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Tju [1.3M]
3 years ago
11

Radar uses radio waves of a wavelength of 2.2 m . The time interval for one radiation pulse is 100 times larger than the time of

one oscillation; the time between pulses is 10 times larger than the time of one pulse. What is the shortest distance to an object that this radar can detect?
Physics
1 answer:
liq [111]3 years ago
8 0

Answer:

<em>The shortest distance to an object that this radar can detect would be </em>

<em>111 m</em>

Explanation:

The shortest distance is the minimum distance that would be detected by the radar, The time of oscillation can be obtained thus;

T = 1/f   ..........1

but f= v/λ substituting f into equation 1 we have;

T = λ/v...............2

Where λ is the wavelength = 2.2 m

v is the velocity of  light  since the radar is an electromagnetic wave

   = 3 x10^{8} m/s

T = 2.2 m / 3 x10^{8} m/s

T = 7.33 x10^{-9} s

<u>Calculating the time interval required by the pulse</u>

Since the interval of the pulse (t) is 100 times greater than the period of oscillation, the time of the pulse is expressed as;

t = 100 x T

t = 100 x 7.33 x10^{-9} s

t = 7.33 x 10^{-7} s

Therefor the time of the pulse is 7.33 x 10^{-7} s

<u>Calculating for the shortest distance of the radar</u>

The shortest distance of the radar can be obtained using the equation below;

λ_s = (t/2) x v ............3

Substituting into equation 3 we have

λ_s = (7.33 x 10^{-7} /2) x 3 x10^{8} m/s

λ_s  =  111 m

Therefore the shortest distance to an object that this radar can detect would be 111 m

   

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Answer:

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Explanation:

The frequency of a wave can be found using the following formula.

f=\frac{v}{\lambda}

where <em>f</em> is the frequency, <em>v</em> is the velocity/wave speed, and λ is the wavelength.

The wavelength is 10 meters and the velocity is 200 meters per second.

  • 1 m/s can also be written as 1 m*s^-1

Therefore:

v= 200 \ m*s^{-1} \\\lambda = 10 \ m

Substitute the values into the formula.

f=\frac{200 \ m*s^{-1}}{10 \ m}

Divide and note that the meters (m) will cancel each other out.

f=\frac{200 \ s^{-1}}{10 \ }

f=20 \ s^{-1}

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f= 20 \ Hz

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Two disks of polaroid are aligned so that they polarize light in the same plane. Calculate the angle through which one sheet nee
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Answer: The unpolarized light's intensity is reduced by the factor of two when it passes through the polaroid and becomes linearly polarized in the plane of the Polaroid. When the polarized light passes through the polaroid with the plane of polarization at an angle \theta with respect to the polarization plane of the incoming light, the light's intensity is reduced by the factor of \cos^2\theta (this is the Law of Malus).

Explanation: Let us say we have a beam of unpolarized light of intensity I_0 that passes through two parallel Polaroid discs with the angle of \theta between their planes of polarization. We are asked to find \theta such that the intensity of the outgoing beam is I_2. To solve this we follow the steps below:

Step 1. It is known that when the unpolarized light passes through a polaroid its intensity is reduced by the factor of two, meaning that the intensity of the beam passing through the first polaroid is

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This beam also becomes polarized in the plane of the first polaroid.

Step 2. Now the polarized beam hits the surface of the second polaroid whose polarization plane is at an angle \theta with respect to the plane of the polarization of the beam. After passing through the polaroid, the beam remains polarized but in the plane of the second polaroid and its intensity is reduced, according to the Law of Malus, by the factor of \cos^2\theta. This yields I_2=I_1\cos^2\theta. Substituting from the previous step we get

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