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zalisa [80]
3 years ago
8

The attraction will vary directly with the separation between the charges.

Physics
1 answer:
Burka [1]3 years ago
4 0
No it won't. It'll vary inversely as the square of the separation.
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A long distance runner sees the finish line and accelerates at a rate in 1.2 m/s2 for
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Answer: he did travel 15 meters.

Explanation:

We have the data:

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Then we start writing the acceleration:

a(t) = 1.2 m/s^2

now for the velocity, we integrate over time:

v(t) = (1.2 m/s^2)*t + v0

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v(t) = (1.2 m/s^2)*t + 3.2 m/s

For the position, we integrate again.

p(t) = (1/2)*(1.2 m/s^2)*t^2 + 3.2m/s*t + p0

Because we want to know the displacementin those 3 seconds ( p(3s) - p(0s)) we can use p0 = 0m

Then the displacement at t = 3s will be equal to p(3s).

p(3s) = (1/2)*(1.2 m/s^2)*(3s)^2 + 3.2m/s*3s = 15m

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