For an extended object, the weight force can be considered to act at which point?
1 answer:
You might be interested in
A) Image position: -19.3 cm
B) Image height: 1.5 cm, upright
Explanation:
A)
In order to calculate the image position, we can use the lens equation:
where
p is the distance of the object from the lens
q is the distance of the image from the lens
f is the focal length
In this problem, we have:
p = 13 cm (object distance)
f = 40 cm (focal length, positive for a converging lens)
So the image distance is
The negative sign means that the image is virtual.
B)
In order to calculate the image height, we use the magnification equation:
where
y' is the image height
y is the object height
In this problem, we have:
y = 1.0 cm (object height)
p = 13 cm
q = -19.3 cm
Therefore, the image heigth is
And the positive sign means the image is upright.
Answer: 31.6ft
Explanation:
Check the attachment for the diagram.
According to the right angle triangle AEC, we will use Pythagoras theorem to get |AC|. Note that |AE| = |AB| - |CD|
that is 20ft - 10ft = 10ft
According to the theorem, the square of the sum of the adjacent side and the opposite side is equal to the square of the hypotenuse.
|AE|^2 + |EC|^2 = |AC|^2
10^2 + 30^2 = |AC|^2
100 + 900 = |AC|^2
|AC| = √1000
|AC| = 31.6ft
Therefore, the wire should be anchored 31.6ft to the ground to minimize the amount of wire needed.
