電鑽發出響度很大的聲音,這聲音所傳播能量的功率為15W。假設聲波均勻地向各個方向傳播。現有一人用聲級計在某距離探測該電鑽的聲強級，聲級計錄得的值為104dB。問
a)聲級計和電鑽的距離是多少﹖(詳細步驟﹗﹗)
b)如何禁機??(型號:fx-3650p)
S3物理[20分]
2010-06-19 7:24 am
回答 (2)
2010-06-19 7:20 pm
✔ 最佳答案
Intensity level(聲強級), h = 10 log (I/Io)104 = 10 log (I / 1 x 10^(-12) )
10.4 = log (I / 1 x 10^(-12) )
10 ^(10.4) = I / 1 x 10^(-12)
Intensity(強度), I = [1 x 10^(-12) ][10 ^(10.4) ]=10^(10.4-12) = 10^(-1.6)
power (功率) = I / A = I / (4 pi r^2)
15 = I / (4 pi r^2)
4 pi r^2 = I / 15 = 10^(-1.6) / 15
距離, r = √[10^(-1.6) / (60 pi)]
= 0.0115 m
如何禁機:
10^(-1.6) (除) 60 (除) (pi) (exe)
√Ans (exe)
註:
(pi) = (SHIFT)(EXP)
2010-06-19 13:58:08 補充：
Sorry I made a big mistake.
I = P/A
I = power (功率) /A = 15 / (4 pi r^2)
10^(-1.6) = 15 / (4 pi r^2)
r = 6.89 m
天同:
Is that 6.74?
2010-06-19 7:50 pm
Intensity of sound I at distance R from the electric drill,
I = 15/(4.pi.R^2) w/m^2 = 1.194/R^2 w/m^2
where pi = 3.14159...
Since by definition, 104 = 10.log(I/Io), where Io is the threshold of hearing, taken to be about 10^-12 w/m^2
hence, 104 = 10.log(I/10^-12)
i.e. I/10^-12 = 10^(10.4)
1.194/R^2 = 10^-12 x 10^(10.4) = 10^(-1.6)
solve for R gives R = 6.74 m
I = 15/(4.pi.R^2) w/m^2 = 1.194/R^2 w/m^2
where pi = 3.14159...
Since by definition, 104 = 10.log(I/Io), where Io is the threshold of hearing, taken to be about 10^-12 w/m^2
hence, 104 = 10.log(I/10^-12)
i.e. I/10^-12 = 10^(10.4)
1.194/R^2 = 10^-12 x 10^(10.4) = 10^(-1.6)
solve for R gives R = 6.74 m
收錄日期: 2021-04-29 17:36:10
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100618000051KK01763