The answer is 167 pounds.
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Answer:
The focal length of the lens is 34.047 cm
The power of the needed corrective lens is 2.937 diopter.
Explanation:
Distance of the object from the lens,u = 26 cm
Distance of the image from the lens ,v= -110 cm
(Image is forming on the other side of the lens)
Since ,lens of the human eye is converging lens,convex lens.
Using a lens formula:


f = 34.047 cm = 0.3404 m
Power of the lens = P

The component of the force in negative z-direction is -0.144 N.
The given parameters;
- <em>current in the wire, I = 2.7 A</em>
- <em>length of the wire, L = (3.2 i + 4.3j) cm</em>
- <em>magnetic filed, B = 1.24 i</em>
The force on the segment of the wire is calculated as follows;

where;
- <em>θ is the angle wire and magnetic field</em>
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The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.
The angle between the wire and the magnetic field is calculated as follows;

The magnitude of the wire length is calculated as follows;

The component of the force in negative z-direction is calculated as;

Thus, the component of the force in negative z-direction is -0.144 N.
Learn more here:brainly.com/question/22719779