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3 years ago

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What is the wavelength of light that is deviated in the first order through an angle of 13.1 ∘ by a transmission grating having

5000 slits/cm? Assume normal incidence.

1 answer:

faltersainse [42]3 years ago

7 0

Answer:

The wavelength of light is $4.53\times10^{-7}\ m$

Explanation:

Given that,

Angle = 13.1°

Number of slits = 5000

We need to calculate the wavelength of light

Diffraction of first order is defined as,

$d \sin\theta=n\lambda$ .....(I)

The separation of the slits

$d = \dfrac{1}{N}$

$d=\dfrac{1}{5000}$

$d=2\times10^{-6}\ m$

Now put the value in equation (I)

$2\times10^{-6}\sin13.1^{\circ}=\lambda$

Here, n = 1

$\lambda=4.53\times10^{-7}\ m$

Hence, The wavelength of light is $4.53\times10^{-7}\ m$

You might be interested in

Approximately what is the smallest detail observable with a microscope that uses ultraviolet light of frequency 1. 72 x 1015 hz?

postnew [5]

The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light , is calculated as follows: 3× $10^{8}$ / 1.72× $10^{15}$ or approximately 1.74× $10^{-7}$ m.

The distance between the two positive, two negative, or two minimal points on the waveform is known as the wavelength of the wave. The following formula expresses the relationship between the frequency and wavelength of light:

f = c / λ

where, f = frequency of light

c = speed of light

λ = wavelength of light

Given data = f = 1.72× $10^{15}$ Hz

Therefore, λ = 3× $10^{8}$ / 1.72× $10^{15}$

λ = 1.74× $10^{-7}$ m

The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light , is calculated as follows: 3× $10^{8}$ / 1.72× $10^{15}$ or approximately 1.74× $10^{-7}$ m.

Learn more about light here;

brainly.com/question/15200315

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7 0

1 year ago

f the magnitude of the acceleration of a propeller blade's tip exceeds a certain value amaxamax, the blade tip will fracture. If

Luden [163]

Answer:

The angular velocity is $w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

Explanation:

Generally the acceleration experienced by the propeller blade's is broken down into

The Radial acceleration which is mathematically represented as

$a_r = \frac{v^2}{r} = w^2r$

And the Tangential acceleration which is mathematically represented as

$a_r = \alpha r$

The net acceleration is evaluated as

$a = \sqrt{a_r^2 + a_t^2}$

Now since angular speed varies directly with angular acceleration so when acceleration is maximum the angular velocity is maximum also and this point if the propeller blade's tip exceeds it the blade would fracture

So at maximum angular acceleration we a have

$a_{max} = \sqrt{a_r^2 + a_t^2}$

$a_{max}^2 = a_r^2 + a_t^2$

$a_{max}^2 = (w^2r)^2 + (\alpha r)^2$

$a_{max}^2 = r^2 w^4 + r^2 \alpha ^2$

$a_{max}^2 = r^2 (w^4 + \alpha^2 )$

$w^4 +\alpha ^2 = \frac{a_{max}^2}{r^2}$

$w^4 = \frac{a_{max}^2}{r^2} - \alpha ^2$

$w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

3 0

2 years ago

Which type of bone is this <br>covalent bond<br>hydrogen bond <br>ionic bond

Fynjy0 [20]

Answer:

Explanation:A covalent bond is formed when electrons are shared between non-metal atoms, and the positive nuclei are attracted towards the pair of negative bonded electrons. ... Hence, the hydrogen bond is weaker than ionic and covalent bonds. Example: Water molecules are held to each other by intermolecular forces of attraction.

5 0

3 years ago

Jack and Jill are on two different floors of their high rise office building and looking out of their respective windows. Jack s

Maru [420]

Answer:

a) speed when Jack sees the pot : 12.92 meters per second

b) height difference 163.115 meters

Explanation:

First to calculate te initial speed we use the acceleration formula:

a= v1-v0/t

Acceleration being gravity's acceleration (9.8 m/s^2)

v1 being the speed when Jill sees the pot

v0 when Jack sees it

and t the time between

Solving for v0 it would be

v1 - a*t = v0

replacing

$58 m/s - 9.8 m/s^2 *4.6 s = v0 = 12.92 m/s$

For the second question we use the position formula setting y0 and t0 as the position and time when jack sees the pot. (and setting the positive axis downward I.E. one meter below jack would be 1m not -1m)

The formula is

$y0 + v0*t + 1/2 g *t^2 = yt$

replacing

$0m + 12.92m/s* 4.6 s + 1/2 * 9.8 m/s^2 * (4.6 s)^2 = 163.115 m$

5 0

2 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

1 year ago