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AleksandrR [38]
2 years ago
14

Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p

lates. What is the magnitude of the electric force on q1?
Physics
1 answer:
ira [324]2 years ago
8 0

Answer:

The force exerted on the q_1 is  F =  2.25*10^{3} \ N

Explanation:

From the question we are told that

     The area is  A =  2.34*10^{-3} \ m^2

     The magnitude of charge placed on them is  q =  7.07 * 10^{-7} C

     The charge placed between the plate is q_1 = 6.62 *10^{-5} C

   

The electric field generated around the plate  is mathematically represented as

           E =  \frac{q}{A \epsilon_o}

Substituting values

          E =  \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}

         E = 34*10^{6} \ V/m

The force exerted the charge q_1 is  mathematically represented as

        F =  q_1 * E

Substituting values  

        F =  6.62 *10^{-5} * 34*10^{6}

        F =  2.25*10^{3} \ N

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Answer:

The answers and workings is in the Explanation section

Explanation:

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According to Ohm’s law V =I*R  

Where V = Voltage, I = Current and R = Resistance

I = V/R =6/3 =2 Amps of current

Answer = 2 Amps

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Since the second lamp is connected in series with the first, the total resistance will R₂ = 3 + 3 = 6Ω

2 resistance in series  R₂ =  6Ω

The current in the circuit with the two lamps connected in series is I₂ =V/R₂ =6/6 = 1 Amps

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Since the third lamp is connected in series with the first and second, the total resistance will R₃ = 3 + 3 + 3 = 9Ω

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The current in the circuit with the three lamps connected in series is

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The current through the 3 lamps I₃ = 0.67 Amps

Answer =  9Ω  and 0.67 Amps

<em>The current through each individual lamp is __________ Amps.  </em>

Since all 3 lamps are connected in series, the same current will flow through each of the  3 lamps, and that current is I₃  

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The formula for power P = I*V =120*3 = 360 Watts

power P  = 360 Watts

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From P =I*V, make I the subject of the formula, I = P/V =90/120 = 0.75

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Answer =  0.75 Amps

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Current I =P/V = 75/120 = 0.625 Amps.

Answer = 0.625 Amps

<em>If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it? </em>

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24 hours make 1 day, so the number of hours the bulb was left on = 24 *30 = 720 hours

The power rating of the bulb is 60W = 60/1000 = 0.06 KiloWatt

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Answer:

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Explanation:

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E=\dfrac{\Delta V}{r}

\Delta V=\dfrac{1}{4}\dfrac{m_pv_p^2}{e}

So

E=\dfrac{1}{4}\dfrac{m_pv_p^2}{e.r_e}

Now by putting the all given values in the questions

E=\dfrac{1}{4}\times \dfrac{5.18\times 10^{-26}\times 180^2}{1.6\times 10^{-19}\times 0.46}

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2 years ago
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Answer:

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Explanation:

 

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Now let's use Newton's equilibrium relationship

         B - W = 0

         B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

         σ = W / A

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The area of ​​a sphere is

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The volume of a sphere is

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Let's replace

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    P = ρ RT

    ρ = P / RT

 

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    r² 4π (P/3RT  r - σ) = W₁

Let's replace the values

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As the independent term is very small we can despise it, to find the solution

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3 0
2 years ago
A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad
maksim [4K]

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

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Replacing this values at our previous equation we have,

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Answer:

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3 0
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