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3 years ago

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In a science fiction story, a microscopic black hole is given an enormous positive charge by firing an un-neutralized ion drive

exhaust (consisting of positively charged xenon ions) into it for six months. The idea behind the charging process is to be able to confine and manipulate this dangerous object with powerful electric fields. Suppose the charge on the black hole is 5740 C. At what distance from it is the electric potential equal to 1.09e 3 V

1 answer:

Olegator [25]3 years ago

6 0

Answer: distance d = 4.73e10m

Explanation: Suppose the charge on the black hole is 5740 C which is a positive charge.

Using electric potential V formula:

V = kq / d

Where K = 9.05×10^9Nm^2/C

And e = 1.6×10^-19C

But you don't need to substitute it.

1090 V = 8.99e9N·m²/C² * 5740C /d

Make d the subject of formula

d = 4.73e10 m

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Answer:

Refer to the attachment for the diagram.

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For $0 < x \le 1$ :

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For $2 < x \le 3$ :

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For $4 < x \le 5$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 4$ : $t = -3t + 14$ ; Hence the point on the segment: $(4, 2)$ .
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2 years ago

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OlgaM077 [116]

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tensa zangetsu [6.8K]

Answer:

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Sorry I didn't see this before...
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Does that help?

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2 years ago

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