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2 years ago

I need help getting started with number 8.

Physics

1 answer:

ratelena [41]2 years ago

4 0

In #8, the distance and the (magnitude of displacement) are equal, because he crawled in a straight line.

Displacement = (straight-line distance from start-point to end-point) in the direction from start to end, regardless of what route was actually followed.

Displacement = 5m, in the negative direction.

In #9 . . . distance will be the same. Displacement is going to be the same magnitude, but in the positive direction.

This is so simple that it's hard to talk about.

In #8, "What was the bug's distance ?". "Distance was 5 meters.". "What was the bug's displacement ?", "Displacement was 5 meters backwards."

In #9, What was the bug's distance ?". "Distance was 5 meters.". "What was the bug's displacement ?", "Displacement was 5 meters forward."

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Answer:

Correct answer: t = 2.86 seconds

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We will calculate the acceleration from this formula

a = (V² - V₀²) / (2 d) = (2.5² - 1²) / (2 · 5) = (6.25 - 1) / 10 = 5.25 / 10

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3 years ago

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We have all the charges for q1, q2, and q3.
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Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.

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(a) On the coil: 20 V, on the resistor: 0 V

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$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

Therefore, the potential difference across the resistor will be

$V_R = V-V_L = 20 V- 0 = 20 V$

(c) Yes

The two voltages will be equal when:

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$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

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Here we have that the current in the circuit is

$I_0 = 3.20 A$

The problem says this current is stable: this means that we are in a situation in which $t>>\tau$ , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

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Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

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We find that

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