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siniylev [52]
3 years ago
9

A thin spherical shell of radius 7.6 cm carries a uniform surface charge density of 6.7 times 10-9 C/m2. The magnitude of the el

ectric field (in N/C) at r = 7.8 cm is approximately:___________.
Physics
1 answer:
zhannawk [14.2K]3 years ago
5 0

Answer:

 E= 1.968 10¹¹  N/C

Explanation:

To calculate the electric field of a sphere we can use Gaussian law

        Ф = ∫ E. dA = q_{int}/ ε₀

In this case we must use a Gaussian surface of spherical shape, since the radii of the sphere and the electric field line have been parallel, consequently the scalar product is reduced to an algebraic product.

         The error of a sphere is

          A = 4π π r²

         E (4π r³) = q_{int} / εo                           (1)

to find the charge inside the sera we use the concept of density

         ρ = quint / V

         qint = rho V

the volume of a sphere is

          V = 4/3 pi r³

we substitute in 1

    E 4π r² = ρ 4/3 π r³

     E = ρ / 3ε₀

now let's change the charge density to the charge

      E = Q 3 / 4π r³    r / 3ε₀

       E = (1 / 4πε₀) Q / r²

       E = k Q /r²

calculate

        E = ρr / 3ε₀

       E = 6.7 10⁻⁹  7.8  10⁻² / 3 8.85 10⁻¹²

         E= 1.968 10¹¹  N/C

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The core of a certain reflected reactor consist of a cylinder 10 ft high 10 ft in diameter The measured maximum-to-average flux
Westkost [7]

Answer:

The maximum power density in the reactor is 37.562 KW/L.

Explanation:

Given that,

Height = 10 ft = 3.048 m

Diameter = 10 ft = 3.048 m

Flux = 1.5

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V =\pi r^2 h

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V=\pi\times(1.524)^2\times 3.048

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Using formula of power density

P=\dfrac{E}{V}

Where, P = power density

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V = volume

Put the value into the formula

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6 0
3 years ago
n explosion breaks an object initially at rest into two pieces, one of which has 2.0 times the mass of the other. If 8400 J of k
ANTONII [103]

Answer:

The heavier piece acquired 2800 J  kinetic energy

Explanation:

From the principle of conservation of linear momentum:

0 = M₁v₁ - M₂v₂

M₁v₁ = M₂v₂

let the second piece be the heavier mass, then

M₁v₁ = (2M₁)v₂

v₁  = 2v₂ and v₂ = ¹/₂ v₁

From the principle of conservation of kinetic energy:

¹/₂ K.E₁ + ¹/₂ K.E₂ = 8400 J

¹/₂ M₁(v₁)² + ¹/₂ (2M₁)(¹/₂v₁)² = 8400

¹/₂ M₁(v₁)² + ¹/₄M₁(v₁)² = 8400

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Now, we determine K.E₁ and note that K.E₂ = ¹/₂K.E₁

1.5 K.E₁ = 8400

K.E₁ = 8400/1.5

K.E₁ = 5600 J

K.E₂ = ¹/₂K.E₁ = 0.5*5600 J = 2800 J

Therefore, the heavier piece acquired 2800 J  kinetic energy

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