The planet Mercury. Hope this helps
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)
Answer:
a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.
b) λ = c / f
Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted,
c) threshold energy
h f =Ф
Explanation:
It's photoelectric effect was fully explained by Einstein by the expression
Knox = h f - fi
Where K is the kinetic energy of the photoelectrons, f the frequency of the incident radiation and fi the work function of the metal
a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.
b) wavelength is related to frequency
λ = c / f
Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted, so there is a wavelength from which electrons cannot be removed from the metal.
c) As the work increases, more frequency radiation is needed to remove the electrons, because there is a threshold energy
h f =Ф
Average velocity = (x( 2.08 ) - x ( 0 )) / ( 2.08 s - 0 s )
x ( 2.08 ) = 1.42 * 2.08² - 0.05 * 2.08³ =
= 1.42 * 4.3264 - 0.443456 = 6.143484 - 0.443456 ≈ 5.7 m
v = ( 5.7 m - 0 m) / (2.08 s - 0 s ) = 5.7 / 2.08 m/s = 27.4 m/s
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