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2 years ago

15

Which type of communications equipment functions as a radio receiver and searches across several frequencies?

1 answer:

krok68 [10]2 years ago

7 0

A <u>scanner</u> is a type of communications equipment that functions as a radio receiver and searches across several frequencies.

A scanner is a kind of a radio receiver that has the ability to receive multiple signals.

There are three modes which a scanner uses for acting as a radio receiver. The scan mode of the radio receiver constantly changes frequencies that helps in transmissions. There is also a manual scan mode that allows the users to search for their interested frequencies. The search mode allows the users to search through two sets of frequencies.

A scanner is a type of communication equipment that is easy to use with various features such as the volume, numeric keypad, trunk tracking etc.

To learn more about scanners, click here:

brainly.com/question/24937533

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3 years ago

Read 2 more answers

An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e

zimovet [89]

Answer:

a) $v = 6.25\times 10^{5} m/s$

b) $v = 1.73\times 10^{4} m/s$

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is $4.55\times 10^5 m/s$

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

$a =\frac{q E}{m}$

$a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}$

$a = 2.58\times 10^{11} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355$

$v = 6.25\times 10^{5} m/s$

b) for proton

$a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}$

$a = -1.41\times 10^{8} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355$

$v = 1.73\times 10^{4} m/s$

3 0

3 years ago

A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C d

Alex73 [517]

Answer

given,

Side of copper plate, L = 55 cm

Electric field, E = 82 kN/C

a) Charge density,σ = ?

using expression of charge density

σ = E x ε₀

ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

now,

σ = 82 x 10³ x 8.85 x 10⁻¹²

σ = 725.7 x 10⁻⁹ C/m²

σ = 725.7 nC/m²

change density on the plates are 725.7 nC/m² and -725.7 nC/m²

b) Total change on each faces

Q = σ A

Q = 725.7 x 10⁻⁹ x 0.55²

Q = 219.52 nC

Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC

7 0

4 years ago