The sound intensity of a normal conversation is 10^(6) times greater than the quietest noise a person can hear. The sound intensity of a jet at takeoff is 10^(14) times greater than the quietest noise a person can hear. How many times more intense is the sound of a jet at takeoff than the sound of a normal conversation? Write your answer as a power.

If the sound of the quietest noise a person can hear is x, then the sound of the normal conversation is x(10^6). Also, the sound of the jet at takeoff is x(10^14). When we divide x(10^14) by x(10^6), we get the answer that is equal to 10^8. Thus, the sound of jet at takeoff is 10^8 times that of the normal conversation.

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ashishdwivedilVT ashishdwivedilVT · Answer 2 · 2022-03-31T12:58:38+02:00

The sound of a jet at takeoff is [tex]10^{8}[/tex] times intense than the sound of a normal conversation.

What is sound intensity?

Sound intensity is the amount of energy flowing per unit time through a unit area that is perpendicular to the direction in which the sound waves are traveling.

Let us say the sound intensity of the quietest noise a person can hear is k

Then, The sound intensity of a normal conversation = [tex]10^{6} k[/tex]

Sound intensity of a jet at takeoff = [tex]10^{14} k[/tex]

Sound intensity of a jet at takeoff/ sound intensity of a normal conversation =[tex]\frac{10^{14} k}{10^{6}k }[/tex] = [tex]10^{8}[/tex]

Therefore, the sound of a jet at takeoff is [tex]10^{8}[/tex] times intense than the sound of a normal conversation.

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