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:
Answer:
1. v = 6.67 m/s
2. d = 9.54 m
Explanation:
1. To find the horizontal velocity of the rock we need to use the following equation:
<u>Where</u>:
d: is the distance traveled by the rock
t: is the time
The time can be calculated as follows:
<u>Where:</u>
g: is gravity = 9.8 m/s²
Now, the horizontal velocity of the rock is:
Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.
2. To calculate the distance at which the projectile will land, first, we need to find the time:

So, the distance is:
Therefore, the projectile will land at 9.54 m of the second cliff.
I hope it helps you!
Answer:
your mom and mark me brainlyist if I was right
Jsisisusuagahannananana this is worth 22 points g that’s crazy 25