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GenaCL600 [577]
2 years ago
6

What is the approximate wavelength of a light whose first-order bright band forms a diffraction angle of 45.0° when it passes th

rough a diffraction grating that has 500.0 lines per mm?
236 nm
353 nm
943 nm
1414 nm
Physics
2 answers:
max2010maxim [7]2 years ago
7 0

Answer:

its D

Explanation:

edg2020

Bond [772]2 years ago
6 0

Answer:

D) 1414 nm

Explanation:

This is correct on Edge.

Using the equation dsin(angle)=n(wavelength), we can solve for wavelength.

First we must convert the 500 lines per mm to nm. We do this by 1/500, giving you 0.002. Then move the decimal over six places to the right, resulting in 2000.

Then by plugging in the other values, we have 2000sin(45)=1(wavelength).

N is one because we are just solving for a first-order band.

So 2000sin(45)=wavelength

By using a calculator, we can see that the wavelength equals approximately 1414nm.

I hope this helped. If it did, I would really appreciate a Brainliest!!

Have a great day:)

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apple, earth. earth more massive than moon. astronauts showed this - armstrong etc

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Bowen’s reaction series illustrates relations between:
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C. Temperature, chemical composition and mineral structure

Explanation:

The Bowen's reaction series illustrates the relationship between temperature, chemical composition and mineral structure.

The series is made up of a continuous and discontinuous end through which magmatic composition can be understood as temperature changes.

  • The left part is the discontinuous end while the right side is the continuous series.
  • From the series, we understand that a magmatic body becomes felsic as it begins to cool to lower temperature.
  • A magma at high temperature is ultramafic and very rich in ferro-magnesian silicates which are the chief mineral composition of olivine and pyroxene. These minerals are predominantly found in mafic- ultramafic rocks. Also, we expect to find the calcic-plagioclase at high temperatures partitioned in the magma.
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5 0
3 years ago
Is there a "real" simple machine that has an efficiency of 100%?
VARVARA [1.3K]

Answer:

NO

Explanation:

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8 0
2 years ago
A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

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Answer:

1.0s

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