A barking dog delivers about 1.0 mW of power, which is assumed to be

  • Thread starter Thread starter chemistrymole
  • Start date Start date
  • Tags Tags
    Power
AI Thread Summary
A barking dog emits approximately 1.0 mW of power, which is uniformly distributed in all directions. To calculate the intensity level in decibels at a distance of 5.00 m, the power must be divided by the surface area of a sphere (A = 4πr²). The intensity (I) is determined by the formula I = P/A, where P is the power and A is the area. The discussion highlights the need to convert 1 mW to watts and emphasizes understanding the relationship between power, area, and intensity. Participants ultimately found clarity in the calculations and the connection between the variables.
chemistrymole
Messages
17
Reaction score
0

Homework Statement



A barking dog delivers about 1.0 mW of power, which is assumed to be uniformly distributed in all directions. What is the intensity level in decibels at a distance 5.00 m from the dog? The threshold of human hearing is 1.0 × 10-12 W/m2.

Homework Equations



I=P/A A is 4pi(r^2)

The Attempt at a Solution



The only part I am missing is converting the 1 mW to something else so I can plug in the equation above. I tried the 10*log(xxxxxxxx) formula no luck.
 
Physics news on Phys.org
chemistrymole said:

Homework Statement



A barking dog delivers about 1.0 mW of power, which is assumed to be uniformly distributed in all directions. What is the intensity level in decibels at a distance 5.00 m from the dog? The threshold of human hearing is 1.0 × 10-12 W/m2.

Homework Equations



I=P/A A is 4pi(r^2)

The Attempt at a Solution



The only part I am missing is converting the 1 mW to something else so I can plug in the equation above. I tried the 10*log(xxxxxxxx) formula no luck.

Homework Statement


Homework Equations


The Attempt at a Solution


The power all goes through each spherical surface surrounding the dog. If the sphere has a surface area of 1m^2, then the power flux density is 1.0mW/m^2. If the sphere is a little bigger, and has a surface area of 2m^2, then the power flux density 1.0mW/2m^2 = 0.5mW/m^2.

Does that help?
 
No still lost and confused.

Can someone please explain equations further I am not seeing how it connects to the variables given in the question.
 
chemistrymole said:
No still lost and confused.

Can someone please explain equations further I am not seeing how it connects to the variables given in the question.

As the sound waves of the bark radiate out from the dog, the 1mW of power (P) gets spread over a progressively larger area (A). At a distance of 5m, what area is it spread over?
 
haruspex said:
As the sound waves of the bark radiate out from the dog, the 1mW of power (P) gets spread over a progressively larger area (A). At a distance of 5m, what area is it spread over?

Do I have to convert 1 mW to W?

So would my setup be I = 1/pir^2?

I think my main formula template is I=P/A
 
Thanks everyone I found a solution that shows me how to do it. Now I actually understand it and know how the units work.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top