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Schach [20]
3 years ago
8

1-Calculate Req

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
3 0

Answer:

1. 21.66 Ohms

2.  3.38 A

3. 6.7 V

Explanation:

1. Req = 6+2 = 8 Ohms (2 and 6 are in a series circuit)

  Req = 1/8 +1/4 = 3/8 = 8/3 = 2.66 Ohms (8 and 4 are parallel, so we will add them using this equation)

  Req = 2.66 + 1 + 9 + 3 + 6 = 21.66 Ohms

2. I = V/R = 9/2.66 = 3.38 A (In a series circuit, the current is the same across the resistors, so we will add them and divided them by 9 volts)

3. V = IR = 3.38 x 2 = 6.7 V (In a series circuit, the voltage is different, so each resistor will have a different voltage.)

I hope this helps.  I am not an expert in physics but its ok :)

<u><em>Note: If the answer benefited u, mark me as the brainliest answer if u can, thx.</em></u>

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Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixe
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Answer:

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}\ J.

Explanation:

Given charges are:

\rm q_1 = +6e.\\q_2 = -6e.

The electric potential energy of a charge due to the electric field of another charge is given by

\rm EPE=\dfrac{kq_1q_2}{r}.

where,

  • k = Coulomb's constant, having value = \rm 9\times 10^9\ Nm^2/C^2.
  • r = distance between the charges.

When the charges are infinite distance apart, \rm r = \infty,

\rm EPE_{initial} = \dfrac{kq_1q_2}{r}=0\ J.

When the charges are \rm 5.61\times 10^{-12}\ m apart, \rm r=5.61\times 10^{-12}\ m,

\rm EPE_{final}=\dfrac{kq_1q_2}{r}\\=\dfrac{(9\times 10^9)\times (+6e)\times (-6e)}{5.61\times 10^{-12}}\\=-5.775\ e^2\times 10^{22}.

Here, e is the charge on one electron, such that, \rm e = -1.6\times 10^{-19}\ C.

Therefore,

\rm EPE_{final}=-5.775\times (-1.6\times 10^{-19})^2\times 10^{22} = -1.478\times 10^{-15}\ J.

Thus,

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}-0=-1.478\times 10^{-15}\ J.

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Answer:

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Explanation:

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