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Viefleur [7K]
4 years ago
9

What must occur for neutral chargers to occur

Physics
1 answer:
Evgesh-ka [11]4 years ago
7 0

Answer:

 Friction charging is a very common method of charging an object. However, it is not the only process by which objects become charged. In this section of Lesson 2, the charging by induction method will be discussed. Induction charging is a method used to charge an object without actually touching the object to any other charged object. An understanding of charging by induction requires an understanding of the nature of a conductor and an understanding of the polarization process.

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32. Which sub-atomic particle has a charge of +1?
mestny [16]

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Protons

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3 years ago
The battery charger for an mp3 player contains a step-down transformer with a turns ratio of 1:29, so that the voltage of 120 v
kifflom [539]
Transformer contains two coils: primary and secondary. They allow change of voltage to lower or higher value. In first case we have step-down and in second case we have step-up transformer.

Formula used for transformer is:
\frac{ N_{1} }{N_{2} } = \frac{ V_{1} }{V_{2} }

Where:
N1 = number of turns on primary coil
N2 = number of turns on secondary coil
V1 = voltage on primary coil
V2 = voltage on secondary coil

In a step-down transformer primary coil has more turns than secondary coil. So the ratio 1:29 means that for each turn on secondary coil we have 29 turns on primary coil.

We can solve the equation for V2:
{V_{2}}=\frac{ V_{1}{N_{2} }  }{N_{1} } \\  {V_{2}}=\frac{ 120*{x }  }{29x } \\  {V_{2}}=4.14V

Secondary coil provides voltage of 4.14V.
8 0
4 years ago
2. Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m. The charges are +7.0μC, -8.0 μC
Scorpion4ik [409]

Answer:

0.53 N, 25.6°

Explanation:

side of triangle, a = 1.2 m

q = 7 μC

q1 = - 8 μC

q2 = - 6 μC

Let F1 be the force between q and q1

By using the coulomb's law

F_{1}=\frac{Kq_{1}q}{a^{2}}

F_{1}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 8\times 10^{-6}}{1.2^{2}}

F1 = 0.35 N

Let F2 be the force between q and q2

By using the coulomb's law

F_{2}=\frac{Kq_{2}q}{a^{2}}

F_{2}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 6\times 10^{-6}}{1.2^{2}}

F2 = 0.26 N

Write the forces in the vector form

\overrightarrow{F_{1}}=0.35\widehat{i}

\overrightarrow{F_{2}}=0.26\left ( Cos60 \widehat{i}+Sin60\widehat{j}\right )

\overrightarrow{F_{2}}=0.13 \widehat{i}+0.23\widehat{j}

Net force

\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}

\overrightarrow{F}=0.48 \widehat{i}+0.23\widehat{j}

Magnitude of the force

F=\sqrt{0.48^{2}+0.23^{2}}

F = 0.53 N

Direction of force with x axis

tan\theta =\frac{0.23}{0.48}

θ = 25.6°

7 0
3 years ago
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