It’s e 2.0 x 10^-4 because it is a fraction
(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


Learn more about intensity of sound here: brainly.com/question/17062836
Explanation:
(a) Draw a free body diagram of the cylinder at the top of the loop. At the minimum speed, the normal force is 0, so the only force is weight pulling down.
Sum of forces in the centripetal direction:
∑F = ma
mg = mv²/RL
v = √(g RL)
(b) Energy is conserved.
EE = KE + RE + PE
½ kd² = ½ mv² + ½ Iω² + mgh
kd² = mv² + Iω² + 2mgh
kd² = mv² + (m RC²) ω² + 2mg (2 RL)
kd² = mv² + m RC²ω² + 4mg RL
kd² = mv² + mv² + 4mg RL
kd² = 2mv² + 4mg RL
kd² = 2m (v² + 2g RL)
d² = 2m (v² + 2g RL) / k
d = √[2m (v² + 2g RL) / k]
Answer:
Hits per second=199 hit/s
Explanation:
#Given the angular velocity,
, radius of the record
and the distance between any two successive bumps on the groove as
.
The linear speed of the record in meters per second is:

#From
above, if the bumps are uniformly separated by 1m, then the rate at which they hit the stylus is:

Hence the bumps hit the stylus at around 199hit/s
The main cause of this is Friction. The more oil that is laid down, the less friction there is between the ball and the lane surface. The less friction, the harder it is for the bowler to send the ball in a curved path imparted by the spin that the bowler puts on the ball at the instant of release.