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liubo4ka [24]
3 years ago
8

. Inside a conducting sphere of radius 1.2 m, there is a spherical cavity of radius 0.8 m. At the center of the cavity is a poin

t charge of -200 nC. There is a charge of +530 nC on the conducting sphere. Find the magnitude of the electric field at distance 1.35 m from the center of the sphere.
Physics
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

 E = 1,873 10³ N / C

Explanation:

For this exercise we can use Gauss's law

        Ф = E. dA = q_{int} / ε₀

Where q_{int} is the charge inside an artificial surface that surrounds the charged body, in this case with the body it has a spherical shape, the Gaussian surface is a wait with radius r = 1.35 m that is greater than the radius of the sphere.

The field lines of the sphere are parallel to the radii of the Gaussian surface so the scald product is reduced to the algebraic product.

        The surface of a sphere is

             A = 4π r²

             E 4π r² = q_{int} /ε₀

  The net charge within the Gauussian surface is the charge in the sphere of q1 = + 530 10⁻⁹ C and the point charge in the center q2 = -200 10⁻⁹ C, since all the charge can be considered in the center the net charge is

           q_{int} = q₁ + q₂

           q_{int} = (530 - 200) 10⁻⁹

           q_{int} = 330 10⁻⁹ C

The electric field is

             E = 1 / 4πε₀   q_{int} / r²

            k = 1 / 4πε₀

            E = k q_{int}/ r²

Let's calculate

           E = 8.99 10⁹   330 10⁻⁹/ 1.32²

           E = 1,873 10³ N / C

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Answer:

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v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s

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s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s

Time taken to reach the ground from the maximum height is 0.45 seconds

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3 years ago
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Answer:

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For a standing wave, the distance between two consecutive nodes is equal to half of the wavelength.

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MA_775_DIABLO [31]

<em>There are some placeholders in the expression, but they can be safely assumed</em>

Answer:

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<u>Sinusoidal Waves </u>

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