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disa [49]
2 years ago
14

Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear

drums, which lie on the surface of the head. And, surprisingly, the sound has nothing to do with the frog's inflated throat. If the emitted sound has a frequency of 262 Hz and a sound level of 84 dB (near the eardrum), what is the amplitude of the eardrum's oscillation? The air density is 1.21 kg/m3 and the speed of sound is 346 m/s.
Physics
1 answer:
Sidana [21]2 years ago
8 0

Answer:

The amplitude of the eardrum's oscillation is 6.65×10^-13 m.

Explanation:

Given data:

The sound has a frequency of 262 Hz

The sound level is 84 dB

The air density is 1.21 kg/m^3

The speed of sound is 346 m/s

Solution:

As, Intensity of sound is given by,

I = Io×10^(s/10 db)

I = 2×π^2×ρ×v×f^2×Sm^2

Thus,

Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)

Now, put the values,

Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )

Sm = √(2.51×10^-4 / 5.66×10^8)

Sm = √0.443×10^-12

Sm = 6.65×10^-13 m.

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If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
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5 0
2 years ago
An astronaut takes what he measures to be a 10-min nap in a space station orbiting Earth at 8000 m/s. A signal is sent from the
svet-max [94.6K]

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longer than

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using the equation of time dilation

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3 0
2 years ago
A uniform metre rule of weight 0.9 N is suspended horizontally by two vertical loops of thread A and B placed at 20cm and 30cm f
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Answer:

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Explanation:

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Taking the sum of the torques about B:

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(b) when loop B is slack, there are three forces acting on the metre rule.

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T at 20 cm mark

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∑τ = Iα

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2 N (x − 20 cm) = -27 Ncm

x − 20 cm = -13.5 cm

x = 6.5 cm

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Answer:

Explanation:D

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