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kirza4 [7]
4 years ago
10

Two flat 4.0 cm 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite each

other. Between the electrodes but not near their edges, the electric field strength is 5.5 106 N/C. What is the magnitude of the charge on each electrode? (? 0 = 8.85 10-12 C2/N m2)
Physics
1 answer:
BabaBlast [244]4 years ago
8 0

Answer:

The magnitude of the charge on each electrode is 77.8 nC.

Explanation:

Given that,

Length = 4.0 cm

Distance = 2.0 mm

Electric field strength E=5.5\times10^{6}\ N/C

We need to calculate the magnitude of the charge on each electrode

Using formula of electric field

E=\dfrac{\sigma}{\epsilon_{0}}

Where, \sigma = charge density

We know that,

The charge density is

\sigma=\dfrac{Q}{A}

Put the value of charge density into the formula of electric field

E=\dfrac{Q}{A\epsilon_{0}}

Q=EA\epsilon_{0}

Put the value into the formula

Q=5.5\times10^{6}\times4.0\times4.0\times10^{-4}\times8.85\times10^{-12}

Q=77.8\times10^{-9}\ C

Q=77.8\ nC

Hence, The magnitude of the charge on each electrode is 77.8 nC.

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A rocket is moving away from the solar system at a speed of 7.5 ✕ 103 m/s. It fires its engine, which ejects exhaust with a spee
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Answer:

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b.mr=48kg/s

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A rocket is moving away from the solar system at a speed of 7.5 ✕ 103 m/s. It fires its engine, which ejects exhaust with a speed of 5.0 ✕ 103 m/s relative to the rocket. The mass of the rocket at this time is 6.0 ✕ 104 kg, and its acceleration is 4.0 m/s2. What is the velocity of the exhaust relative to the solar system? (B) At what rate was the exhaust ejected during the firing?

velocity of the exhaust relative to the solar system

velocity of the rocket -velocity of the exhaust relative to the rocket.

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2.5x 10^3 m/s

. b  we will look for the thrust of the rocket

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5 0
3 years ago
An interference pattern is produced by light with a wavelength 520 nm from a distant source incident on two identical parallel s
Goshia [24]

Answer:

1) θ = 0.00118 rad, 2)  θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4)  I / Io = 0.216

Explanation:

In the double-slit interference phenomenon it is explained for constructive interference by the equation

          d sin θ = m λ

1) the first order maximum occurs for m = 1

           sin θ = λ  / d

           θ = sin⁻¹ λ  / d

let's reduce the magnitudes to the SI system

           λ  = 520 nm = 520 10⁻⁹  θ = 0.00118 radm

           d = 0.440 mm = 0.440 10⁻³ m ³

let's calculate

           θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)

            θ = sin⁻¹ (1.18 10⁻³)

            θ = 0.00118 rad

2) the second order maximum occurs for m = 2

            θ = sin⁻¹ (m λ  / d)

            θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)

            θ = 0.00236 rad

3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains

          I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )

where the function sinc = sin x / x

and b is the width of the slits

we caption the values

             x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)

             x = 2.21

            I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²

remember angles are in radians

            I / I₀ = cos² (3.0945) [0.363] 2

            I / I₀ = 0.9978 0.1318

            I / I₀ = 0.1738

4) the maximum second intensity is

            I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)

            x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)

            x = 4.41

            I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²

            I / Io = cos² 6.273    0.216

            I / Io = 0.216

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