The number of lines per mm in the diffraction grating is 326.
<h3>What is diffraction grating?</h3>
A diffraction grating is a type of optical instrument obtained with a continuous pattern. The pattern of the diffracted light by a grating depends on the structure and number of elements present.
The given data in the problem is
is the angle formed between the path of the incident light and the diffracted light = 9. 2°
λ is the wavelength of the light=490nm=4.9
N is the number of lines per mm in the diffraction grating=?
n is ordered = 1
The formula for the diffraction grating is;

The number of lines per mm is found as;

Hence the number of lines per mm in the diffraction grating is 326.
To learn more about diffraction grating refer to the link;
brainly.com/question/1812927
Infrared waves. The other 3 would hurt you because they are below the UV scale of 400 nanometers. Infrared is light that wouldn't harm you
Answer:
A)A=0.075 m
B)v= 0.21 m/s
Explanation:
Given that
m = 0.75 kg
K= 13.5 N
The natural frequency of the block given as

The maximum speed v given as

A=Amplitude


A=0.075 m
A= 0.75 cm
The speed at distance x



v= 0.21 m/s
Answer:
v = 8.57 m/s
Explanation:
As we know that the wagon is pulled up by string system
So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane
So as per work energy theorem we know that
work done by tension force + work done by force of gravity = change in kinetic energy

so we have


m = 38.2 kg
d = 85.4 m
so now we have


it would be C laminated soda lime glass