Diagram B .... light shines through at an angle
The answer would be 54 m/s as the maximum speed
Answer:
4
Explanation:
We are given that

K.E at x=0 m=20 J
K.E at x=3 m=11 J
We have to find the value of c.
By work energy theorem
Work done=Change in kinetic energy
W=
![W=[\frac{cx^2}{2}-x^3]^{3}_{0}](https://tex.z-dn.net/?f=W%3D%5B%5Cfrac%7Bcx%5E2%7D%7B2%7D-x%5E3%5D%5E%7B3%7D_%7B0%7D)






(a) -48.0 cm, diverging
We can use the lens equation:

where
f is the focal length
p = 16.0 cm is the object distance
q = -12.0 cm is the image distance (with a negative sign because the image is on the same side as the object, so it is virtual)
Solving for f, we find the focal length of the lens:


The lens is diverging, since the focal length is negative.
(b) 6.38 mm, erect
We can use the magnification equation:

where
y' is the size of the image
y = 8.50 mm is the size of the object
Substituting p and q that we used in the previous part of the problem, we find y':

and the image is erect, since the sign is positive.
(c)
See attached picture.
Answer:
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