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3 years ago

How do the charges compare when two objects are charged through friction?

Physics

1 answer:

spayn [35]3 years ago

3 0

Answer:

The objects become oppositely charged and have equal amounts of charge.

Explanation:

There are three methods for charging objects:

- Conduction: a charged object is brought in contact with a neutral object. Electrons are transferred from the charged object to the neutral one, which also becomes charged

- Induction: a charged object is brought close (but not in contact) to a neutral object. The charges inside the neutral object redistribute, such that those of opposite sign to the charge in the charged object migrate on the side closer to the charged object, while the charges of same sign migrate towards the opposite side. If the neutral object is then grounded, the charges on the opposite side flow to the ground, leaving the neutral object charged as well

- Friction: two objects initially neutral are rubbed against each other. Electrons move from one object to the other one: therefore, one object becomes positively charged while the other one becomes negatively charged. Since the charge gained by one object is equal to charge lost by the other object, it follows that the two objects have same magnitude of charge, but with opposite sign.

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3 years ago

Ocean waves of wavelength 26 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall,

Genrish500 [490]

Answer:

a) the distance between her and the wall is 13 m

b) the period of her up-and-down motion is 6.5 s

Explanation:

Given the data in the question;

wavelength λ = 26 m

velocity v = 4.0 m/s

a) How far from the wall is she?

Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;

x = λ/2

we substitute

x = 26 m / 2

x = 13 m

Therefore, the distance between her and the wall is 13 m

b) What is the period of her up-and-down motion?

we know that the relationship between frequency, wavelength and wave speed is;

v = fλ

hence, f = v/λ

we also know that frequency is expressed as the reciprocal of the time period;

f = 1/T

Hence

1/T = v/λ

solve for T

Tv = λ

T = λ/v

we substitute

T = 26 m / 4 m/s

T = 6.5 s

Therefore, the period of her up-and-down motion is 6.5 s

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