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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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Answer:

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Answer:

Explanation:

Given that,

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Answer:  75,242.9 m/s

Explanation:

from the question we are given the following parameters

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velocity of lion (VL) = 777,377.7 m/s

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mass of Lion and Gazelle (M) = 200,900.7 kg

velocity of Lion and Gazelle (V) = ?

The first figure below shows the motion of the Lion and Gazelle with their direction.

The second diagram shows the motion of the Lion and Gazelle with their directions rearranged to form a right angle triangle.

from the triangle formed we can get the velocity of the Lion and Gazelle immediately after collision using their momentum and Phytaghoras theorem

momentum = mass x velocity

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now applying Phytaghoras theorem we have

13,089,840,463.7 + 2,026,506,942.46 =  200,900.7 x V

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Answer:

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c)

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