Question

1 week streak level 4 1/4 skills normal conversation 40 db 0.002 pa washing machine 50 db 0.00632 pa television (tv) 60 db 0.02 pa handheld electric mixer 70 db 0.0632 pa vacuum cleaner 80 db 0.2 pa lawnmower 90 db 0.632 pa jackhammer 100 db 2 pa the data points were taken to show an example of the relationship between sound level, measured in decibels (db); and sound pressure, measured in pascals (pa). what sound pressure (in pa) corresponds to 110 db?

Explanation:

Step1: Analyze the pattern

Looking at the given data:

• When \(dB = 40\), \(Pa=0.002\)
• When \(dB = 50\), \(Pa = 0.00632\)
• When \(dB=60\), \(Pa = 0.02\)
• When \(dB = 70\), \(Pa=0.0632\)
• When \(dB=80\), \(Pa = 0.2\)
• When \(dB=90\), \(Pa=0.632\)
• When \(dB = 100\), \(Pa=2\)

We can observe that for each increase of \(10\) \(dB\), the sound pressure \(Pa\) is multiplied by approximately \(3.16\) (since \(\frac{0.00632}{0.002}=3.16\), \(\frac{0.02}{0.00632}\approx3.16\), \(\frac{0.0632}{0.02}=3.16\), \(\frac{0.2}{0.0632}\approx3.16\), \(\frac{0.632}{0.2}=3.16\), \(\frac{2}{0.632}\approx3.16\))

Step2: Calculate for \(110\) \(dB\)

Since for \(100\) \(dB\) the pressure is \(2\) \(Pa\), and we need to find for \(110\) \(dB\) (an increase of \(10\) \(dB\) from \(100\) \(dB\))
Multiply the pressure at \(100\) \(dB\) by \(3.16\)
\(2\times3.16 = 6.32\) \(Pa\)

Answer:

\(6.32\) \(Pa\)