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gizmo_the_mogwai [7]
3 years ago
7

A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glo

w, sending the light through a diffraction grating, and measuring the positions of first-order spectral lines on a detector 15.0 cm behind the grating. Unfortunately, she has lost the card that gives the specifications of the grating. Fortunately, she has a known compound that she can use to calibrate the grating. She heats the known compound, which emits light at a wavelength of 501 nm, and observes a spectral line 9.95 cm from the center of the diffraction pattern. PART A:
What is the wavelength emitted by compound A that have spectral line detected at position 8.55 cm?


PART B:


What is the wavelength emitted by compound B that have spectral line detected at position and 12.15 cm?
Physics
1 answer:
padilas [110]3 years ago
4 0

Answer:

 a)    λ = 189.43 10⁻⁹ m  b)    λ = 269.19 10⁻⁹ m

Explanation:

The diffraction network is described by the expression

      d sin θ= m λ

Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

        tan θ = y / L

The diffraction spectrum is measured at very small angles, therefore

      tan θ = sin θ / cos θ = sin θ

We replace

      d y / L = m λ

Let's place in the first order m = 1

Let's look for the separation of the lines (d)

     d = λ  L / y

     d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²

     d = 332.33 10⁻⁹ m

Now we can look for the wavelength of the other line

     λ  = d y / L

    λ  = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²

    λ = 189.43 10⁻⁹ m

Part B

The compound wavelength B

      λ  = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

      λ = 269.19 10⁻⁹ m

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An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Dete
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Answer:43.34 m

Explanation:

Given

acceleration(a)=2 m/s^2

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Velocity after 6 s

v=u+at

v=0+2\times 6=12 m/s

After this object will start moving under gravity

height reached in first 6 s

s=ut+\frac{at^2}{2}

s=0+\frac{2\times 6^2}{2}

s=36 m

After fuel run out distance traveled in upward direction is

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here v=0

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How does an increase in temperature generally affect the rate of a reaction?
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A 2.8 kg grinding wheel is in the form of a solid cylinder of radius 0.1 m. a) What constant torque will bring it from rest to a
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Answer:

a) τ =  0.672 N m , b) θ = 150 rad , c) W = 100.8 J

Explanation:

a) for this part let's start by finding angular acceleration, when the angular velocity stops it is zero (w = 0)

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The moment of inertia of a cylinder is

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Let's calculate the torque

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b) we look for the angle by kinematics

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c) work in angular movement

      W = τ θ

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