Physicscels gtfih (Easy question)

LiterallyNightwing

LiterallyNightwing

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The sound level from an engine is 102 dB from one meter away.

What intensity does the sound have at that distance?

I didn't read anything from this chapter and don't rly have time rn so any help is appreciated.
 
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To determine the sound intensity from the given sound level, we can use the formula:

I = (10^(L/10)) * I0

Where:
I is the sound intensity,
L is the sound level in decibels (dB), and
I0 is the reference sound intensity, which is typically set at 10^(-12) watts per square meter (W/m²).
 
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lmfao you're so lazy you can't just look up a simple equation in your textbook and plug in the values, so sad, imagine if you had to take a real physics course like stat mech - you would fail
 
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lmfao you're so lazy you can't just look up a simple equation in your textbook and plug in the values, so sad, imagine if you had to take a real physics course like stat mech - you would fail
I've been extremely busy the last couple weeks
 
lmfao you're so lazy you can't just look up a simple equation in your textbook and plug in the values, so sad, imagine if you had to take a real physics course like stat mech - you would fail
It took me 50+ pages on avg to finish a stat mech assignment. Fuck partition functions.
 
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The sound level from an engine is 102 dB from one meter away.

What intensity does the sound have at that distance?

I didn't read anything from this chapter and don't rly have time rn so any help is appreciated.
The sound intensity at a distance of one meter from the engine is approximately 158489.3192 watts per square meter (W/m²).
 
The sound intensity at a distance of one meter from the engine is approximately 158489.3192 watts per square meter (W/m²).
Gpt said its 0,01 but it says its the same from 30 m so i think its not that smart
 
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Gpt said its 0,01 but it says its the same from 30 m so i think its not that smart
considering the inverse square law, the sound intensity at a distance of 30 meters would be the same as at a distance of 1 meter if there are no other factors affecting the sound propagation. According to the inverse square law, the sound intensity remains constant when the distance is squared. Therefore, the sound intensity at 30 meters would also be approximately 1.584893192 x 10^22 watts per square meter (W/m²).
 
considering the inverse square law, the sound intensity at a distance of 30 meters would be the same as at a distance of 1 meter if there are no other factors affecting the sound propagation. According to the inverse square law, the sound intensity remains constant when the distance is squared. Therefore, the sound intensity at 30 meters would also be approximately 1.584893192 x 10^22 watts per square meter (W/m²).
I think i'm supposed to use P = A * I though bcuz thats a shit ton of watts
 
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I think i'm supposed to use P = A * I though bcuz thats a shit ton of watts
L = 102 dBI₀ = 10^(-12) W/m²

I = 10^((102 - 10 * log10(10^(-12))) / 10)= 10^((102 + 120) / 10)= 10^(222 / 10)≈ 1.584893192 x 10^22 W/m²
 

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