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erik [133]
2 years ago
10

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4cm apart with a 28kV potential d

ifference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.
1. What is the electric field strength between the plates?
2. With what speed does an electron exit the electron gun if its entry speed is close to zero?
Physics
1 answer:
yulyashka [42]2 years ago
3 0

Answer:

a)   F = 3.2 10⁻¹⁰ N , b)       v = 9.9 10⁷ m / s

Explanation:

a) The electric force is

       F = q E

The electric field is related to the potential reference

     V = E d

     E = V / d

Let's replace

    F = e V / d

Let's calculate

    F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²

    F = 3.2 10⁻¹⁰ N

b) For this part we can use kinematics

          v² = v₀ + 2 a d

          v = √ 2 ad

Acceleration can be found with Newton's second law

        e V / d = m a

        a = e / m V / d

        a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²

        a = 3,516 10⁻¹⁷ m / s²

Let's calculate the speed

       v = √ (2 3,516 10¹⁷ 1.4 10⁻²)

       v = √ (98,448 10¹⁴)

       v = 9.9 10⁷ m / s

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Explanation:

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Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 2
Mila [183]

Answer:

The angular acceleration is 11.66 rad/s²

Explanation:

Step 1: Given data

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Step 2: Find the magnitude of the angular acceleration

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τ = F2*R2 - F1*R1 = 24N*0.22m - 19N*0.10m = 3.38 N*m

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I = ½mr² = 0.29 kg*m²

τ = I*α  =  3.38 N*m

OR

0.29 kg*m² * α = 3.38 N*m

α = 11.655 rad/s²  ≈11.66 rad/s²

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Answer:

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Explanation:

The angle of incidence is given as 32.7°

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We have to find the angle at which the light leave the water means angle of refraction

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