In a gear train with two gears, the gear ratio is defined as follows
where

is the angular velocity of the input gear while

is the angular velocity of the output gear.
This can be rewritten as a function of the number of teeth of the gears. In fact, the angular velocity of a gear is inversely proportional to the radius r of the gear:

But the radius is proportional to the number of teeth N of the gear. Therefore we can rewrite the gear ratio also as
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)


B - A theory seems to be the closest
Answer:
Find the set of value of x Find the set of value of x which satisfy the inequality 2r2- 5x 2 18which satisfy the inequality 2r2- Find the set of value of x which satisfy the inequality 2r2- 5x 2 185x 2 18
m = mass of the ice added = ?
M = mass of water = 1.90 kg
= specific heat of the water = 4186 J/(kg ⁰C)
= specific heat of the ice = 2000 J/(kg ⁰C)
= latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg
= initial temperature of ice = 0 ⁰C
= initial temperature of water = 79 ⁰C
T = final equilibrium temperature = 8 ⁰C
using conservation of heat
Heat gained by ice = Heat lost by water
m
(T -
) + m
= M
(
- T)
inserting the values
m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)
m = 1.53 kg