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Liono4ka [1.6K]
3 years ago
14

In Part 6.2.2, you will determine the wavelength of the laser by shining the laser beam on a "diffraction grating", a set of reg

ularly spaced lines. Suppose the pattern is displayed on a screen a distance L from the grating and the spots are separated by s. If the screen is 11 m away, the spots are 4.5 cm apart, and the lines of the grating are separated by 0.5 mm, what is λ (mm)?
Physics
1 answer:
harkovskaia [24]3 years ago
3 0

Answer:

λ = 2042 nm

Explanation:

given data

screen distance d = 11 m

spot s = 4.5 cm = 4.5 ×10^{-2} m

separation L = 0.5 mm = 0.5 ×10^{-3} m

to find out

what is λ

solution

we will find first angle between first max and central bright

that is tan θ = s/d

tan θ = 4.5 ×10^{-2}  / 11

θ = 0.234

and we know diffraction grating for max

L sinθ  = mλ

here we know m = 1  so put all value and find λ

L sinθ  = mλ

0.5 ×10^{-3}  sin(0.234)  = 1 λ

λ = 2042.02 ×10^{-9}  m

λ = 2042 nm

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a student holds a 3.0kg mass in each hand while sitting on a rotating stool. when his arms are extended horizontally, the masses
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Answer:

Explanation:

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Initial moment of inertia I₁ = M R² = 3  x 1 ² = 3 kg m²

Final moment of inertia I₂ = M R² = 3  x .3 ² = 0.27  kg m²

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Putting the values ,

3 x .75 = .27 x ω₂

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8 0
3 years ago
Does the distance a person kicks a soccer ball
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Answer:

Independent variable: how far the soccer ball is kicked.

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The distance the ball is kicked is what the scientist can change. On the other hand, in this experiment, how well the person does on their test supposedly relies on how far the soccer ball is kicked.

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3 years ago
Physics Question:<br> (ignore the part crossed in red) 14459688
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3 years ago
The intensity of a 70-dB whistle is quintupled. What is the new decibel level
hoa [83]

Multiplying the power of any signal by 5 can be described as
an increase of  6.99 dB .

If the whistle blew at 70 dB initially, and its sound power became
multiplied by 5, and the whistle and the listener both stayed in
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7 0
4 years ago
Consider a stone in free fall on a planet with gravitational acceleration 3.4 m/s^2. Suppose you would like the stone to experie
Stella [2.4K]

Answer:

Angle of incline is 20.2978°

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Gravitational acceleration on a planet a = 3.4 m/s²

Gravitational acceleration on Earth g = 9.8 m/s²

Angle of incline = ∅

Mass of the stone = m

Force on the stone along the incline will be;

F = mgSin∅

F = ma

The stone has the same acceleration as that of the gravitational acceleration on the planet.

so

ma = mgSin∅

a = gSin∅

Sin∅ = a / g

we substitute

Sin∅ = (3.4 m/s²) / (9.8 m/s²)

Sin∅ = 0.3469

∅ = Sin⁻¹( 0.3469 )

∅ = 20.2978°

Therefore, Angle of incline is 20.2978°

8 0
3 years ago
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