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Mashcka [7]
3 years ago
8

If the distance between slits on a diffraction grating is 0.50 mm and one of the angles of diffraction is 0.25°, how large is th

e path difference?
nm




How many orders of bright lines does this equal for red light with a wavelength of 650 nm?


wavelengths
Physics
2 answers:
Trava [24]3 years ago
8 0

Answer:

-  path differnce = 2.18*10^-6

-  1538 lines

Explanation:

- The path difference for the waves that produce the pattern of diffraction, is given by the following formula:

path\ difference\ = dsin\theta           (1)

d: separation between slits = 0.50mm = 0.50*10^-3 m

θ: angle of a diffraction = 0.25°

Then, the path difference is:

path\ difference\ =(0.50*10^{-3}m)sin(0.25\°)=2.18*10^{-6}m

- The maximum number of bright lines are calculated by using the following formula:

m\lambda = dsin\theta           (2)

m: order of the bright

λ: wavelength = 650nm

The maximum bright is calculated for an angle of 90°:

m=\frac{(0.50*10^{-3}m)sin90\°}{650*10^{-9}m} \approx 769

The maxium number of bright lines are twice the previous result, that is, 1538 lines

Masja [62]3 years ago
5 0

Answer:

a. 2200

b. 3

Explanation:

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The sun appears to move through the background stars. This apparent motion would not exist if:_________.
otez555 [7]

Answer:

b) the earth did not orbit the sun

Explanation:

The sun appears to move through the background stars due to parallax . Parallax is a phenomenon when the near object appears to move faster than the distant object . If we travel in a train , the near object like electric poles on rail track or trees and fields nearby appear to move faster against the distant background .

Hence when the earth moves around the sun , the sun appears to move against the background the stars which are far away . Had earth remained stationary at a place on its orbit around the sun , the sun would have appeared stationary against the background the   stars .

3 0
2 years ago
the police department determined that the force required to drag a 130 N(29 lb) car tire across the pavement at a constant veloc
ki77a [65]
The force of friction is given by:
f = μR, where μ is the friction coefficient and R is the reaction force, which will be equal to the weight.
100 = μ x 130
μ = 0.77
4 0
3 years ago
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
Viefleur [7K]

Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If  the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N  

4 0
3 years ago
A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa
pav-90 [236]

Answer

given,

current (I) = 16 mA

circumference of the circular loop (S)= 1.90 m

Magnetic field (B)= 0.790 T

S = 2 π r

1.9 = 2 π r

r = 0.3024 m

a) magnetic moment of loop

    M= I A

    M=16 \times 10^{-3} \times \pi \times r^2

   M=16 \times 10^{-3} \times \pi \times 0.3024^2

   M=4.59 \times 10^{-3}\ A m^2

b)  torque exerted in the loop

 \tau = M\ B

 \tau = 4.59 \times 10^{-3}\times 0.79

 \tau = 3.63 \times 10^{-3} N.m

8 0
2 years ago
(f) What do you mean by: (i) 1 J energy (ii) 1 W power​
max2010maxim [7]

Explanation:

Watts are defined as 1 Watt = 1 Joule per second (1W = 1 J/s)

which means that 1 kW = 1000 J/s. A Watt is the amount of energy (in Joules) that an electrical device (such as a light) is burning per second that it's running.

3 0
2 years ago
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