In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
![y_n= \frac{n \lambda D}{a}](https://tex.z-dn.net/?f=y_n%3D%20%5Cfrac%7Bn%20%5Clambda%20D%7D%7Ba%7D)
(1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
![\lambda](https://tex.z-dn.net/?f=%5Clambda)
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,
![D=37.0 cm=0.37 m](https://tex.z-dn.net/?f=D%3D37.0%20cm%3D0.37%20m)
![\lambda=530 nm=5.3 \cdot 10^{-7} m](https://tex.z-dn.net/?f=%5Clambda%3D530%20nm%3D5.3%20%5Ccdot%2010%5E%7B-7%7D%20m)
while the distance between the first and the fifth minima is
![y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m](https://tex.z-dn.net/?f=y_5-y_1%20%3D%200.500%20mm%3D0.5%20%5Ccdot%2010%5E%7B-3%7D%20m)
(2)
If we use the formula to rewrite
![y_5, y_1](https://tex.z-dn.net/?f=y_5%2C%20y_1)
, eq.(2) becomes
![\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%20%5Clambda%20D%7D%7Ba%7D%20-%20%5Cfrac%7B1%20%5Clambda%20D%7D%7Ba%7D%20%3D%5Cfrac%7B4%20%5Clambda%20D%7D%7Ba%7D%3D%200.5%20%5Ccdot%2010%5E%7B-3%7D%20m%20%20)
Which we can solve to find a, the width of the slit:
I believe the answer is B.
Answer:
the initial velocity is 20 m/s and the acceleration is 2 m/s²
Explanation:
Given equation of motion, v = 20 + 2t
If V represents the final velocity of the object, then the initial velocity and acceleration of the object is calculated as follows;
From first kinematic equation;
v = u + at
where;
v is the final velocity
u is the initial velocity
a is the acceleration
t is time of motion
If we compare (v = u + at) to (v = 20 + 2t)
then, u = 20 and
a = 2
Therefore, the initial velocity is 20 m/s and the acceleration is 2 m/s²
Assuming Adam is on earth g= 9.8 m/s and m= weight/ gravity = 667/9.8 = 68 kg