In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by

(1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern

is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit
In our problem,


while the distance between the first and the fifth minima is

(2)
If we use the formula to rewrite

, eq.(2) becomes

Which we can solve to find a, the width of the slit:
Centripetal force is given by F= mv²/r.
Given: m = 0.5 kg, v = 3 m/s, r = 0.5 m
Putting values,
F= mv²/r = 0.5× 3²/0.5 = 9 N
You'd get an extra 40/60 of the energy, or 2/3. Multiply 5/3 by the required energy to get the actual consumption.
Answer:
3125 N
Explanation:
diameter /2 =radius
so r1 =14cm , r2 =35cm
f1/A1 =f2/A2.
f2 = f1 × A2 / A1
=500×1225 pi cm² / 96 pi cm²
f2 =3125N