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GarryVolchara [31]
3 years ago
15

What is the wavelength of light falling on double slits separated by 2.00 μm if the third-order maximum is at an angle of 60.0∘?

Physics
1 answer:
Ksju [112]3 years ago
3 0

Answer:

λ = 5.773 x 10⁻⁷ m = 577.3 nm

Explanation:

In order to solve this problem we will use the grating equation:

mλ = d Sin θ

where,

m = order = 3

λ = wavelength of light = ?

d = slit separation = 2 μm = 2 x 10⁻⁶ m

θ = angle = 60°

Therefore,

(3)λ = (2 x 10⁻⁶ m)Sin 60°

λ = 1.732 x 10⁻⁶ m/3

<u>λ = 5.773 x 10⁻⁷ m = 577.3 nm</u>

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A student claims an object in motion must experience a force to stay in motion. Do you agree or disagree?
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Answer:

agree because there is always a force that causes motion..

4 0
3 years ago
A typical ten-pound car wheel has a moment of inertia of about 0.35kg⋅m2. The wheel rotates about the axle at a constant angular
QveST [7]

Answer:

The rotational kinetic energy of the rotating wheel is 529.09 J  

Explanation:

Given;

moment of inertia I = 0.35kg⋅m²

number of revolutions = 35.0

time of revolution, t = 4.00 s

Angular speed (in revolution per second), ω = 35/4 = 8.75 rev/s

Angular speed (in radian per second), ω = 8.75 rev/s x 2π = 54.985 rad/s

Rotational kinetic energy, K = ¹/₂Iω²

Rotational kinetic energy, K = ¹/₂ x 0.35 x (54.985)²

Rotational kinetic energy, K = 529.09 J  

Therefore, the rotational kinetic energy of the rotating wheel is 529.09 J  

7 0
3 years ago
1) When making a digital animation of a person running on a sidewalk in a scene, which parameter would be an initial condition?
lbvjy [14]

Answer:

The position of the person's feet. a running shoe company studying how different surfaces affect the life of a shoe tread

Explanation:

100%

4 0
3 years ago
What is cheetah speed and strength?
pav-90 [236]

Answer:

Speed: 109.4–120.7 km/h (68.0–75.0 mph)

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Explanation:

3 0
3 years ago
A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1
GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

\large{\maltese{\textsf{\underline{To find :-}}}}

The X and Y coordinates of the rocket relative of firing

⠀

⠀

\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s

⠀

⠀

\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}

⠀

⠀

<u>The</u><u> </u><u>horizontal</u><u> </u><u>range</u><u> </u><u>of</u><u> </u><u>projectile</u><u> </u><u>at</u><u> </u><u>x</u><u>.</u><u> </u>

⠀

\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}

⠀

⠀

\large\textsf{\underline{Now substituting the required values}}

⠀

⠀

\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}

⠀

⠀

The vertical position of projectile at y.

⠀

⠀

\sf \large  y = v_{yi} \times t -  (\frac{1}{2}  \times g  \times {t}^{2}) \\  \\   \sf  \large y = v_i \times  \cos \theta  \times t -  \frac{1}{2} g {t}^{2}

⠀

⠀

\textsf{ \large {\underline{Now substituting the required values}}  }

⠀

⠀

\sf y = 100 \times  \cos55{ \degree} \times 12 -  \frac{1}{2}   \times 9.80 \times  {12}^{2} \\  \\  \sf  y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\  \\  \sf y = 983.04 - 705.6 \\  \\  \underline{ \boxed{ \bold{y = 277.44m}}}

⠀

⠀

⠀

<h3><u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>distance</u><u> </u><u>at</u><u> </u><u>horizon</u><u> </u><u>is</u><u> </u><u>6</u><u>8</u><u>8</u><u>.</u><u>3</u><u>2</u><u>m</u><u> </u><u>and</u><u> </u><u>at</u><u> </u><u>vertical</u><u> </u><u>is</u><u> </u><u>2</u><u>7</u><u>7</u><u>.</u><u>4</u><u>4</u><u>m</u><u>.</u></h3>

8 0
2 years ago
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