A) 4.7 cm
The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

where
n is the order of the maximum
is the wavelength
is the distance between the slits
In this problem,
n = 5


So we find

And given the distance of the screen from the slits,

The distance of the 5th bright fringe from the central bright fringe will be given by

B) 8.1 cm
The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

And the distance of the 8th dark fringe from the central bright fringe will be given by

Answer:
The resulting, needed force for equilibrium is a reaction from a support, located at 2.57 meters from the heavy end. It is vertical, possitive (upwards) and 700 N.
Explanation:
This is a horizontal bar.
For transitional equilibrium, we just need a force opposed to its weight, thus vertical and possitive (ascendent). Its magnitude is the sum of the two weights, 400+300 = 700 N, since weight, as gravity is vertical and negative.
Now, the tricky part is the point of application, which involves rotational equilibrium. But this is quite simple if we write down an equation for dynamic momentum with respect to the heavy end (not the light end where the additional weight is placed). The condition is that the sum of momenta with respect to this (any) point of the solid bar is zero:

Where momenta from weights are possitive and the opposed force creates an oppossed momentum, then a negative term. Solving our unknown d:

So, the resulting force is a reaction from a support, located at 2.57 meters from the heavy end (the one opposed to the added weight end).
Brownian motion<span> or pedesis is the </span>random motion<span> of particles suspended in a fluid </span>
Answer:

Explanation:
= Actual wavelength = 
= Relative permittivity = 1.44
The observed wavelength in the glass is given by

The wavelength lies in the range of green light.
Hence, the observed color of light is 
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