Answer: (a) Z-score are 1 and -1.2 for northern and southern regions, respectively.
Explanation: <u>Z-score</u> is how many standard deviations a data is from the population mean or how far a data point is from the mean.
The z-score is calculated by the following:
where
x is the data point
μ is population mean
σ is standard deviation
For the <u>northern</u> <u>region</u> birds:
μ = 10, σ = 3, x = 13
z = 1
The z-score for birds living in the northern region is 1, which means it is 1 standard deviation <em>above the mean</em>.
For the southern region:
μ = 16, σ = 2.5, x = 13
z = -1.2
The z-score for southern living birds is -1.2, meaning it is 1.2 standard deviations <em>below the mean</em>.
The impulse (the variation of momentum of the ball) is related to the force applied by
where
is the variation of momentum, F is the intensity of the force and
is the time of application of the force.
Using F=1000 N and
, we can find the variation of momentum:
This
can be rewritten as
where
and
are the final and initial momentum. But the ball is initially at rest, so the initial momentum is zero, and
from which we find the final velocity of the ball:
Answer:
The intensity of light from the 1mm from the central maximu is 
Explanation:
From the question we are told that
The wavelength is 
The width of the slit is 
The distance from the screen is 
The intensity at the central maximum is 
The distance from the central maximum is 
Let z be the the distance of a point with intensity I from central maximum
Then we can represent this intensity as
![I = I_o [\frac{sin [\frac{\pi * w * sin (\theta )}{\lambda} ]}{\frac{\pi * w * sin (\theta )}{\lambda } } ]^2](https://tex.z-dn.net/?f=I%20%3D%20I_o%20%5B%5Cfrac%7Bsin%20%5B%5Cfrac%7B%5Cpi%20%2A%20w%20%2A%20sin%20%28%5Ctheta%20%29%7D%7B%5Clambda%7D%20%5D%7D%7B%5Cfrac%7B%5Cpi%20%2A%20w%20%2A%20sin%20%28%5Ctheta%20%29%7D%7B%5Clambda%20%7D%20%7D%20%5D%5E2)
Now the relationship between D and z can be represented using the SOHCAHTOA rule i.e
if the angle between the the light at z and the central maximum is small
Then 
Which implies that
substituting this into the equation for the intensity
![I = I_o [\frac{sin [\frac{\pi w}{\lambda} \cdot \frac{z}{D} ]}{\frac{\pi w z}{\lambda D\frac{x}{y} } } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20%5B%5Cfrac%7Bsin%20%5B%5Cfrac%7B%5Cpi%20w%7D%7B%5Clambda%7D%20%5Ccdot%20%5Cfrac%7Bz%7D%7BD%7D%20%20%5D%7D%7B%5Cfrac%7B%5Cpi%20w%20z%7D%7B%5Clambda%20D%5Cfrac%7Bx%7D%7By%7D%20%7D%20%7D%20%5D)
given that 
We have that
![I = I_o [\frac{sin[\frac{3.142 * 0.45*10^{-3}}{(620 *10^{-9})} \cdot \frac{1*10^{-3}}{3} ]}{\frac{3.142 * 0.45*10^{-3}*1*10^{-3} }{620*10^{-9} *3} } ]^2](https://tex.z-dn.net/?f=I%20%3D%20I_o%20%5B%5Cfrac%7Bsin%5B%5Cfrac%7B3.142%20%2A%200.45%2A10%5E%7B-3%7D%7D%7B%28620%20%2A10%5E%7B-9%7D%29%7D%20%5Ccdot%20%5Cfrac%7B1%2A10%5E%7B-3%7D%7D%7B3%7D%20%5D%7D%7B%5Cfrac%7B3.142%20%2A%200.45%2A10%5E%7B-3%7D%2A1%2A10%5E%7B-3%7D%20%7D%7B620%2A10%5E%7B-9%7D%20%2A3%7D%20%7D%20%5D%5E2)
![=I_o [\frac{sin(0.760)}{0.760}] ^2](https://tex.z-dn.net/?f=%3DI_o%20%5B%5Cfrac%7Bsin%280.760%29%7D%7B0.760%7D%5D%20%5E2)
