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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 .

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An object traveling at a constant

mel-nik [20]

Answer:

pi / 2 radians / s

Explanation:

One revolution = 2 pi Radians in 4 seconds

2 pi / 4 = pi/2 radians / s

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2 years ago

The equation that is used to solve second law problems is # F= ma.

maw [93]

F= Force
M=Mass
A= acceleration
F=N
Mass= in grams or kilo grams (mostly kg)
A= m/s

8 0

3 years ago

A vertically polarized beam of light of intensity 100 W/m2 passes through two ideal polarizers. The transmission axis of the fir

TEA [102]

To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

$I=I_0 cos^2\theta$

Where,

$I_ {0}$ indicates the intensity of the light before passing through the polarizer,

I is the resulting intensity, and

$\theta$ indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

Since we have two objects the law would be,

$I=I_0cos^2\theta_1*cos^2(\theta_2-\theta_1)$

Replacing the values,

$I=100*cos^2(20)*cos^2(40-20)$

$I=100*cos^4(20)$

$I=77.91W/m^2$

Therefore the intesity of the light after it has passes through both polarizers is $77.91W/m^2$

7 0

3 years ago

A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th

zhannawk [14.2K]

Answer: $V_{f}=2.96m/s$

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight $w$ of the block has an x-component and y-component:

$w_{x}=wsin(\theta)$ (1)

$w_{y}=wcos(\theta)$ (2)

As well as the Normal Force $N$ :

$N_{x}=Nsin(\theta)$ (3)

$N_{y}=Ncos(\theta)$ (4)

In addition, we know $N=w$ , then $\sum F_{y}=0$

In the X-component:

$\sum F_{x}=m.a$

$m.a=w_{x}$ (5)

Substituting (1) in (5):

$wsin(\theta)=m.a$ (6)

In addition, we know $w=m.g$ , where $m$ is the mass of the block and $g$ the gravity acceleration, which is equal to $9.8m/{s}^{2}$

So:

$m.g.sin(\theta)=m.a$ (7)

$a=g.sin(\theta)$ (8)

$a=5.88m/{s}^{2}$ (9) >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance $d$ with the acceleration $a$ and time $t$ :

$d=\frac{1}{2}a{t}^{2}$ (10)

We already know the value of $d$ and calculated $a$ , we have to find $t$ :

$t=\sqrt{\frac{2d}{a}}$ (11)

$t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}$ (12)

$t=0.50s$ (13) >>>This is the time it takes to the block to go from the initial velocity $V_{o}$ to its final velocity $V_{f}$

If the acceleration is the variation of the velocity in time, we can use the following equation to find $V_{f}$ :

$V_{f}-V_{o}=a.t$ (13)

If $V_{o}=0$

$V_{f}=a.t$ (14)

$V_{f}=(5.88m/{s}^{2})(0.50s)$ (15)

Finally we get the value of the Final Velocity of the block:

$V_{f}=2.96m/s$

6 0

3 years ago

How does adding more of a substance affect it's density?

fomenos

I’m going to use molasses as an example of a substance.

The mass and volume both change when changing the amount of molasses.
However, the density does not change. This is because the mass and volume increase at the same rate/proportion!

Even though there is more molasses (mass) in test tube A, the molasses also takes up more space (volume). Therefore, the spacing between those tiny particles that make up the molasses is constant (does not change).

The size or amount of a material/substance does not affect its density.

5 0

4 years ago

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