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:
Answer:
Explanation:
b) Gravity reduces the initial upward velocity to zero in a time of
t = v/g = 40/10 = 4 s
a) h = v₀t + ½gt² = 40(4) + ½(-10)4² = 80 m
or
v² = u² + 2as
h = (0² - 40²) / 2(-10) = 80 m
Answer:
4.45×10¯¹¹ N
Explanation:
From the question given above, the following data were obtained:
Mass of ball (M₁) = 4 Kg
Mass of bowling pin (M₂) = 1.5 Kg
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Distance apart (r) = 3 m
Force of attraction (F) =?
The force of attraction between the ball and the bowling pin can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 4 × 1.5 / 3²
F = 4.002×10¯¹⁰ / 9
F = 4.45×10¯¹¹ N
Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N
All of the elements in a period have the same number of atomic orbitals. For example, every element in the top row (the first period) has one orbital for its electrons. All of the elements in the second row (the second period) have two orbitals for their electrons. As you move down the table, every row adds an orbital.