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3 years ago

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What is the role of the part in the diagram labeled Y?

1 answer:

Vinil7 [7]3 years ago

7 0

Question is incomplete and image is not attached ti the question. The required image is attached below, so the complete question is:

The diagram shows a device that uses radio waves.

What is the role of the part in the diagram labeled Y?

modulate, amplify, and send out waves
capture, amplify, and demodulate waves
change the amplitude and frequency of waves
change the pulse and phase of waves

Answer:

2. capture, amplify, and demodulate waves

Explanation:

The part Y labeled in the diagram refers to radio receiver which capture, amplify and demodulate the radio waves.

The radio receiver seperates required radio frequency signals through antenna and consist of an amplifier that amplify or increase the power of receiving signal. At the end, demodulators present in receivers recover the information from the modulated wave.

Hence, the correct option is 2.

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What does addition of two vectors give you?

mash [69]

Answer:

Explanation:

To add or subtract two vectors, add or subtract the corresponding components. Let →u=⟨u1,u2⟩ and →v=⟨v1,v2⟩ be two vectors. The sum of two or more vectors is called the resultant. The resultant of two vectors can be found using either the parallelogram method or the triangle method .

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3 years ago

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

3 years ago

A suspended object A is attracted to a charged object B, can one conclude that A is charged? Explain

irinina [24]

Explanation:

In my view, when the Object A is attracted to a Charged object B. Object B should be Negatively or Positively charged. So Object B should be the Opposite charged according to the Object B

Example =

If Object B is Negatively Charged, the Object A should be Positively Charged

If the Object B is Positively Charged, the Object A should be Negatively Charged

Sometimes it can Mix as a Neutral as well

Hope this Helps

5 0

3 years ago

What forces are those that act on an object causing the net force to be something other than zero?

Arisa [49]

Gravity is all ways pulling down and the normal force acting on top of the object and for it to have to push or pull to the object

3 0

3 years ago

Consider the table below. Which of the following data sets depict an accelerating object? Mark all that apply.

Gre4nikov [31]

Answer:

A and B

Explanation:

The data sets that depict an accelerating object is Data Set A & Data Set B.

The both data sets show that the body is accelerating. Also, they show that the body started from rest (0m/s) at a 0sec.

Data Set A shows a non-constant acceleration which has changing amount of velocity with change in time. While Data Set B shows a constant acceleration which has constant amount of velocity with change in time.

7 0

3 years ago