A simple pendulum with length L is swinging freely with

A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

Part A

I = 1.4 mW/m²

Part B

β = 91.46 dB

Explanation:

Part A

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

4 0

3 years ago

Nvm i got the answer but free points

faust18 [17]

Answer:

thank you 谢谢

Explanation:

8 0

3 years ago

Read 2 more answers

A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the

sp2606 [1]

From the question,
$u = 0m {s}^{ - 1}$
$v = 6m {s}^{ - 1}$

$t = 3s$
$F=12N$

Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.

$Ft =m(v-u)$

$12(3.0)=m(6.0- \: 0)$
This implies that

$36.0 = 6m$
$m = \frac{36.0}{6.0}$
$\therefore \: m = 6.0kg$

You can also use the equation of linear motion,
$v = u + at$
$6 = 0 + a(3)$
$6 = 3a$
$a = \frac{6}{3}$

$a = 2 {ms}^{ - 2}$
But
$F=ma$
$12 = m(2)$
$12 = 2m$
$\frac{12}{2} = m$
$\therefore \: m = 6kg$

4 0

3 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s$

3 0

3 years ago

Which formula is used to find fluctuation of the shape of body

Sladkaya [172]

Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

5 0

3 years ago