The focal length of given concave lens will be -26.85 cm
The height of an image to the height of an object is the ratio that is used to determine a lens' magnification. Additionally, it is provided in terms of object and image distance. It is equivalent to the object distance to image distance ratio.
Given concave lens creates a virtual image at -47.0 cm and a magnification of +1.75.
We have to find focal length
The focal length can be found out by following way:
Magnification = m = +1.75
m = hi/h
hi = -47 cm
1.75 = -47/h
h = -26.85 cm
So the focal length of given concave lens will be -26.85 cm
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By Newton's second law, the net vertical force acting on the object is 0, so that
<em>n</em> - <em>w</em> = 0
where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that
80 N = <em>µ</em> (196 N) → <em>µ</em> = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is
40 N = <em>ν</em> (196 N) → <em>ν</em> = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: <em>µ</em> and <em>ν</em> are the Greek letters mu and nu)
Active Optics.
Hope that helps, Good luck! (:
