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3 years ago

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A car left skid marks 78 m long on a road as it slid to a stop. If the car's acceleration was -3.9 m/s2, what was its velocity b

efore it began to skid, to the nearest m/s?

1 answer:

spayn [35]3 years ago

7 0

It is 25 I’m sure of it

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Answer:

$q=1\times10^{-8}C$

Explanation:

Let the charge on the ball bearing is q.

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By use of Coulomb's law in electrostatics

$F=\frac{KQq}{d^{2}}$

By substituting the values

$0.018=\frac{9\times10^{9}\times20\times10^{-9}q}{0.01^{2}}$

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Thus, the charge on the ball bearing is $q=1\times10^{-8}C$

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