Answer:
Explanation:
We shall apply Stefan's formula
E = AσT⁴
When T = 300
I₁ = Aσ x 300⁴
When T = 400K
I₂ = Aσ x 400⁴
I₂ / I₁ = 400⁴ / 300⁴
= 256 / 81
= 3.16
I₂ = 3.16 I₁ .
Answer:
The thickness is 
Explanation:
From the question we are told that
The wavelength is 
The first order of the dark fringe is 
The second order of dark fringe considered is 
Generally the condition for destructive interference is mathematically represented as

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe
So the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

=> 
=> 
=> 
Answer:
A) 
B) 
C) 
Explanation:
Given:
- mass of flywheel,

- diameter of flywheel,

- rotational speed of flywheel,

- duration for which the power is off,

- no. of revolutions made during the power is off,

<u>Using equation of motion:</u>



Negative sign denotes deceleration.
A)
Now using the equation:


is the angular velocity of the flywheel when the power comes back.
B)
Here:

Now using the equation:


is the time after which the flywheel stops.
C)
Using the equation of motion:


revolutions are made before stopping.
D. all of the above
Hope this helps!
This is due to the tilt of the earth on its axis. Although the Sun shines on Earth, because of how the Earth is tilted, the equator is more directly hit compared to places found on the poles. The poles are hit at an angle, therefore the sunlight they receive is lesser than the places at the equator.