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dimulka [17.4K]
3 years ago
15

If you decrease the distance between successive crests of a wave, this changesA. The frequency.B. The wavelength.C. The speed.D.

Both the wavelength and the speed.E. Both the wavelength and the frequency
Physics
1 answer:
Feliz [49]3 years ago
8 0

Answer:

option E

Explanation:

The correct answer is option E

Writing the relation between wavelength and frequency

      f = \dfrac{v}{\lambda}

f is the frequency

v is the velocity of wave

λ is the wavelength

From the above expression we can clearly see that frequency is inversely proportional to wavelength.

When the distance between the two successive crest is decreased then wave length of the wave also decrease.

If wavelength of the wave decreases then frequency of the wave increase.

hence, we can say that both wavelength and the frequency changes.

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A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius
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Explanation:

Gauss Law relates the distribution of electric charge to the resulting electric field.

Applying Gauss's Law,

                              EA = Q / ε₀

Where:

E is the magnitude of the electric field,

A is the cross-sectional area of the conducting sphere,

Q is the positive charge

ε₀ is the permittivity

We be considering cases for the specified regions.

<u>Case 1</u>: When r < R

The electric field is zero, since the enclosed charge is equal to zero

                                            E(r) = 0

<u>Case 2</u>: When R < r < 2R

The enclosed charge equals to Q, then the electric field equals;

                              E(4πr²) = Q / ε₀

                              E = Q / 4πε₀r²

                              E = KQ /r²

Constant K = 1 / 4πε₀ = 9.0 × 10⁹ Nm²/C²

<u>Case 3</u>: When r > 2R    

The enclosed charge equals to Q, then the electric field equals;

                               E(4πr²) = 2Q / ε₀

                               E = 2Q / 4πε₀r²

                               E = 2KQ /r²        

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3 years ago
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