A microphone has an area of 3.8 cm2. It receives during a 2.87 s time period a sound energy of 4.2x10-11 J. What is the intensity of the sound?
(Sound Intensity) = (Energy)/[(Area)*(Time)]
It can be converted to decibels using a logarithmic formula, if that is what they want.
How would I do that?
Do the indicated division first, to get 3.85*10^-8 W/m^2
Zero dB corresponds to 10^-12 W/m^2
Each factor of ten in intensity above the zero dB reference level corresponds to a 10 dB increase.
I get 46 dB for your case
See http://www.animations.physics.unsw.edu.au/jw/dB.htm
for a better explanation
To find the intensity of the sound, we can use the formula:
Intensity = Power / Area
Where Intensity is measured in watts per square meter (W/m^2), Power is the energy received per second (in watts), and Area is the area over which the sound is spread (in square meters).
First, let's convert the given sound energy to power (watts). Since the time period is given as 2.87 s, we can use the formula:
Power = Energy / Time
Substituting the given values:
Power = 4.2x10^(-11) J / 2.87 s
Now, we can solve for power:
Power = 1.462 x 10^(-11) W
Next, we need to convert the given microphone area to square meters. Since the area is given as 3.8 cm^2, we can convert it to square meters by dividing by 10000 (since there are 10000 square centimeters in a square meter):
Area = 3.8 cm^2 / 10000
Area = 3.8 x 10^(-4) m^2
Now, we can substitute the values of power and area into the formula to find the intensity:
Intensity = 1.462 x 10^(-11) W / 3.8 x 10^(-4) m^2
Calculating the intensity:
Intensity = 3.847 x 10^(-7) W/m^2
Therefore, the intensity of the sound is approximately 3.847 x 10^(-7) W/m^2.