Answer with Explanation:
We are given that
Angle of incidence,
Angle of refraction,
a.Refractive index of air,
We know that


b.Wavelength of red light in vacuum,

Wavelength in the solution,

c.Frequency does not change .It remains same in vacuum and solution.
Frequency,
Where 
Frequency,
d.Speed in the solution,

Answer:
Francium has fewer valence electrons, but they are in a higher energy level
Answer:

Explanation:
Given that,
The distance between two spheres, r = 25 cm = 0.25 m
The capacitance, C = 26 pF = 26×10⁻¹² F
Charge, Q = 12 nC = 12 × 10⁻⁹ C
We need to find the work done in moving the charge. We know that, work done is given by :

Put all the values,

So, the work done is
.
To convert 2030 rad into rev, divide 2030 by 2pie. So final answer will be
2030/2 pie =323.08 revolutions.
Answer: Energy can neither be created nor destroyed, rather it is converted from one form to another
Explanation:
The principle of conversation of energy explains how energy is conserved in nature by being converted from one form to another such that no energy is created nor destroyed.
Practical examples include:
- electrical pressing iron that converts electrical energy to heat energy
- solar panels that converts solar energy to electrical energy
- Car batteries that converts chemical energy to light energy etc