Is any force exerted on the vertical sides of the loop that you used in the experiment and how does it affect the apparent mass?
1 answer:
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Answer:
Hz
Explanation:
We know that
1 cm = 0.01 m
= Length of the human ear canal = 2.5 cm = 0.025 m
= Speed of sound = 340 ms⁻¹
= First resonant frequency
The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as
for first resonant frequency, we have n = 1
Inserting the values
Hz
Answer:
r₂ = 0.316 m
Explanation:
The sound level is expressed in decibels, therefore let's find the intensity for the new location
β = 10 log
let's write this expression for our case
β₁ = 10 log \frac{I_1}{I_o}
β₂ = 10 log \frac{I_2}{I_o}
β₂ -β₁ = 10 ( )
β₂ - β₁ = 10
log \frac{I_2}{I_1} = = 3
= 10³
I₂ = 10³ I₁
having the relationship between the intensities, we can use the definition of intensity which is the power per unit area
I = P / A
P = I A
the area is of a sphere
A = 4π r²
the power of the sound does not change, so we can write it for the two points
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
we substitute the ratio of intensities
I₁ r₁² = (10³ I₁ ) r₂²
r₁² = 10³ r₂²
r₂ = r₁ / √10³
we calculate
r₂ =
r₂ = 0.316 m
Answer:
Explanation:
As we know the , equation of time period for simple pendulum ,
T = 2*pi*
hence putting values we get ,
the solution is in picture ,
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