Answer:
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Explanation:
D. dopamine
Answer:
D) θ₂= 36. 6º
Explanation:
In this diffraction experiment it is described by the equation
sin θ = m λ
The first dark strip occurs for m = 1 and since the angle is generally small we can approximate sine to the value of the angle
θ₁ = λ/ a
This equation is valid for linear slits, in the case of a circular slit the problem must be solved in polar coordinates, so the equation changes slightly
θ₂ = 1.22 λ / a
In the proposed exercise we start with a linear slit of width a, where tes1 = 30º and end with a circular slit of the same diameter
θ₂ = 1.22 λ / a
Let's clear (Lam/a) of equalizing the two equations
θ₁ = θ₂/ 1.22
θ₂ = 1.22 θ₁
θ₂ = 1.22 30
θ₂= 36. 6º
When reviewing the correct results is D
For a circular aperture, the first minima (n=1) as an angular separation from the peak of the central maxima given by
Sinθ = 1.22λ / d
Where,
d is the aperture or pupil diameter
d = 4.69 mm = 4.69 × 10^-3m
λ is the wavelength
λ = 545 nm = 545 × 10^-9 m
Then,
Sinθ = 1.22λ / d
Sinθ = 1.22 × 545 × 10^-9 / 4.69 × 10^-3
Sinθ = 1.418 × 10^-4 rad
Then, the head light sources have the same angular separation θ from the eye as the image have inside the eye.
For the headlight
Sinθ ≈ light separation / distantce for the eye
Light separation is give as x = 0.659 m
And let the distance of the eye be D
Then,
Sinθ = x / D
Make D subject of formula
D = x / Sinθ
D = 0.695 / 1.418 × 10^-4
D = 4902.316m
To km, 1km = 1000m
D ≈ 4.9 km
The force on the box is:
F = mgsin∅
If we multiply by this with the distance it traveled, we will know the work done by the box.
W = dmgsin∅
This work will be converted to elastic potential energy in the spring which is:
1/2 kx². Equating these and substituting values:
1/2 * 170 * x² = 4 * 13 * 9.81 * sin(30)
x = 1.73 m
The box's maximum speed will at the point right before contact with the spring, when the compression is 0.
