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:
2 N
Explanation:
From the question, it's given that
Mass m = 0.2 kg
Acceleration a = 10 m/s^2
The force a soccer goalie experience when stopping a ball will be equal to the force at which the ball is being kicked. This is
F = ma
Substitute all the parameters into the formula
F = 0.2 × 10
F = 2 Newton.
Answer:
Explanation:
Due to first charge , electric field at origin will be oriented towards - ve of y axis.
magnitude
Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j
= - 31.6 j N/C
Due to second charge electric field at origin
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.2²+ .6²
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.8
= 18 N/C
It is making angle θ where
Tanθ = .6 / 1.2
= 26.55°
this field in vector form
= - 18 cos 26.55 i - 18 sin26.55 j
= - 16.10 i - 8.04 j
Total field
= - 16.10 i - 8.04 j + ( - 31.6 j )
= -16.1 i - 39.64 j .
Ex = - 16.1 i
Ey = - 39.64 j .