All categories

3 years ago

What will be the length of a half wavelength dipole antenna operating at 10GHz?

Physics

1 answer:

iVinArrow [24]3 years ago

4 0

Given :

Frequency , $\nu = 10\ GHz$ .

To Find :

The length of a half wavelength dipole .

Solution :

Now , wavelength is given by :

$\lambda=\dfrac{c}{\nu}\\\\\lambda=\dfrac{3\times 10^8\ m/s}{10\ GHz}\\\\\lambda=\dfrac{3\times 10^8}{10\times 10^9}\ m\\\\\lambda=0.03\ m\\\\\lambda=3 \ cm$

Now , half length is given by :

$L=\dfrac{\lambda}{2}\\\\L=\dfrac{3}{2}\ cm\\\\L=1.5 \ cm$

Therefore , length of a half wavelength dipole is 1.5 cm .

Hence , this is the required solution .

You might be interested in

The wavelengths of radio waves are much _______________ than the wavelength of microwaves. therefore, radio waves carry much ___

nydimaria [60]

The wavelengths of radio waves are much "Longer" than the wavelength of microwaves therefore, radio waves carry much "Lower" <span>energy than a microwave.

Hope this helps!</span>

7 0

3 years ago

Read 2 more answers

Help me get a answer

daser333 [38]

Answer:

I think D sorry if I'm wrong

6 0

2 years ago

Assume the Earth is a ball of perimeter 40, 000 kilometers. There is a building 20 meters tall at point a. A robot with a camera

torisob [31]

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

No intersection: There's nothing that blocks the camera's view of the top of the building.
Two intersections: The planet blocks the camera's view of the top of the building.
One intersection: The point at which the top of the building appears or disappears.

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle $\angle \mathrm{B\hat{C}D}$ which corresponds to this minor arc.

This angle comes can be split into two parts:

$\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}$ .

Also,

$\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}$ .

The radius of this circle is:

$\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}$ .

The lengths of segment DC, AC, BC can all be found:

$\rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m$ ;
$\rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m$ ;
$\rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m$ .

In the two right triangles $\triangle\mathrm{DAC}$ and $\triangle \rm BAC$ , the value of $\angle \mathrm{B\hat{C}A}$ and $\angle \mathrm{A\hat{C}D}$ can be found using the inverse cosine function:

$\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}$

$\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}$

$\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}$ .

The length of the minor arc will be:

$\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km$ .

5 0

3 years ago

Name of impurities present in zinc plate used in simple cell.

raketka [301]

Mercury is present in zinc plate which is used in simple cell

8 0

3 years ago

If the wave being sent is transmitted via an electromagnetic wave explain how the digital signal would be perceived if the signa

Andrej [43]

Answer:

amplitude is commonly for transmitting messages with a radio carrier wave, the amplitude (signal strength) of the carrier wave is varied in proportion to that of the message signal. at the receiving end, the message signal is extracted from the modulated carrier by demodulation. frequency is the encoding of information in a carrier wave with instantaneous frequency. with digital data, the frequency of the carrier is shifted among a set of frequencies, using digits like 1 and 0

6 0

3 years ago