Respuesta :
Answer: 5 m
Explanation:
We have the following data:
[tex]I_{1}=0.11 W/m^{2}[/tex] is the intensity of the sound at 7.5 m from the source
[tex]r_{1}=7.5 m[/tex] is the distance at which the intensity [tex]I_{1}[/tex] was measured
[tex]I_{2}=1 W/m^{2}[/tex] is the intensity of the sound at [tex]r_{2}[/tex] from the source
We have to find [tex]r_{2}[/tex]
Since the object is radiating the signal uniformly in all directions, we can use the Inverse Square Law for Intensity:
[tex]\frac{I_{1}}{I_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}[/tex]
Isolating [tex]r_{2}[/tex]:
[tex]r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}}}[/tex]
[tex]r_{2}=7.5 m\sqrt{\frac{0.11 W/m^{2}}{1 W/m^{2}}}[/tex]
[tex]r_{2}=2.48 m[/tex] This is the distance at which the intensity is the "threshold of pain"
Now, we have to substract this value to [tex]r_{1}[/tex] to find how much closer to the source can we move:
[tex]r_{1}-r_{2}=7.5 m - 2.48 m=5.02 m \approx 5 m[/tex]
Answer:
d = 5 m
Explanation:
Given:
- The initial distance from UFO r_1 = 7.5 m
- The initial intensity I_1 = 0.11 W / m^2
- The intensity of "threshold of pain" I_2 = 1.0 W / m^2
Find:
How much closer to the source can you move before the sound intensity reaches this threshold?
Solution:
- For waves that spread out in 3 dimensions , its intensity I is inversely proportional to square of the distance from the source. The expression is given as :
I = k / r^2
Where,
k: Proportionality constant
r: The distance from the source.
- Using the relation above the amount of distance r_2 from source that is required before I_1 --> I_2 is :
I_1 / I_2 = (r_2 / r_1)^2
Re- arrange to get r_2:
r_2 = r_1*sqrt(I_1 / I_2)
Plug in the given values:
r_2 = 7.5*sqrt(0.11/1)
r_2 = 2.5 m
- So the amount of distance from source is 2.5 m. So from initial position we have moved distance d:
d = 7.5 - 2.5
d = 5 m
