Two light waves will interfere constructively if the path-length difference between them is a whole number.
<h3>
SUPERPOSITION</h3>
The principle of superposition state that, when two or more waves meet at a point, the resultant displacement at that point is equal to the sum of the displacements of the individual waves at that point.
Interference of waves can either be constructive, or destructive.
The two light waves, initially emitted in phase, will interfere constructively with maximum amplitude if the path-length difference between them is a whole number of wavelenght 1λ, 2λ, 3λ, 4λ etc
The equivalent phase differences between the waves will be 2
or 360 degrees, 4 or 720 degrees, 6 1080 degrees etc
Therefore, the two light waves, initially emitted in phase, will interfere constructively with maximum amplitude if the path-length difference between them is a whole number.
Learn more about Interference here: brainly.com/question/25310724
Answer:
Explanation:
Given data
Length of tube L=0.632 m
Speed of sound v=344 m/s
To find
Fundamental frequency f
Solution
The fundamental frequency of the tube can be given as:
Answer:
<em>Answer: positive velocity & negative acceleration</em>
Explanation:
<u>Accelerated Motion</u>
Both the velocity and acceleration are vectors because they have magnitude and direction. When the motion is restricted to one dimension, i.e. left-right or up-down, the direction is marked with the sign according to some preset reference.
The locomotive is moving at a certain speed with a (so far) unknown sign but the acceleration has a negative sign. Since the locomotive comes to a complete stop it means the velocity and the acceleration are of opposite signs.
Thus the velocity is positive.
Answer: positive velocity & negative acceleration
Answer:
a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s
Explanation:
It is angular momentum given by
L = r x p
Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum
One of the easiest ways to make this vector product is with the use of determinants
![{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]](https://tex.z-dn.net/?f=%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%26z%5C%5Cpx%26py%26pz%5Cend%7Barray%7D%5Cright%5D)
Let's apply this relationship to our case
Let's start by breaking down the speed
v₀ₓ = v₀ cosn 45
voy =v₀ sin 45
v₀ₓ = 9 cos 45
voy = 9 without 45
v₀ₓ = 6.36 m / s
voy = 6.36 m / s
a) at launch point r = 0 whereby L = 0
. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero
vfy² = voy²- 2 g y
y = voy² / 2g
y = (6.36)²/2 9.8
y = 2.06 m
Let's calculate the angular momentum
L= ![\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%260%5C%5Cpx%260%260%5Cend%7Barray%7D%5Cright%5D)
L = -px y k ^
L = - (m vox) (2.06) k ^
L = - 20 6.36 2.06 k ^
L = 262 k ^ Kg m² / s
The angular momentum is on the z axis
c) At the point of impact, at this point the height is zero and the position on the x-axis is the range
R = vo² sin 2θ / g
R = 9² sin (2 45) /9.8
R = 8.26 m
L = ![\left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%260%260%5C%5Cpx%26py%260%5Cend%7Barray%7D%5Cright%5D)
L = - x py k ^
L = - x m voy
L = - 8.26 20 6.36 k ^
L = 1020.7 k^ kg m² /s
Answer:
1.03
Explanation:
0.02 has two significant decimal figures, so start by shortening 1.0090 to 2 decimal places (3 significant figures). Round up the decimal to become 1.1 and add .02 to get 1.3.
