Solving Intensity and Power Homework: 100 m, 126 mW

So your answer of 0.126 watts is actually 126 milliwatts.In summary, a spherical sinusoidal source radiates sound uniformly in all directions with a sound intensity level of 1.00 × 10−4 W/m2 at a distance of 10.0m. At a distance of 100m, the sound intensity level is 1.00 × 10−6 W/m2. The power radiated by this source is 126 milliwatts.
  • #1
sonutulsiani
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Homework Statement



A spherical sinusoidal source radiates sound uniformly in all directions. At a distance 10.0m the sound intensity level is 1.00 × 10−4 W/m2.

1. At what distance from the source is the intensity 1.00 × 10−6 W/m2?
A. 1 m
B. 10 m
C. 100 m
D. 1 km


2. What is the power radiated by this source?
A. Pav = 208 mW
B. Pav = 126 mW
C. Pav = 332 mW
D. Pav = 412 mW



Homework Equations





The Attempt at a Solution



I got 1 as 100 m and 2 as 126 watt, but I am not sure about the 2nd answer. So please help me. Thanks
 
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  • #2
Both answers are correct. Why did you doubt your second answer?
 
  • #3
I doubt the 2nd answer because I get 0.126 watt, but the answer is in megawatt
 
  • #4
Did you get what i amsaying
 
  • #5
The answers are in milliwatts, not megawatts.
 

1. How do I calculate intensity and power in this scenario?

To calculate the intensity, divide the power (126 mW) by the area (100 m). The resulting unit will be in mW/m^2. To calculate the power, multiply the intensity by the area.

2. What are the units for intensity and power?

The unit for intensity is mW/m^2 and the unit for power is mW.

3. How do I convert mW to W?

To convert from mW to W, divide the value by 1000. In this case, the power would be 0.126 W.

4. What is the relationship between intensity and power?

Intensity and power are directly proportional to each other. This means that as the intensity increases, the power also increases and vice versa.

5. How can I use this information in real-life applications?

The calculation of intensity and power is important in many fields such as optics, physics, and engineering. It can be used to determine the amount of energy in a given area, which is useful in designing and optimizing devices such as lasers and solar panels.

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