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saw5 [17]
1 year ago
12

Because the sensitivity of the human ear varies over the audio spectrum, how do many sound systems handle very high frequencies

and very low frequencies, respectively, to compensate?
1. dampen; amplify
2. amplify; amplify
3. Sound systems do nothing to compensate.
4. amplify; do nothing to
5. dampen; do nothing to
6. amplify; dampen
7. do nothing to; amplify
8. dampen; dampen
9. do nothing to; dampen
Physics
1 answer:
antoniya [11.8K]1 year ago
3 0

Answer:

Option 2.

Amplify; Amplify

Explanation:

Amplification of sound simply means increasing the loudness of the sound by modifying its electro-acoustic signals. To make very high and very low frequencies of sound audible by the human ear,  the sounds need to be amplified by the sound system amplifier.

Sound systems modify high frequencies by amplifying them, and also, they also modify low frequencies by amplifying them. This changes their wave form and brings them into the hearing range of humans.

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Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t  (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= (m+ \frac{1}{2} \lambda)

replacing ;

Δx = 2t   ; we have:

2t = (m+ \frac{1}{2} \lambda)

Given that thickness t = 700 nm

Then

2× 700 = (m+ \frac{1}{2} \lambda)     --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = (m+ \frac{1}{2} \lambda)     ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus;  the wavelength λ of the light when it is traveling in air = 560 nm

b)  

For the smallest thickness t_{min} ; \ \ \ m =0

∴ 2t_{min} =\frac{\lambda}{2}

t_{min} =\frac{\lambda}{4}

t_{min} =\frac{560}{4}

t_{min} =140 \ \  nm

Thus, the smallest thickness t of the air film = 140 nm

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<u>Explanation</u>:

The continuous random variable is random variable where the data can take infinite variables. For example random variable is taken for measuring "speed of automobiles" on the highways. The radar instrument depicts time taken by automobile in particular what speed. They are the generalization of discrete random variables not the real numbers as a random data is created. It gives infinite sets of all possible outcomes. It is obvious that outcomes of the instrument depend on some "physical variables" those are not predictable as depends on the situation.

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Answer:

#4 is the accurate answer.

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