Answer:
2f
Explanation:
The formula for the object - image relationship of thin lens is given as;
1/s + 1/s' = 1/f
Where;
s is object distance from lens
s' is the image distance from the lens
f is the focal length of the lens
Total distance of the object and image from the lens is given as;
d = s + s'
We earlier said that; 1/s + 1/s' = 1/f
Making s' the subject, we have;
s' = sf/(s - f)
Since d = s + s'
Thus;
d = s + (sf/(s - f))
Expanding this, we have;
d = s²/(s - f)
The derivative of this with respect to d gives;
d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²
Equating to zero, we have;
(2s/(s - f)) - s²/(s - f)² = 0
(2s/(s - f)) = s²/(s - f)²
Thus;
2s = s²/(s - f)
s² = 2s(s - f)
s² = 2s² - 2sf
2s² - s² = 2sf
s² = 2sf
s = 2f
By looking at the potential energies before and after the reaction, we can tell that the reaction is exothermic (final < initial) or endodermic (final > initial).
Also, the amount of activation energy gives an idea of the external energy required to initiate the reaction (for example, by heating the reactants).
Furthermore, by the same principle, we can also deduce the activation energy for the reverse reaction.
If a catalyst is available, the diagram will show a reduced activation energy, compared to a reaction without catalyst. However, it will also show that the catalyst does not alter the initial and final energies of the reaction.
1) v = gt = 10*1.5 = 15 m/s
2) r = gt^2 /2 = 10*(1.5)^2 / 2 = 11.25 meters
