We know that the measure of an incident ray is: α 1 = 40°.
The index of refraction:
- for the air : n 1 = 1.00,
- for the water: n 2 = 1.33
Snell`s Law of Refraction :
n 1 · sin α 1 = n 2 · sin α 2
sin α 2 = n 1 · sin α 1 / n 2 =
= 1.00 · sin 40° / 1.33 = 0.64278 / 1.33 = 0.4833
α 2 = sin ^(-1) 0.4833
α 2 = 28.9 °
Answer: The angle relative to the water`s surface of the rays when beneath the surface is 28.9°.
Answer:
he maximum frequency occurs when the denominator is minimum
f’= f₀ 
Explanation:
This is a doppler effect exercise, where the sound source is moving
f = fo
when the source moves towards the observer
f ’=f_o
Alexandrian source of the observer
the maximum frequency occurs when the denominator is minimum, for both it is the point of maximum approach of the two objects
f’= f₀ 
1 watt = 1 joule/sec
2,000 watts = 2,000 joules/sec
(2,000 joule/sec) x (120 sec)
= (2,000 x 120) (joule-sec/sec)
= 240,000 joules .
Answer:
Magnetic field at the center of the loop 
Explanation:
It is given that total length of wire is 2 m and number of circular loop is 5 turns.
Therefore ,

We know , magnetic field at the center of loop is given by :

Putting all values in above equation we get :

Hence , this is the required solution.
The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:

(1)
where
P is the power
A is the area
In our problem, the intensity is

. At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:

And so if we re-arrange (1) we find the power emitted by the source: