distance from the Sun of 2.77 astronomical units or about 414 million km 257 million miles and orbiting period of 4.62 years
Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
Answer:
s = 6.25 10⁻²² m
Explanation:
Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by
p = q s
s = p / q
let's calculate
s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹
s = 0.625 10⁻²¹ m
s = 6.25 10⁻²² m
We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.
Answer:
0.34 m/s^2
Explanation:
force=mass × acceleration
400 =1200 × acceleration
acceleration=400/1200
=0.34 m/s^2
