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natita [175]
3 years ago
11

An isloated point charge produce an electric field with magnitude E at a point 2 m away. At a point 1 m from the charge magnitud

e of the charge is .. A. E B. 2E C. 4E D. E/2
Physics
1 answer:
Nina [5.8K]3 years ago
8 0

Answer:

the correct answer is C,      E’= 4E

Explanation:

In this exercise you are asked to calculate the electric field at a given point

         E = k \frac{q}{r^2}

indicates that the field is E for r = 2m

         E = \frac{ k q}{4}                  (1)

the field is requested for a distance r = 1 m

         E ’= k \frac{q}{r'^2}

         E ’= k q / 1

 

from equation 1

         4E = k q

       

we substitute

        E’= 4E

so the correct answer is C

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When a liquid is cooled, the kinetic energy of the particles . The force of attraction between the particles , the space between
anzhelika [568]

Answer:

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Explanation:

For the first blank, the answer is decreases. For the second blank, the answer is increases. And finally for the third blank, the answer is decreases.

When a liquid is cooled, the kinetic energy of the particles decreases. The force of attraction between the particles increases, the space between the particles decreases, and the matter changes its state to solid.

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3 years ago
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Which form of energy is equal to the sum of an object’s kinetic and potential energy?
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Answer:

Mechanical Energy

Explanation:

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3 years ago
Why is freshly distilled or deionized water used in this standardization?
sukhopar [10]

Answer:

The amount of carbon dioxide is little in deionized water.

Explanation:

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An object with kinetic energy k explodes into two pieces, each of which moves with twice the speed of the original object.
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6 0
3 years ago
7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf
Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to  1 μV/m

Intrinsic impedance of air = 120 π  Ω

Intrinsic impedance of  water = \dfrac{120\pi}{\epsilon_r}

Using equation to solve the problem

  E(z) = E_0 e^{-\alpha\ z}

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}

\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}

\alpha =21.07\ Np/m

now ,

  1 \times 10^{-6} = 20 e^{-21.07\times z}

  e^{21.07\times z}= 20\times 10^{6}

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0
3 years ago
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