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maria [59]
4 years ago
15

consider a wave approaching a barrier with a small hole. what change is the wave likely to undergo as it passes through the hole

?
Physics
2 answers:
Ratling [72]4 years ago
6 0
Diffraction, that is the change in the direction of the wave, when it passes through a gap or around an edge. Diffraction is most noticable when the wavelength of the wave is approximately the same size as the hole, the gap
Nimfa-mama [501]4 years ago
3 0

Answer: The correct answer is Diffraction.

Explanation:

Diffraction is the phenomenon in which a beam of light will get bend when it passes through a narrow aperture. It is accompanied by the interference between the wave form produced.

Consider a wave approaching a barrier with a small hole. The diffraction will occur when the wave passes through the hole. There is a bending of light as it passes through the edge of the object or through the hole.

The amount of bending light depends on the relative size of wavelength of the light in comparison to the size of the aperture.

Therefore, the correct answer is Diffraction.

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Zookeepers carry a stretcher that holds a sleeping lion. The total mass of the stretcher and lion is 175 kg. The lion's forward
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Answer:

350 N

Explanation:

Newton's second law:

∑F = ma

∑F = (175 kg) (2 m/s²)

∑F = 350 N

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5 0
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8 0
4 years ago
A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos
lukranit [14]

Answer:

The inductor contains N = 523962.32 loops  

Explanation:

From the question we are told that

     The capacitance of the capacitor is  C =  286nF = 286 * 10^{-9} \  F

      The resonance frequency is  f = 18.0 kHz =  18*10^{3} Hz

       The diameter is  d =  1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m

       The  of the air-core inductor is l = 12 \ m

        The permeability of free space is  \mu_o = 4 \pi *10^{-7} \ T \cdot m/A

 

Generally the inductance of this air-core inductor is mathematically represented as

              L =  \frac{\mu_o * N^2 \pi d^2}{4 l}

This inductance can also be mathematically represented as

               L = \frac{1}{w^2}

Where w is the angular speed mathematically given as

             w = 2 \pi f

So

            L =  \frac{1}{4 \pi ^2 f^2}

Now equating the both formulas for inductance

         \frac{\mu_o * N^2 \pi d^2}{4 l}  =  \frac{1}{4 \pi ^2 f^2}

making N the subject of  the formula

              N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C}  }

              N =  \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }

             

 Substituting value

            N =  \frac{1}{ 3.142  * 18*10^{3} * 0.00011 }  \sqrt{\frac{12}{ 3.142  * 4 \pi *10^{-7}* 286 *10^{-9}} }

              N = 523962.32 loops  

4 0
3 years ago
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