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mel-nik [20]
3 years ago
10

A grating has 320 lines/mm. How many orders of the visible wavelength 551 nm can it produce in addition to the m = 0 order

Physics
1 answer:
Anastaziya [24]3 years ago
4 0

Answer:

6

Explanation:

Given that

dsinθ = mλ,

now, if sinθ = 1, then

m = d / λ, where

m = order of interference

d = distance between the slits

λ = wavelength of light

this is the formula we would use to solve the question

d = 1 / 320 lines/mm

d = 1 / 320*10^3

d = 3.125*10^-6 m

At λ = 551 nm, we have

m = 3.125*10^-6 / 551*10^-9

m = 5.67

5.67 ~ 6

thus, we can say that the orders of visible wavelength 551 nm, can produce is 6

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1 year ago
Astronomers observe the motion of four planets that orbit a star similar to the Sun. Each planet follows an elliptical orbit aro
Sergeu [11.5K]

Answer:

planet that is farthest away is planet X

kepler's third law

Explanation:

For this exercise we can use Kepler's third law which is an application of Newton's second law to the case of the orbits of the planets

          T² = (\frac{4\pi ^2}{ G M_s} a³ = K_s a³

           

Let's apply this equation to our case

          a = \sqrt[3]{ \frac{T^2}{K_s} }

for this particular exercise it is not necessary to reduce the period to seconds

Plant W

             10² = K_s  a_{w}^3

             a_w = \sqrt[3]{ \frac{100}{ K_s} }

             a_w = \frac{1}{ \sqrt[3]{K_s} }  4.64

Planet X

             a_x = \sqrt[3]{ \frac{640^3}{K_s} }

             a_x = \frac{1}{ \sqrt[3]{K_s} } 74.3

Planet Y

              a_y = \sqrt[3]{ \frac{80^2}{K_s}  }

              a_y = \frac{1}{ \sqrt[3]{K_s} } 18.6

Planet z

              a_z = \sqrt[3]{ \frac{270^2}{K_s} }

              a_z = \frac{1}{ \sqrt[3]{K_s} } 41.8

From the previous results we see that planet that is farthest away is planet X

where we have used kepler's third law

3 0
3 years ago
A tension force of 175 N inclined at 20.0° above the horizontal is used to pull a 40.0 - kg packing crate a distance of 6.00 m o
Murljashka [212]

Answer:

(a) Work done by the tension force is 987 J

(b) Coefficient of kinetic friction between the crate and surface is 0.495

Explanation:

(a)

Work done by any force F in moving an object by a distance d making an angle \Theta with the direction of force  is given by

W=Fd\cos (\Theta )

W=175 \times 6 \times  \cos (20 )J=987J

Thus work done by the tension force is 987 J

(b)

Normal force on the crate is given by

N= mg-F\sin \Theta  =(40\times 9.8-175\times \sin 20)Newton

=>N=332 Newtons

Since crate is moving with constant speed . Therefore using Newtons second law .

Fcos(\Theta ) -\mu_k N=0

Where \mu_k=coefficient of kinetic friction

\therefore \mu_k=\frac{F\cos (\Theta )}{N}

=>\mu_k=\frac{175\times \cos 20}{332}

=>\mu _k= 0.495

Thus coefficient of kinetic friction between the crate and surface is 0.495

4 0
3 years ago
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