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elena-s [515]
3 years ago
12

If a = 7 × 10−6 C/m4 and b = 1 m, find E at r = 0.6 m. The permittivity of a vacuum is 8.8542 × 10−12 C 2 /N · m2 . Answer in un

its of N/C
Physics
1 answer:
mart [117]3 years ago
7 0

Answer:

The electric field potential (E) is  1.75 X 10⁵ N/C

Explanation:

Electric field potential (E)  is force per unit charge.

Electric field potential (E) = force/charge = f/q, unit is Newton (N) per Coulomb (C); N/C

E = \frac{q}{4 \pi \epsilon r^2}

where;

q is charge in coulomb (C)

ε is permittivity of free space = 8.8542 X 10⁻¹² C²/Nm²

k = \frac{1}{4 \pi \epsilon} = \frac{1}{4\pi (8.8542 X 10^{-12})} = 8.99 X10^9 Nm^2/C^2

if a = 7 X 10⁻⁶ C/m⁴  and b = 1m

a in "C" = 7 X 10^-{6} \frac{C}{m^4} X (1m)^4 = 7 X 10^-{6} C

Electric field potential at r = 0.6m = \frac{ka}{r^2}

E = \frac{(8.99 X10^9)(7X 10^{-6})}{0.6^2} = 1.75 X 10⁵ N/C

Therefore, the electric field potential (E) is  1.75 X 10⁵ N/C

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A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d
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Answer:

2.09\ \text{m/s}

22298.4\ \text{J}

Explanation:

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As the momentum is conserved in the system we have

mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}

Speed of the three coupled cars after the collision is 2.09\ \text{m/s}.

As energy in the system is conserved we have

K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}

The kinetic energy lost during the collision is 22298.4\ \text{J}.

6 0
3 years ago
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Answer:

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Atom A and atom C are the same element.

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