Question

Joey works at the airport repairing airplane engines, where the decibel level can reach over 100 dB. Wearing protective earwear, the decibel level reacing Joey's ear only reaches 80 dB. How many times less intense is the sound with his ear protection when compared to no protection? a. 1 b. 10 c. 100 d. 1000

Answer 1

c.100. the ear plugs block out the sound

Answer 2

c). 100

Step-by-step explanation:

As we know that the relation between level of sound in dB and intensity of sound is given as

`L = 10 Log(I)/(I_0)`

now in order to compare two different level of sounds we will have

`L_2 - L_1 = 10 Log(I_2)/(I_0) - 10 Log(I_1)/(I_0)`

now here we have

`L_2 - L_1 = 10 Log(I_2)/(I_1)`

here we know that without wearing protective earwear the sound level is 100 dB and after wearing it the sound level decreased to 80 dB

so now we have

`100 - 80 = 10 Log(I_2)/(I_1)`

`20 = 10 Log(I_2)/(I_1)`

`(I_2)/(I_1) = 10^2 = 100`

So intensity of sound is decrease by 100 times by wearing protective earwear