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:
the answer to this question is 2,4,3,1
Answer:
The answer is given below
Explanation:
u is the initial velocity, v is the final velocity. Given that:
a)
The final velocity of cart 1 after collision is given as:
The final velocity of cart 2 after collision is given as:
b) Using the law of conservation of energy:
Answer:
The box displacement after 6 seconds is 66 meters.
Explanation:
Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object (
), in meters, can be determined by the following expression:
(1)
Where:
- Initial velocity, in meters per second.
- Time, in seconds.
- Acceleration, in meters per square second.
If we know that 
, 
and 
, then the box displacement after 6 seconds is:
The box displacement after 6 seconds is 66 meters.
