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2 years ago

What happens to the diffraction pattern when the number of lines per centimeter of a diffraction grating is increased?

1 answer:

3 0

The separation between two neighboring vital maxima increments. As the quantity lines per centimeter expand, the detachment d between neighboring openings ends up noticeably littler. The point that determines primary maxima of a diffraction grinding is given by transgression = m/d, where is the wavelength and m = 0, 1, 2, 3. For settled estimations of m and the wavelength, this connection demonstrates that as d abatements, increments. For a settled screen area, this implies the separation between two neighboring chief maxima additionally increments.

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6. 498.82 mg comverted to kg

Ivanshal [37]

Answer:

0.00049882 kg

........................

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2 years ago

Read 2 more answers

A sailboat ends up 7 miles to the west of a pier. Which statement is true about this situation?

Natali [406]

Answer: which statement

Explanation:

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2 years ago

A 2.0m long pendulum is released from rest when the support string is at an angle of 25 degrees with the vertical. What is the s

Blizzard [7]

Answer:

$v=1.92m/s$

Explanation:

Given data

Length h=2.0m

Angle α=25°

To find

Speed of bob

Solution

From conservation of energy we know that:

$P.E=K.E\\mgh=(1/2)mv^{2}\\ gh=(1/2)v^{2}\\v^{2}=\frac{gh}{0.5}\\ v=\sqrt{\frac{gh}{0.5}}\\ v=\sqrt{\frac{(9.8m/s^{2} )(2.0-2.0Cos(25^{o} ))}{0.5}}\\v=1.92m/s$

8 0

2 years ago

IP A spark plug in a car has electrodes separated by a gap of 6.5×10−2 in . To create a spark and ignite the air-fuel mixture in

emmainna [20.7K]

Answer:

a.) $V=5283.2 \ V$

b.) $V=4064\ V$

Explanation:

Electric Field and Potential Difference

There are several conditions that must be met for a spark to be created into an air gap. Once the physical conditions are fixed, a minimum electric field is necessary for the spark to be initiated. Let s be the separation between the electrodes and V their potential difference. The electric field is

a.)

$\displaystyle E=\frac{V}{s}$

Solving for V

$V=E.s$

The separation is

$s= 5.5\cdot 10^{-2}\ in=0.001651 \ m$

Thus the potential difference is

$V=3.2\cdot 10^{6}\ V/m\times 0.001651 \ m$

$V=5283.2 \ V$

b.) If the separation was greater, the applied voltage needs to be greater if the electric field has to be constant. One possible measure to keep electrodes as close as possible is to build them as sharp edges. It gives the spark an easier path to travel to.

If the separation is 0.05 inches =0.00127 m

$V=3.2\cdot 10^{6}\ V/m\times 0.00127 \ m$

$V=4064\ V$

7 0

2 years ago

CuCl2

MArishka [77]

Answer:

Copper ( II ) chloride

Explanation:

According to the nomenclature ,

The metal ion name is followed by the non-metal ion
The oxidation state of the metal is to be mentioned in the brackets .
As $CuCl_{2}$ has two $Cl^-$ ions , and the net charge on the compound is zero , the charge on copper is :

$Cu^{x}+2Cl^-=0\\x-2=0\\x=2+$

thus ,

⇒ $Cu^{2+}$

∴ ,

The proper nomenclature is :

Copper ( II ) chloride

6 0

3 years ago