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nlexa [21]
2 years ago
8

Classify each of the following statements as a characteristic (a) of electric forces only, (b) of magnetic forces only, (c) of b

oth electric and magnetic forces, or (d) of neither electric nor magnetic forces. (viii) The magnitude of the force depends on the charged object's direction of motion.
Physics
1 answer:
kakasveta [241]2 years ago
5 0

(b) Magnetic forces only

When an object is placed in an electric field it experiences an electric force given by :

F_{E} = |q| × E

Similarly when it is placed in a magnetic field it experiences a magnetic force given by :

F_{B} = |q|×v×Bsinθ

When a particle is placed in electric field the particle gets accelerated.

The electric field has a direction, positive to negative. This is the direction that the electric field will cause a positive charge to accelerate. For example If a positive charge is moving in the same direction as the electric field vector the particle's velocity will increase. If it is moving in the opposite direction it will decelerate.  ( No change in direction of motion in this case)

When a particle is moved in magnetic field ,the magnetic force is perpendicular to velocity and magnetic field . Since force is perpendicular to velocity ,it only changes the direction of motion.

To know more about electric and magnetic forces kindly refer to the below link:

brainly.com/question/1594186

#SPJ4

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shutvik [7]

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer:  3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

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Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance  d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f =  5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

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