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
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)

C. making fun of a peer because she is Asian
hope this helps
It’s B, the sum of protons and neutrons
Edwin Hubble discovered the cosmos , he discovered that there were other galaxies other than the Milky Way.
Answer: 0.02kg
Explanation:
The quantity of heat energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = 50 joules
Mass of iron sample = ? (let unknown value be Z)
C = 100 J/kg°C
Φ = 25°C
Then, Q = MCΦ
50J = Z x 100 J/kg°C x 25°C
50J = 2500J/kg x Z
Z = (50J/2500J/kg)
Z = 0.02kg
Thus, the mass of the iron sample is 0.02kg