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4 years ago

1. An asteroid is traveling in a straight line at a constant speed of 327 m/s. How long will it take

Physics

1 answer:

Lady bird [3.3K]4 years ago

4 0

Answer:

t = 37.3 minute.

Explanation:

Given that,

The speed of an asteroid is 327 m/s

Displacement, d = 732 km

We need to find time. Speed is equal to distance divided by time. So,

$t=\dfrac{732\times 10^3\ m}{327\ s}\\\\t=2238.53\ s$

1 minute = 60 seconds

So,

t = 37.3 minute

So, it will take 37.3 minute.

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$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

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$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

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$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

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