Answer:
3.16 ×
W/![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
Explanation:
β(dB)=10 × ![log_{10}(\frac{I}{I_{0} })](https://tex.z-dn.net/?f=log_%7B10%7D%28%5Cfrac%7BI%7D%7BI_%7B0%7D%20%7D%29)
=
W/![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
β=55 dB
Therefore plugging into the equation the values,
55=10
})[/tex]
5.5=
})[/tex]
= ![\frac{I}{10^{-12} }](https://tex.z-dn.net/?f=%5Cfrac%7BI%7D%7B10%5E%7B-12%7D%20%7D)
316227.76×
= I
I= 3.16 ×
W/![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
Work equals force × displacement (distance between initial point and end point is displacement)
if u follow this it becomes
work = 50 × 2 which is equal to 100
comment if u have more questions
ANSWER: d) 8
EXPLANATION: Two sets of two shared electrons (4 electrons total shared) = one set of a double covalent bond.
Therefore, 8 electrons total shared = two sets of double covalent bonds
Answer:
B) 1.5 m/s
Explanation:
The apparent frequency will be enhanced due to Doppler effect
If f be the apparent frequency , F be the real frequency , V be the velocity of sound and v be the velocity of approaching submarine then f is given by
f = F \frac{V+v}{V-v}\\
\frac{f}{F} =\frac{V+v}{V-v}\\
\frac{f}{F}-1 =\frac{V+v}{V-v}-1\\
\Delta f = \frac{2vf}{V-v}\\
200=\frac{2\times v\times 100\times 1000}{1482-v}\\
v=1.48 m/s
<span>sound waves are a type of wave sometimes called compression waves, vibrations with enough of an amplitude can compress and decompress the air adjacent to the object causing the waves to form.</span>