Question

A howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 5.0 km. Assume the acoustic output of a howler to be uniform in all directions and that the threshold of hearing is 1.0 × 10-12 w/m2. The acoustic power emitted by the howler is closest to

Answer 1

The surface area of a sphere with radius of 5.0 km is

A = 4 · π · r²

A = (4·π) · (5,000m)²

A = (4·π) · (2.5 x 10⁷) m²

A = 3.14 x 10⁸ m²

If the sound intensity over the surface of that sphere is exactly the threshold of human hearing (1.0 x 10⁻¹² w/m²), then the howler's power output has to be

P = (1.0 x 10⁻¹² w/m²)  x (3.14 x 10⁸ m²)

P = (3.14 x 10⁻¹²⁺⁸) watt

P = 3.14 x 10⁻⁴ watt of sound

That's 0.000314 of a watt ... only about -5 dBm.  Gosh, that's 12 dB less than the telephone industry's standard level out of a telephone earpiece straight into your ear.  I'm not happy with this answer at all, and even though I'm happily going to take your points for it, I don't think you should use it, until somebody else comes along and either confirms it or shows both of us where my mistake is.  I'll be the first to admit that this is a ridiculous number for an isotropic monkey at 5 km.

Answer 2

The acoustic power emitted by the howler is $3.142 \times 10^{-4} \ W$.

The given parameters:

• distance the sound can be heard, d = 5.0 km = 5,000 m
• intensity of the sound, I = 1 x 10⁻¹² W/m²

The area of the howler is calculated as follows;

$A = 4\pi r^2\\\\A = 4 \pi \times (5,000)^2\\\\A = 3.142 \times 10^8 \ m^2$

The acoustic power emitted by the howler is calculated as follows;

$P = IA\\\\P = (1.0 \times 10^{-12} ) \times (3.142 \times 10^8)\\\\P = 3.142 \times 10^{-4} \ W$

Learn more about acoustic power  here: https://brainly.com/question/21052445