(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².
(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².
(1.c) The relative intensity of the sound as heard by the listener is 103 dB.
(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.
(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.
<h3>
Surface area being vibrated</h3>
The surface area being vibrated by the time the sound reaches the listener is calculated as follows;
A = 4πr²
A = 4π x (20)²
A = 5,026.55 m²
<h3>Intensity of the sound</h3>
The intensity of the sound is calculated as follows;
I = P/A
I = (100) / (5,026.55)
I = 0.02 W/m²
<h3>Relative intensity of the sound</h3>

<h3>Speed of sound at the given temperature</h3>

<h3>Frequency of the sound</h3>
The frequency of the sound heard is determined by applying Doppler effect.

where;
- -v₀ is velocity of the observer moving away from the source
- -vs is the velocity of the source moving towards the observer
- fs is the source frequency
- fo is the observed frequency
- v is speed of sound


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Different elements require different levels of energy to make or break a bond
Incomplete question.The Complete question is here
A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.
a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.
b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.
Answer:
(a)ω = 1 rad/s
(b)t = 2.41 s
Explanation:
(a) initial angular momentum = final angular momentum
0 = L for disk + L............... for runner
0 = Iω² - mv²r ...................they're opposite in direction
0 = (MR²/2)(ω²) - mv²r
................where is ω is angular speed which is required in part (a) of question
0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)
0=200ω²-200
200=200ω²
ω = 1 rad/s
b.)
lets assume the "starting point" is a point marked on the disk.
The person's angular speed is
v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s
As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.
(angle) + (angle disk turns) = 2π
(1.6 rad/s)(t) + ωt = 2π
t[1.6 rad/s + 1 rad/s] = 2π
t = 2.41 s
The angles in the triangle are 91 degrees, 53 degrees and 36 degrees respectively.
<h3>What is the cosine rule?</h3>
From the cosine rule we know that;
c^2 = a^2 + b^2 - 2abcosC
Since;
a = 0.47 m
b = 0.62 m
c = 0.78 m
Then;
(0.78)^2 = (0.47)^2 + (0.62)^2 - 2(0.47 * 0.62)cosC
0.61 = 0.22 + 0.38 - 0.58 cosC
0.61 - ( 0.22 + 0.38) = - 0.58 cosC
0.01 = - 0.58 cosC
C = cos-1(0.01/-0.58)
C = 91 degrees
Using the sine rule;
b/Sin B = c/Sin C
0.62/sinB = 0.78/sin 91
0.62/Sin B = 0.78
B = sin-1 (0.62//0.78)
B = 53 degrees
Angle A is obtained from the sum of angles in a triangle;
180 - (91 + 53)
A = 36 degrees
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