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Crank
2 years ago
6

Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc

tor, with respect to the outer conductor, is +600 V. An electron is released from rest at the surface of the outer conductor. What is the speed of the electron as it reaches the inner conductor? (e = 1.60 × 10-19 C, m el = 9.11 x 10-31 kg, k = 1/4π ε0 = 8.99 × 109 N · m2/C2)
Physics
1 answer:
xxMikexx [17]2 years ago
3 0

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is  equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2  v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

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Answer:

The child's mass is 14.133 kg

Explanation:

From the principle of conservation of linear momentum, we have;

(m₁ + m₂) × v₁ + m₃ × v₂ = (m₁ + m₂)  × v₃ - m₃ × v₄

We include the negative sign as the velocities were given as moving in the opposite directions

Since the child and the ball are at rest, we have;

v₁ = 0 m/s and v₂= 0 m/s

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0 = m₁ × v₃ - m₂ × v₄

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Where:

m₁ = Mass of the child

m₂ = Mass of the scooter = 2.4 kg

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m₃ = Mass of the ball = 2.4 kg

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Plugging the values gives;

(m₁ + 2.4)× 0.45 = 2.4 × 3.1

(m₁ + 2.4) = 16.533

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Option C is the correct answer which says; KE is has a maximum value at position 2. EPE has a maximum value at position 3. GPE has a maximum value at position 1.

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Option C fulfills that and therefore will be the correct answer.

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