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3 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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The charge within a small volume dv is dq=ρdv. the integral of ρdv over a cylinder of length l is the total charge q=λl within t

azamat

The volume of the shell that you described would be:
$dV=2L\pi r dr$
Now we can rewrite the given integral:
$\lambda L=\int\rho dV=L\rho\int2\pi r dr \\ \lambda L =L\rho \pi r^2\\ \rho=\frac{\lambda}{\pi r^2}$
I have attached the picture explaining how we got the formula for the volume.
On the picture, I marked the rectangle. You can of this rectangle as the base, and the height would be the circumference of the cylinder.

7 0

3 years ago

Read 2 more answers

The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 3.00 mm, are charged by a 6.00

Vera_Pavlovna [14]

Answer:

(a). The charge is $1.5045\times10^{-11}\ C$

(b). The initial stored energy is $4.5135\times10^{-11}\ J$

(c). The final stored energy is $12.036\times10^{-11}\ J$

(d). The work required to separate the plates is $7.5225\times10^{-11}\ J$

Explanation:

Given that,

Area = 8.50 cm²

Distance = 3.00 mm

Potential = 6.00 V

Distance without discharge = 8.00 mm

(a). We need to calculate the capacitance

Using formula of capacitance

$C_{1}=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C_{1}=\dfrac{8.85\times10^{-12}\times8.5\times10^{-4}}{3.00\times10^{-3}}$

$C_{1}=2.5075\times10^{-12}\ F$

We need to calculate the charge

Using formula of charge

$Q=CV$

Put the value into the formula

$Q=2.5075\times10^{-12}\times6.00$

$Q=1.5045\times10^{-11}\ C$

(b). We need to calculate the initial stored energy

Using formula of initial energy

$E_{i}=\dfrac{1}{2}\times CV^2$

$E_{i}=\dfrac{1}{2}\times2.5075\times10^{-12}\times36$

$E_{i}=4.5135\times10^{-11}\ J$

(c). We need to calculate the capacitance

Using formula of capacitance

$C_{2}=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C_{2}=\dfrac{8.85\times10^{-12}\times8.5\times10^{-4}}{2\times8.00\times10^{-3}}$

$C_{2}=9.403\times10^{-13}\ F$

We need to calculate the final stored energy

Using formula of initial energy

$E_{f}=\dfrac{1}{2}\times \dfrac{Q^2}{C}$

$E_{f}=\dfrac{1}{2}\times\dfrac{(1.5045\times10^{-11})^2}{9.403\times10^{-13}}$

$E_{f}=12.036\times10^{-11}\ J$

(d). We need to calculate the work done

Using formula of work done

$W=E_{f}-E_{i}$

Put the value in the formula

$W=12.036\times10^{-11}-4.5135\times10^{-11}$

$W=7.5225\times10^{-11}\ J$

Hence, (a). The charge is $1.5045\times10^{-11}\ C$

(b). The initial stored energy is $4.5135\times10^{-11}\ J$

(c). The final stored energy is $12.036\times10^{-11}\ J$

(d). The work required to separate the plates is $7.5225\times10^{-11}\ J$

5 0

4 years ago

1. How many grams are in a dekagram?

zepelin [54]

Answer: 10

Explanation:

none just search it up

4 0

3 years ago

Read 2 more answers

15) What is the frequency of a pendulum that is moving at 30 m/s with a wavelength of .35 m?

____ [38]

A pendulum is not a wave.

-- A pendulum doesn't have a 'wavelength'.

-- There's no way to define how many of its "waves" pass a point
every second.

-- Whatever you say is the speed of the pendulum, that speed
can only be true at one or two points in the pendulum's swing,
and it's different everywhere else in the swing.

-- The frequency of a pendulum depends only on the length
of the string from which it hangs.

If you take the given information and try to apply wave motion to it:

   Wave speed = (wavelength) x (frequency)

   Frequency = (speed) / (wavelength) ,

you would end up with

   Frequency = (30 meter/sec) / (0.35 meter) = 85.7 Hz

Have you ever seen anything that could be described as
a pendulum, swinging or even wiggling back and forth
85 times every second ? ! ?   That's pretty absurd.

This math is not applicable to the pendulum.

6 0

3 years ago

What is the speed of a car that travels 0.200 km in 95 seconds? State your answer in meters per second.

WARRIOR [948]

Answer:

<h2>2.10 meters per second</h2><h2 />

Explanation:

speed (velocity) = distance / time

= <u>0.2 km (1000 m/km) </u>

95 secs.

= 2.10 meters per second

8 0

3 years ago