Speed = wavelength × frequency
giving that frequency is 0, wavelength and speed are directionally proportional. wavelength decrease = speed decrease
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
Answer:
a) fr = 224.3 N
, b) fr = 224.3 N
, c) v = 198.0 m/s
Explanation:
a) For this exercise let's start by calculating the acceleration in the fall
v² = v₀² - 2 a (y-y₀)
When it jumps the initial vertical speed is zero
a = -v² / 2 (y-y₀)
a = -68 2/2 (1000-2000)
a = 2,312 m / s²
Let's use the second net law to enter the average friction force
fr = m a
fr = 97 2,312
fr = 224.3 N
b) let's look for acceleration
v² = v₀² - 2 a y
a = (v² –v₀²) / 2 (y-y₀)
a = (4² - 68²) / 2 (0-1000)
a = 2,304 m / s²
fr = m a
fr = 97 2,304
fr = 223.5 N
c) the speed of the wallet is searched with kinematics
v² = v₀² - 2 g (y-y₀)
v = √ (0-2 9.8 (0-2000))
v = 198.0 m/s
Answer:

and

Explanation:
Given:
- first charge,

- second charge,

- position of first charge,

- position of second charge,

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.
<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

- since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.





Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

and
