Answer:
a) 2.4 mm
b) 1.2 mm
c) 1.2 mm
Explanation:
To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:

a: width of the slit
λ: wavelength
m: order of the minimum
for little angles you have:

y: height of the mth minimum
a) the width of the central maximum is 2*y for m=1:

b) the width of first maximum is y2-y1:
![w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm](https://tex.z-dn.net/?f=w%3Dy_2-y_1%3D%5Cfrac%7B%28500%2A10%5E%7B-9%7Dm%29%281.2m%29%7D%7B0.50%2A10%5E%7B-3%7Dm%7D%5B2-1%5D%3D1.2mm)
c) and for the second maximum:
![w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm](https://tex.z-dn.net/?f=w%3Dy_3-y_2%3D%5Cfrac%7B%28500%2A10%5E%7B-9%7Dm%29%281.2m%29%7D%7B0.50%2A10%5E%7B-3%7Dm%7D%5B3-2%5D%3D1.2mm)
Electric energy into light energy and sound energy
The answer is A. Hope this helps!!!
Answer:
Explanation:
Since the equation for the illumination of an object, i.e. the brightness of the light, is <em>inversely proportional to the square of the distance from the light source</em>, the form of the function is:
Where x is the distance between the object and the light force, k is the constant of proportionality, and f(x) is the brightness.
Then, if you move halfway to the lamp the new distance is x/2 and the new brightness (call if F) is :

Then, you have found that the light is 4 times as bright as it originally was.
Answer:

Explanation:


If the sun considered as x=0 on the axis to put the center of the mass as a:

solve to r1


Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

where T = orbital period
then L'(x',y') = L(x) by conservation of angular momentum. So that means

Since
then
