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STatiana [176]
3 years ago
14

A coaxial cable consists of an inner conductor with radius ri = 0.20 cm and an outer radius of ro = 0.4 cm and has a length of 1

3 meters. Plastic, with a resistivity of ρ = 2.00 ✕ 1013 Ω · m, separates the two conductors. What is the resistance (in Ω) of the plastic when current is flowing between the two conductors of the coaxial cable?
Physics
1 answer:
N76 [4]3 years ago
4 0

Answer:

R = 1.69*10^{11} ohm

Explanation:

ri = 0.20cm

ro = 0.4 cm

length L = 13m

resistivity \rho = 2.00*10^13 ohm m

resistance can be determine by using following relation

R = \frac{\rho}{2\pi L} ln\frac{ro}{ri}

R = \frac{2.00*10^{13}}{2\pi *10} ln\frac{.004}{.002}

R = 1.69*10^{11} ohm

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Answer:

t_1 = \frac{v_i}{a_i}

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Explanation:

As v(t) = v_i + at, when the car is making full stop, v(t_1) = 0 . a = -a_i . Therefore,

0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}

Apply the same formula above, with v(t_2) = v_i and a = a_i, and the car is starting from 0 speed,  we have

v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}

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s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}

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After t time, the train would have traveled a distance of s(t) = v_i(t_1 + t_2) = 2v_it_1

Therefore, Δd would be 2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i

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