Sound intensity of 1 baby, I = 8*10^-8 W/m^2
The sound heard should be higher by:
10*log (n) where for 5 babies, n = 5. Then
10*log (n) = 10*log (5) ≈ 7 dB
Also give is the reference sound, Io = 1.0*10^-12 W/m^2
Therefore,
Sound intensity, L1 = 10*log (I/I1) = 10*log [(8*10^-8)/(1*10^-12)] ≈ 49 dB
Therefore, total intensity for the five babies is:
Total intensity = 49+7 = 56 dB
Answer:
f_2 = 436 Hz
Explanation:
Given:
- Frequency of the note played f_1 = 440 Hz
- Number of beats heard per second b = 4 Hz
Find:
What was the frequency of the mis-tuned A string?
Solution:
- The beat is the difference in frequency of the notes played and is given by the expression:
| f_1 - f_2 | = beats
- we heard 4 beats, so
| f_1 - f_2 | = 4
- Plug in the value: | 440 - f_2 | = 4
Hence, f_2 = 436 Hz
Answer:
A.
Explanation:
An electrical circuit is a path in which electrons flow because of voltage.
A. electrical
Answer:
The attractive or repulsive interaction between any two charged object is an electric force.It is exerted between any two charged objects.It shows that as the amount of the charge on the objects increases, the force between them increases. and as the distance between the charged object increases, the force between them increases.
Average i just took the text:)