Answer:
The nodes and anti nodes would reverse roles.
Explanation:
I believe it has to do with the path differences. If waves are in phase, then the path differences are such that the waves reach the screen with crests superimposing crests and troughs superimposing troughs. This happens when the periods of each wave are equal or the paths themselves differ by a whole number multiple of the wavelength (λ, 2λ, 3λ, ...).
Now make these waves out of phase. Then half of the waves will travel half a wavelength farther than the rest. So the path difference will be 0.5λ, 1.5λ, 2.5λ, ....
Answer:
6.77 m/s
Explanation:
First, in the x direction:
Given:
Δx = 3.17 m
v₀ = v cos 30.8° = 0.859 v
a = 0 m/s²
Δx = v₀ t + ½ at²
(3.17 m) = (0.859 v) t + ½ (0 m/s²) t²
3.17 = 0.859 v t
3.69 = v t
Next, in the y direction:
Given:
Δy = 0.432 m
v₀ = v sin 30.8° = 0.512 v
a = -9.81 m/s²
Δy = v₀ t + ½ at²
(0.432 m) = (0.512 v) t + ½ (-9.81 m/s²) t²
0.432 = 0.512 v t − 4.905 t²
Two equations, two variables. Solve for t in the first equation and substitute into the second equation:
t = 3.69 / v
0.432 = 0.512 v (3.69 / v) − 4.905 (3.69 / v)²
0.432 = 1.89 − 66.8 / v²
66.8 / v² = 1.458
v² = 45.8
v = 6.77
Answer:
y = 128.0 km
Explanation:
The minimum separation of two objects is determined by Rayleygh's diffraction criterion, which establishes that two bodies are solved if the first minino of diffraction of one coincides with the central maximum of the second, with this criterion the diffraction equation remains
the diffraction equation for the first minimum is
a sin θ = λ
In the case of circular openings, the equation must be solved in polar coordinates, leaving the expression, we use the approximation that the sine of tea is very small.
θ = 1.22 λ / d
d = 15 cm
to find the distance we can use trigonometry
tan θ = y / L
tan θ = sin θ / cos θ = θ
substituting
y / L = λ / d
y = L λ /d
let's calculate
y = 384 10⁸ 500 10⁻⁹ / 0.15
y = 1.28 10⁵ m
Let's reduce to km
y = 1.28 10⁵ m (1km / 10³ m)
y = 128.0 km
the correct answer is 120 km away
Answer:
Part a)
Part b)
Part c)
Explanation:
Part a)
The height of the diving board is given as
now the speed of the diver is given as
when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board
So we will have
Part b)
plug in the values in the above equation
Part c)
Horizontal distance moved by the diver is given as
so the distance from the edge of the pool is given as
