Question
A concert loudspeaker suspended high off the ground emits 26.0 W of sound power. A small microphone with a 1.00 cm area is 50.0 m from the speaker:CorrectPart BHow much sound energy impinges on the microphone each second? Express your answer with the appropriate units_HA150dBSubmitPrevious Answers Request AnswerIncorrect; Try Again; 6 attempts remaining
A concert loudspeaker suspended high off the ground emits 26.0 W of sound power. A small microphone with a 1.00 cm area is 50.0 m from the speaker: Correct Part B How much sound energy impinges on the microphone each second? Express your answer with the appropriate units_ HA 150 dB Submit Previous Answers Request Answer Incorrect; Try Again; 6 attempts remaining
Answers
A concert loudspeaker suspended high above the ground emits 35 W of sound power. A small microphone with a $1.0 \mathrm{cm}^{2}$ area is $50 \mathrm{m}$ from the speaker. a. What is the sound intensity at the position of the microphone? b. How much sound energy impinges on the microphone each second?
This question you have this set up microphone is 50 m away from the loudspeaker in a concert. The power of the loudspeaker is 35. What? We need to find, uh, some intensity and some intensity level at the position off the microphone. Okay, so in part A, we need to find a sound intensity. So you're using. I equals two p, thereby four pi R square. So the intensity with some intensity is equal to 35. Divided by Opie 10 50 square. So we calculate these, you get 1.1 times 10 to the ministry. What's for square? Yeah. So this is the answer for part A. And then be we need to find a sound intensity level. You'll be using the formula. It goes to 10 times long base 10. I have a I know so sound intensity level close to 10 times longer. 10. 1. 21 times 10 to the ministry Divide by one times 10 to the minus. Trump. So this is equal to 90. That's the books. Okay, so this is the answer for party, and that's all
Hi here in this game problem. Sound power emitted by the loudy speaker that is given us 35 bodies. So in the first part of the problem we are the dispense of the microphone that is given as 50 m. So intensity of sound at that point will be i is equal to however, per unit area. So to find this intensity, we consider a sphere surrounding a source of sound and having a radius equal to the distance of observation point. So this area will be the area of surface area of that hypothetical sphere. So this intensity of sound will be given by 35 What divided by four bye or pie. This is 3.14 into square off distance which is 50 m square. So finally this intensity here comes out to be 1.11 into 10 raised to the power minus three what four m square. Which is the answer for the first part of the problem? No. In the second part of the problem we have to find sound intensity level at that point, at the point of at the position of microphone. So that sound intensity level will be given by beta is equal to 10 times off log of the ratio of sound intensity at that point. With the reference sound intensity there, this reference sound intensity is one in 2, 10 for minus 12 what per meter squared. So here this vita will be 10 times of log of 1.11 into 10 days, par minus three. What perimeter is Squire divided by one in 2, 10 par minus 12. What dormitory square. So canceling this war, permit every square we get better is equal to 10 times off. Look at them off 1.11 into. And this two part nine. Now using product rule over the located them here it becomes. Look at them off 1.11 plus nine times off. Look carried them of 10 at best 10 which will come out to be one means it will become nine into one. So finally this is data is equal to 10 times off. Well the value of this log 1.11 will come out to be 0.45 plus nine. So finally this is 90.4 five. Their c bell sound intensity level at that point at the position of microphone, which is the answer for the second part of the problem. Thank you.
Yes, this question covered the concept of problem. So for part A power is intense it into the area and force with spherical distribution area is Opie artist square or we can break the Nigerian voters intensities. 6.5. What problem test square into four. Hi. And the radius is 2.5. You need us square Or the Nigeria. It is 511. No party. The intensity is the power upon food by are square. And from this, the intensity is the power here calculated as 511 votes form for pine and 37 m square. All the intensity in the space is 0.8-9 what he does now in part C. The energy is saved by the world. For secondaries is The power that is 500 levin. Work into the time. There wasn't one second other energies. 511 tools