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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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In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is T=0.200 s. Using this data, we can solve the previous formula to find L:
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3 years ago
An insulating sphere is 8.00 cm in diameter and carries a 6.50 µC charge uniformly distributed throughout its interior volume.
Kobotan [32]

Explanation:

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                       = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}

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Now, calculate the charge as follows.

            q_{in} = \rho(\frac{4}{3} \pi r^{3})

                      = 2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}

                      = 10.106 \times 10^{-8} C

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                q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}

                          = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}

                          = 7.454 \mu C

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egoroff_w [7]
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