As aids in conversions, note that 10 ft = 3.05 m, 100 ft² 0.0002 μbar 10.8 ft². = 20 μPa, 1 ft² = = 9.29 m², 1000 ft³ = 28.3 m³, 1 metric sabin = 0.093 m², and 1 m² 1 English sabin =
As aids in conversions, note that 10 ft = 3.05 m, 100 ft² 0.0002 μbar 10.8 ft². = 20 μPa, 1 ft² = = 9.29 m², 1000 ft³ = 28.3 m³, 1 metric sabin = 0.093 m², and 1 m² 1 English sabin =
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
The reverberation time in a small reverberation chamber 9 ft x 10 ft x 11 ft is 4.0 s.
(a) What is the effective absorptivity of the surfaces of the chamber? (b) When 50 ft^2
of one wall is covered with acoustic tile, the reverberation time is reduced to 1.3 s.
What is the effective absorptivity of the tile? (c) What would be the reverberation
time if all surfaces of the chamber were covered with this tile?
Transcribed Image Text:As aids in conversions, note that 10 ft = 3.05 m, 100 ft²
0.0002 μbar
10.8 ft².
=
20 μPa, 1 ft²
=
=
9.29 m², 1000 ft³
=
28.3 m³,
1 metric sabin
=
0.093 m², and 1 m²
1 English sabin =
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