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r-ruslan [8.4K]
2 years ago
13

In this dark-colored forest, you started with 50% light moths and 50% dark moths. At the end of your simulation, was there a hig

her percentage of light or dark moths?

Physics
1 answer:
krek1111 [17]2 years ago
5 0
Dark moths because they use their color to blend in with the trees to hide from birds
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The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?
Tju [1.3M]

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 =  Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2

For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 =  Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2

The second distance, r₂, can be determined from sound intensity formula given as;

I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 =  \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 =   \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m

Therefore, the second distance of the sound from the source is 431.78 m.

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3 years ago
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2i+3j+4k

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A water wave with a wavelength of 35.6 cm approaches a wall with a large opening in it. The hole begins to close. What size does
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A particle's position along the x-axis is described by. x(t)= At+Bt^2where t is In seconds: x is in meters: and the constants A
DanielleElmas [232]

Answer

given,

position of particle

x(t)= A t + B t²

A = -3.5 m/s

B = 3.9 m/s²

t = 3 s

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\dfrac{dx}{dt}=\dfrac{d}{dt}(-3.5t + 3.9t^2)

v= -3.5 + 7.8 t------(1)

velocity of the particle at t = 3 s

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b) velocity of the particle at origin

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   t = 0 , 0.897 s

speed of the particle at t = 0.897 s

from equation (1)

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6 0
4 years ago
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