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Kruka [31]
3 years ago
13

What is

itle=" \frac{1}{4} = \frac{?}{8} " alt=" \frac{1}{4} = \frac{?}{8} " align="absmiddle" class="latex-formula">
​
Physics
2 answers:
Finger [1]3 years ago
8 0
X=2

First, cross multiply

4*x=1*8

4x=8

Divide by 4 on both sides

x=2

Hope this helped! :)
katen-ka-za [31]3 years ago
7 0

\frac{2}{8}

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A beam of light of wavelength lambda = 541nm is normally incident on a diffraction grating which has 600 lines m * m ^ - 1 Deter
MrMuchimi

Answer:

Angle of diffraction for second order maxima is θ = 18.941°

Explanation:

From the question it is given that

wavelength of incident light = λ = 541 nm = 541 x 10^-^9 m

order of maxima = n =2

diffraction grating has 600 lines per mm

⇒ distance between two slit is \frac{1mm}{600} = 1.66 x 10^-^6 m

using the relation of Braggs diffraction formula i.e.,

2dsinθ = nλ ..................................(1)

where, d = distance between two lines of grating

θ is the angle of diffraction

n= order of maxima

λ is the intensity of incident photon

on substituting the respected values in relation (1) we get,

2 x 1.66 x 10^-^6m sinθ = 2 x 541 x 10^-^9 m

⇒ sinθ = 0.3246

⇒ θ = sin^-^1(0.3246) = 18.941 °

8 0
3 years ago
How is a pareto chart different from a standard vertical bar​ graph?
Alona [7]

Answer:

A Pareto chart, named after an Italian economist, combines a bar chart with a line graph. The bar chart is different from a histogram in more than one way. For example, the vertical bars need not touch one another as per a histogram

Explanation:

5 0
4 years ago
The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
Pete is driving down a highway and sees a brown sign. All of the brown signs
pychu [463]

Answer:

C

Explanation:

He was using visual clues to read the signs.

6 0
3 years ago
Explain what is happening in this<br> induction example.
Semmy [17]

Answer:

An electrical field duh

Explanation:

None needed self explanitory

4 0
2 years ago
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