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Deffense [45]
3 years ago
9

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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If two forces act on an object in the same direction, the net force is the sum of the two forces. In this case, the net force is always greater than either of the individual forces.

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3 examples of newtrons 2 law
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1) Kicking a ball.

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3 years ago
A pole-vaulter just clears the bar at 5.31 m and falls back to the ground. The change in the vaulter's potential energy during t
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Answer:

Weight = 734.46 N

Explanation:

Given:

Initial height of the pole-vaulter is, h_i=5.31\ m

Final height of the pole-vaulter is, h_f=0\ m

Change in the vaulter's potential energy is, \Delta U=-3900\ J

We know that, change in potential energy is given as:

\Delta U=mg(h_f-h_i)

Where, 'm' is the mass of the object, 'g' is acceleration due to gravity and has a value of 9.8 m/s².

Now, weight of the object is given as the product of its mass and acceleration due to gravity. So, replace 'mg' by weight 'w'. So, the equation becomes,

\Delta U = w(h_f-h_i)

Now, rewriting in terms of 'w', we get:

w=\dfrac{\Delta U}{(h_f-h_i)}

Now, plug in all the given values and solve for 'w'. This gives,

w=\dfrac{-3900\ J}{(0-5.31)\ m}\\\\\\w=\dfrac{3900}{5.31}\ N\\\\\\w=734.46\ N

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8 0
3 years ago
Tripling the displacement from equilibrium of an object in simple harmonic motion will bring about a change in the magnitude of
Anni [7]

Answer:

acceleration will be tripled.

Explanation:

We know, when an object is performing Simple harmonic motion, the force

experience by it is directly proportional to its displacement from its mean position.

Also, F = ma , therefore, acceleration is also proportional to its displacement .

Now, F = kx

Therefore, a=\dfrac{k\ x}{m}

If we triple the displacement i.e, 3x.

Acceleration a'=\dfrac{k(3x)}{m}=3a.

Therefore, acceleration is also tripled.

Hence, this is the required solution.

7 0
3 years ago
A positive charge (q = +6.0 µC) starts from point A in a constant electric field and accelerates to point B. The work done by th
riadik2000 [5.3K]

Vb - Va = -366.7 V.

Vab = Va - Vb, the potential of a with respect to b, is equal to the work done by the electric force when a unit of charge moves from a to b, it is given by:

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So, in order to determinate the potential difference Vb - Va we have to multiply by -1 both side of the equation above:

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Given a positive charge q = 6.0μC = 6.0x10⁻⁶C, Wab = 2.2x10⁻³J. Determine Vb - Va.

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8 0
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