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Butoxors [25]
2 years ago
10

The denizens of a distant planet live under an atmosphere made predominantly of CO2 which is a dispersive medium while attending

a concert which of the following would be true for a spectator seated in the back of a large amphitheater?
I. They would hear the low notes and high notes at different times
II. They would hear the low notes and high notes at the same time
Physics
1 answer:
LuckyWell [14K]2 years ago
5 0

Since the frequency of sound in a medium is constant, therefore, the concert-goers would hear the low notes and high notes at the same time.

<h3>What is a dispersive medium?</h3>

A dispersive medium is a medium which spreads out or disperses a substance passing through it.

Since CO2 is a dispersive medium, it means sound waves passing through it would be dispersed based on wavelength.

The note of a sound depends on its frequency, the higher the frequency, the higher the note.

Frequency of sound is constant, therefore, the concert-goers would hear the low notes and high notes at the same time.

Learn more about dispersion of sound at: brainly.com/question/781734

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Answer:

id say the first option.

Explanation:

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8 0
2 years ago
Type the correct answer in the box. What is the resistance of a circuit with a voltage of 10 volts (V) and a current of 5 amps (
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Resistance = Voltage/Current

                   = 10/5

Resistance = 2 ohms

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3 years ago
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a sound wave travels at 330 m/sec and has a wavelenght of 2meters. calcuate the frequency and period
Sergio039 [100]

The frequency of any wave is  (speed) / (wavelength).

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7 0
3 years ago
In a double-slit interference experiment, the wavelength is λ = 432 nm, the slit separation is d = 0.100 mm, and the screen is d
andrew-mc [135]
In the double-slit interference experiment, the distance of the nth-maximum from the center of the screen is given by
y= \frac{n \lambda D}{d}
where
\lambda is the wavelength
D is the distance between the screen and the slits
d is the distance between the slits

In our problem, 
\lambda=432 nm= 432 \cdot 10^{-9} m
D=42.0 cm=0.42 m
d=0.100 mm=0.1 \cdot 10^{-3} m

By applying the previous formula, we can calculate the distance of the 4th maximum from the center of the screen:
y_4 =  \frac{(4)(432 \cdot 10^{-9} m)(0.42 m)}{(0.1 \cdot 10^{-3}m)}=7.25 \cdot 10^{-3} m

Similarly, the distance of the 8th- maximum is
y_8 = \frac{(8)(432 \cdot 10^{-9} m)(0.42 m)}{(0.1 \cdot 10^{-3}m)}=14.5 \cdot 10^{-3} m

Therefore, the distance between the two maxima is
\Delta y=y_8- y_4 = 14.5 \cdot 10^{-3} m- 7.25 \cdot 10^{-3} m =7.25 \cdot 10^{-3} m = 7.25 mm
5 0
3 years ago
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