= Heat released to cold reservoir
= Heat released to hot reservoir
= maximum amount of work
= temperature of cold reservoir
= temperature of hot reservoir
we know that
eq-1
maximum work is given as
= -
using eq-1
= 
-
Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
The difference between speed and velocity is that the speed is a scalar quantity which means that you can say that this object has a speed of x m/s but you don't have to define its direction
while the velocity is a vector quantity which means that you have to express the velocity by which it moves in x,y and z directions and its norm is the speed
It is a chemical change and a physical change
For t1:
t1 = square root of 2h1 / g = square root of 2 * 0.5 / 9.8 = 0.319 sec
For t2:
t2 = sqaure root of 2h2 / g = square root of 2 * 1.0 / 9.8 = 0.451 sec
Wherein:
t = time(s) for the vertical movement
h= height
g = gravity (using the standard 9.8 m/sec measurement)
d1 = 1*0.319 = 0.319 m
d2 = 0.5 * 0.451 = 0.225 m
Where:
d = hor. distance
ratio = d1:d2
= 0.319 : 0.225
=3.19 : 2.25
The answer is 3.19 : 2.25
