Question

how much louder is 103dB than 100dB?

Answer 1

okay, so I know that to produce a sound that is twice as loud we need a wave that is about ten times the intensity.

Answer 2

really, i didn't know that.

Answer 3

so 103dB is 3 times as loud?

Answer 4

lol yes, really.. wait.. so maybe i should find the intensity of each and compare?

Answer 5

yes. that's waht i was thinking.

Answer 6

*what.

Answer 7

brb.

Answer 8

ok

Answer 9

i think is is approx 10 times louder.

Answer 10

they said it should be 2 times greater

Answer 11

i used IL=10log(I/I0) & L=k log I to find that out. What do u think?

Answer 12

Then, i have no idea how this works. :\

Answer 13

i used I=(1x10^-12)(10^10.3) and I2=(1x10^-12)(10^10) then divided..

Answer 14

okay, then we'll just leave that one out.

Answer 15

Ahhh.. i think ur approach is better.

Answer 16

okay, i don't have my calculator.. if you work out the result.. i got 0.3

Answer 17

Heyyy, i got 2 when i divided them as u said!!!

Answer 18

really! how do you divide without a calulator?

Answer 19

<< i did use a calculator..>>

Answer 20

lol but if we didn't.. lemme right them better

Answer 21

\[I _{1}= (1x 10^{-12})(10^{10.3}) = \]

Answer 22

do i just add the powers?

Answer 23

No, when u divide I1 by I2, 10^-12 cancels out. So, keep it as it is.

Answer 24

ahh, so then we just subtract the powers with the tens?\[\frac{ 10^{10.3} }{ 10^{^10}}\]

Answer 25

yep.

Answer 26

so 1^.3?

Answer 27

10^0.3

Answer 28

ahh!! used the calculator but why didn't we divide the tens?

Answer 29

for example: (2 x10^2)/(4 x 10^3)= we divide the 2 and 4

Answer 30

No, we don't.

Answer 31

really the what do we do?