency=\frac{Bandwidth}{2}

 F=\frac{B}{2}

 F=\frac{1420}{2}

 F=710KHZ

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Radio Frequency IDentification (RFID) tags and readers are a category of low-end wireless devices that people may not recognize
Fittoniya [83]

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.

8 0
3 years ago
To read signs you need good focal vision
kow [346]

Answer:eyesight

Explanation:

7 0
3 years ago
Read 2 more answers
) An ammonia nitrogen analysis performed on a wastewater sample yielded 30 mg/L as nitrogen. If the pH of the sample was 8.5, de
Delicious77 [7]

Answer:

NH_4^+ = 2.5 mg/lt

Explanation:

Given data:

Ammonia Nitrogen 30 mg/L

pH = 8.5

-log[H +] = 8.5

[H +] = 10^{-8.5}

NH_4 ^{+} ⇄ H^{+} + NH_3

Rate constant is given as

K_a = \frac{[H^{+}] [NH_3]}{NH_4^{+}} ...........1

K_a = 5.6 \times 10^{-10}

Total ammonia as NItrogen is given as 30 mg/l

\%NH_4^{+} = \frac{ [NH_4] \times 100}{[NH_4^{+}] + [NH_3]}

                    = \frac{100}{\frac{NH_4^+}{NH_4^+} +\frac{NH_3^+}{NH_4^+}}

                    = \frac{100}{1+ \frac{NH_3^+}{NH_4^+}} .....2

from equation 1 we have

\frac{NH_3^+}{NH_4^+} =\frac{K_a}{[H^+]} = \frac{5.6\times 10^{-10}}{10^{8.5}}

plug this value in equation 2 we get

\%NH_4^{+} = 84.96 \%

Total ammonia as N = 30 mg/lt

NH_4^+ = \frac{84.96}{100} \times 30 = 25.5 mg/lt

7 0
3 years ago
Karl and Susan have agreed to come to our party, _______ has made Maria very happy.   that   which   what   who​
Aleks04 [339]

Answer:

that

I am not sure that this is the answer

but i hopethis will help you

4 0
3 years ago
Assume a 100-m tape measures 3 mm too short. Five measurements of a distance give results of 253.294, 253.287, 253.289, 253.284,
mariarad [96]

Answer:

attached below

Explanation:

5 0
3 years ago
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