The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
F = 8.98755e9N·m²/C² * -(7e-6C)² / (0.03m)² = -489N, the negative sign denotes attraction.

4 0

3 years ago

A softball player moving 3.89 m/s

Ahat [919]

Answer:

0.119 s

Explanation:

Given that

$U=3.89\ m/s\\a=-1.44\ m/s^2\\S=4.8\ m$

Lets take final speed of the softball after covering 4.8 m = V

We know that

$V^2=U^2+2aS\\V^2=3.89^2-2\times 1.44\times 4.8\\V=3.7\ m/s$

Also We know that

$V=U+at\\$

Putting the value of V ,U and a in the previous equation We get

$3.7=3.89-1.44\times t\\t=0.119\ s$

Therefore slide time will be 0.119 s

3 0

3 years ago

Action reaction forces are in which one of newtons laws

Klio2033 [76]

Answer:

the third law

Explanation:

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4 0

3 years ago