The decibel level of a sound is defined to be $100 (7) B= 10 log where I is the intensity of the sound and the constant / -1.0 x 10-12 W/m2 is the reference intensi Solving this equation for the sound intensity gives 1-1(108/10). At r= 10 km = 1.0 x 104 m from the train horn, the sound level is 51 dB. Therefore, the intensity of t - (1.0 x 10-12 W/m²)10 (1.0 x 10-12 W/m²) 10 1.26 ✔ 1.26*10-7W/m². 1.26 Step 2 The average power generated by the horn is then given by P = IA 1(4nr²) 51/10 × 10-7 W/m²) [4 * * 10² W. Enter a number. 104 m)²]
The decibel level of a sound is defined to be $100 (7) B= 10 log where I is the intensity of the sound and the constant / -1.0 x 10-12 W/m2 is the reference intensi Solving this equation for the sound intensity gives 1-1(108/10). At r= 10 km = 1.0 x 104 m from the train horn, the sound level is 51 dB. Therefore, the intensity of t - (1.0 x 10-12 W/m²)10 (1.0 x 10-12 W/m²) 10 1.26 ✔ 1.26*10-7W/m². 1.26 Step 2 The average power generated by the horn is then given by P = IA 1(4nr²) 51/10 × 10-7 W/m²) [4 * * 10² W. Enter a number. 104 m)²]
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The decibel level of a sound is defined to be $100 (7) B= 10 log where I is the intensity of the sound and the constant / -1.0 x 10-12 W/m2 is the reference intensi Solving this equation for the sound intensity gives 1-1(108/10). At r= 10 km = 1.0 x 104 m from the train horn, the sound level is 51 dB. Therefore, the intensity of t - (1.0 x 10-12 W/m²)10 (1.0 x 10-12 W/m²) 10 1.26 ✔ 1.26*10-7W/m². 1.26 Step 2 The average power generated by the horn is then given by P = IA 1(4nr²) 51/10 × 10-7 W/m²) [4 * * 10² W. Enter a number. 104 m)²]
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