Question
An 8 hour exposure to a sound level of 95.0 dB may cause hearing damage. What energy in joules falls on a 0.850 cm diameter eardrum so exposed?
I try to use
k = IAT = (1*10^-3)(0.0085^2)(8*3600)=0.00208
but answer is not correct.
I try to use
k = IAT = (1*10^-3)(0.0085^2)(8*3600)=0.00208
but answer is not correct.
Answers
L=10 log₁₀ (I/I₀)
I₀= 10⁻¹² W/m²
I/I₀ = 10^(L/10)=
=10^(L/10)=
I= I₀•10^(L/10) =
=10⁻¹²•10^(95/10) =3.16•10⁻³W/m²
E= I•A•t=I•πr²•t=
=3.16•10⁻³•π•0.0085²•8•3600 =2.07•10⁻²J
I₀= 10⁻¹² W/m²
I/I₀ = 10^(L/10)=
=10^(L/10)=
I= I₀•10^(L/10) =
=10⁻¹²•10^(95/10) =3.16•10⁻³W/m²
E= I•A•t=I•πr²•t=
=3.16•10⁻³•π•0.0085²•8•3600 =2.07•10⁻²J
Elena, thank you for your help, but the answer is not correct.
I found the answer:
E=(3.16^-3)[(π*0.0085/2)^2](8*3600)=0.00516 J
E=(3.16^-3)[(π*0.0085/2)^2](8*3600)=0.00516 J