The howler monkey is the loudest land animal and can be heard up to a distance of 9.8 km.Assume the acoustic output of a howler to be uniform in all directions.The distance at which the intensity level of a howler's call is 25 dB,in SI units,is closest to:

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We can use the formula for intensity level in decibels:
IL = 10log(I/I₀)

Where I₀ is the reference intensity (10^(-12) W/m^2).

We can rearrange this formula to find the intensity I at a given IL:
I = I₀ * 10^(IL/10)

To find the distance at which the intensity level of the howler's call is 25 dB, we need to find the intensity at that level and then use the inverse square law for sound:

I₁/I₂ = (r₂/r₁)^2

Where I₁ and I₂ are the sound intensities at distances r₁ and r₂ from the source.

Assuming the howler monkey has an acoustic power output of 1 W (which may be an overestimate), we can use the fact that sound spreads out uniformly in all directions to find the sound intensity at a distance of 9.8 km:

I = P/(4πr^2) = 1/(4π(9.8×10^3)^2)

Using the formula above, we find that at a distance of 9.8 km, the intensity level is:

IL = 10log(I/I₀) = 10log(1/(4π(9.8×10^3)^2)/I₀) ≈ 3.3 dB

Therefore, the intensity level at 25 dB can only be achieved at a much closer distance:

25 = 10log(I/I₀) ⟹ I = I₀ * 10^(25/10)

I₁/I₂ = (r₂/r₁)^2 ⟹ I₁/I₀ = (9.8 km / r₁)^2

I₁ = I₀ * (9.8 km / r₁)^2 * 10^(25/10)

I₁ = P/(4πr₁^2) ⟹ r₁ = sqrt(P/(4πI₁))

r₁ = sqrt(1/(4πI₀) / (9.8 km / r₁)^2 * 10^(25/10))

Solving for r₁ gives r₁ ≈ 550 m, which is closest to choice A (550) in the given options.

The Howler Monkey Is the Loudest Land Animal and Can