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weqwewe [10]
3 years ago
15

The sound level measured in a room by a person watching a movie on a home theater system varies from 40 dB during a quiet part t

o 80 dB during a loud part. Approximately how many times louder is the latter sound
Physics
1 answer:
UNO [17]3 years ago
7 0

Answer:

\alpha=-3.01dB

Explanation:

From the question we are told that:

Sound level intensity

 \triangle I=40dB-80dB

Generally the equation for  intensity level  is mathematically given by

 \alpha=10log_{10}(I/I_x)dB

Where

 I= Intensity measured

 I_x=Threshold\ of\ audibility

 I_x= 10-12 W / m2

 \alpha= 10 log10 \frac{I_1}{I_x} - 10 log10 \frac{}I_2{I_x}

 \alpha= 10 log10 \frac{I_1}{I_2}

 \alpha=10 log10\frac{40}{80}

 \alpha=-3.01dB

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Answer:

distance difference would a) increase

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Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.

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This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.

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