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
Answer:
This is due to impulse
Explanation:
Impulse equal to mΔv and FΔt
You can set these equal as mΔv = FΔt
When a boxer punches a tissue, it is like punching a cushion or a pillow. The time that the hit takes is much grater than if they were to hit something solid. In addition, the change in velocity of the boxer's arm would be much greater when they hit a punching bag. In this equation, the greater the time, the less force that is needed.
<span>A. frustration-aggression theory.
Good Luck!!</span>
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