Answer:
The sound level of the 26 geese is 
Explanation:
From the question we are told that
The sound level is 
The number of geese is 
Generally the intensity level of sound is mathematically represented as
The intensity of sound level in dB for one goose is mathematically represented as
![Z_1 = 10 log [\frac{I}{I_O} ]](https://tex.z-dn.net/?f=Z_1%20%3D%2010%20log%20%5B%5Cfrac%7BI%7D%7BI_O%7D%20%5D)
Where I_o is the threshold level of intensity with value 
is the intensity for one goose in 
For 26 geese the intensity would be
Then the intensity of 26 geese in dB is
![Z_{26} = 10 log[\frac{26 I }{I_o} ]](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%5B%5Cfrac%7B26%20I%20%7D%7BI_o%7D%20%5D)
![Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%20%28%5C%20%5C%2026%20%2A%20%20%5B%5Cfrac%7B%20I%20%7D%7BI_o%7D%20%5D%5C%20%5C%20%29)
![Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%20%28%5C%20%5C%2026%20%20%5C%20%5C%20%29%20%2A%20%20%20%28%5C%20%5C%20%2010%20log%20%5B%5Cfrac%7B%20I%20%7D%7BI_o%7D%20%5D%5C%20%5C%20%29)
From the law of logarithm we have that
![Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%2026%20%2B%20%2010%20log%20%5B%5Cfrac%7BI%7D%7BI_0%7D%20%5D)
Primary Circular Reactions (1-4 months): This substage involves coordinating sensation and new schemas. For example, a child may suck his or her thumb by accident and then later intentionally repeat the action. These actions are repeated because the infant finds them pleasurable.
The Kelvin scale has no negatives on it.
Zero Kelvin is 'Absolute Zero', and nothing can get colder than that.
Answer:
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
Explanation:
For this exercise let's use the electric field expression
E = k q / r²
where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee
let's calculate the field for each charge
Q = 24 pC = 24 10⁻¹² C
E₁ = 9 10⁹ 24 10⁻¹² / 0.20²
E₁ = 5.4 N / C
Q = 32 pC = 32 10⁻¹² C
E₂ = 9 10⁹ 32 10⁻¹² / 0.2²
E₂ = 7.2 N / C
let's find the difference between these two fields
ΔE = E₂ -E₁
ΔE = 7.2 - 5.4
ΔE = 1.8 N / C
the minimum detection field is
E_minimum = 0.77 N / C
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
