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Taya2010 [7]
3 years ago
12

You are using a diffraction grating with 2230 lines per centimeter. White light passes through it, and the grating spreads the l

ight out into its spectral components. At what angle does green light of wavelength 508 nm appear in second order
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

The angle of diffraction is 13.09°

Explanation:

Given:

Wavelength of light \lambda = 508 \times 10^{-9} m

Order of diffraction n = 2

No. lines per centimeter N = 2230

First find the grating element,

  d = \frac{1}{N}

  d = 4.484 \times 10^{-6} m

From the formula of diffraction,

  d \sin \theta = n \lambda

    \sin \theta = \frac{2 \lambda}{d}

    \sin \theta = \frac{2 \times 508 \times 10^{-9} }{4.484 \times 10^{-6} }

    \theta = \sin ^{-1} (0.2265)

    \theta = 13.09°

Therefore, the angle of diffraction is 13.09°

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Which statement is true about the atom?
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3 years ago
Find the value of currents through each branch
Irina-Kira [14]

Answer:

the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

  top center: I2+I3 = 2.500 A

  bottom center: I2+I3-I1 = 0 A

  right: I3 = 1.875 A

Explanation:

You can write the KVL equations:

Top left loop:

  I2(4) +(I2 +I3)(2) +I1(1) = 10

Bottom left loop:

  (I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10

Right loop:

  (I2+I3)(2) +(I2+I3-I1)(2) = 5

In matrix form, the equations are ...

  \left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]

These equations have the solution ...

  \left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]

This means the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

  top center: I2+I3 = 2.500 A

  bottom center: I2+I3-I1 = 0 A

  right: I3 = 1.875 A

_____

This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.

When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.

Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.

Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.

  (I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)

__

It helps to be familiar with the formulas for resistors in series and parallel.

8 0
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zepelin [54]

Answer:

Here is the answer.

Explanation:

Balanced forces- they are those forces that produce 0 resultant forces.

therefore, on applying a balanced force on the object, it wouldn't result in any change, as resultant force would be 0.

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