To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".
The overall magnification of microscope is

Where
N = Near point
l = distance between the object lens and eye lens
= Focal length
= Focal of eyepiece
Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm
Replacing,


Therefore the correct answer is C.
The force exerted by the magnetic in terms of the magnetic field is,

Where B is the magnetic fied strength and F is the force.
Thus, if the magnetic A has twice magnetic field strength than the magnet B,
Then,

Thus, the force exerted by the magnet B is,

Thus, the force exerted by the magnet B on magnet A is 50 N.
The force exerted by the magnet A exerts on the magnet B is exactly 100 N as given.
Hence, the option B is the correct answer.
Answer:
(i) The angular speed of the small metal object is 25.133 rad/s
(ii) The linear speed of the small metal object is 7.54 m/s.
Explanation:
Given;
radius of the circular path, r = 30 cm = 0.3 m
number of revolutions, n = 20
time of motion, t = 5 s
(i) The angular speed of the small metal object is calculated as;

(ii) The linear speed of the small metal object is calculated as;

If she has a choice and the wiring details are stated on the packaging,
then Janelle should look for lights that are wired in parallel within the
string, and she should avoid lights that are wired in series within the string.
If a single light in a parallel string fails, then only that one goes out.
The rest of the lights in the string continue to shimmer and glimmer.
If a single light in a series string fails, then ALL of the lights in that string
go out, and it's a substantial engineering challenge to determine which light
actually failed.