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Tems11 [23]
3 years ago
12

A point charge of 4 µC is located at the center of a sphere with a radius of 25 cm. Find the electric flux through the surface o

f the sphere. The Coulomb constant is 8.98755 × 109 N · m2 /C 2 and the acceleration due to gravity is 9.8 m/s 2 .Answer in units of N · m2 /C.
Physics
1 answer:
givi [52]3 years ago
5 0

Answer:

Ф = 4.5176x10⁵ N . m² / C

Explanation:

In this case, we need to use two expressions in order to calculate the electric flux of the sphere.

First the Electric flux is calculated using this expression:

Ф = E * A  (1)

Where:

E: Electric field

A: Area of the sphere

To get the electric field E, we use this expression:

E = K * q / r²    (2)

If we replace (2) into (1) we have the following:

Ф = K * q * A / r²    (3)

Finally, we need to know the expression to get the area of a sphere which is the following:

A = 4πr²   (4)

Replacing into (3):}

Ф = K * Q * 4πr² / r²       discarting r²:

Ф = K * Q * 4π   (5)

Now, all we need to do is replace the given values and solve for the electric flux of the sphere:

Ф = 8.98755x10⁹ * 4x10⁻⁶ * 4 * π

<h2>Ф = 4.5176x10⁵ N . m² / C</h2>

Hope this helps

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Answer:

A) Emin = eV

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Explanation:

A)

Energy of electron is the product of electron charge and the applied potential difference.

The energy of an electron in this electric field with potential difference V will be eV. Since this is the least energy that the electron must reach to break out, then the minimum energy required by this electron will be;

Emin = eV

B)

The maximum stopping potential energy is eVo,

The energy of the electron due to the light is E_light.

If the minimum energy electron must posses is Φ, then the minimum energy electron must have to reach the detectors will be equal to the energy of the light minus the maximum stopping potential energy

Φ = E_light - eVo

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eVo = E_light - Φ

Vo = (E_light - Φ) ÷ e

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What are three types of variables in a controlled experiments?
blsea [12.9K]

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A sphere of volume 1.20×10−3m3 hangs from a cable. When the sphere is completely submerged in water, the tension in the cable is
KATRIN_1 [288]

Answer:

B = 62.9 N

Explanation:

This is an exercise on Archimedes' principle, where the thrust force equals the weight of the  liquid

         B = ρ g V

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         T + B -W = 0

         B = W- T               (1)

use the density to write the weight

         ρ = m / V

        m = ρ V

         W = ρ g V

substitute in  1

         B = m g -T

         B = \rho_{body} g V - T

To finish the calculation, the density of the material must be known, suppose it is steel  \rho_{body} = 7850 kg / m³

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         B = 7850 9.8 1.20 10⁻³ - 29.4

          B = 92.3 - 29.4

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3 years ago
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
Pavel [41]

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

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(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

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From law of conservation of energy,

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(negative next to g because it's going against gravity)

Thus;

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Plugging in the relevant values;

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(V_A) = √981

(V_A) = 31.32 m/s

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