Answer:
B
Explanation:
Heat flows from hot to cold to lower the temperature of hot areas and increase temperature of cold areas. The end result is that the 2 areas have the same temperature, thus increasing entropy.
Answer:

Explanation:
Here two charges are placed at distance "d" apart
now the net value of electric field at some position between two charges will be ZERO
so we will have
electric field due to charge 1 = electric field due to charge 2

Let the position where net field is zero will lie at distance "r" from q1

now we will have

now square root both sides

now we have

so we have

Answer:
IDC
Explanation:
I DON'T UNDERSTAND........
PART A)
By Snell's law we know that

here we know that



now from above equation we have


so it will refract by angle 39.3 degree
PART B)
Here as we can see that image formed on the other side of lens
So it is a real and inverted image
Also we can see that size of image is lesser than the size of object here
Here we can use concave mirror to form same type of real and inverted image
PART C)
As per the mirror formula we know that



so image will form at 30 cm from mirror
it is virtual image and smaller in size