Answer:
<em> -11,813.87N </em>
Explanation:
According to coulombs law, the Force between the two charges is expressed as;
F = kq1q2/d²
k is the coulombs constant = 9*10⁹kg⋅m³⋅s⁻²⋅C⁻².
q1 = -0.00067 C
q2 = 0.00096 C
d = 0.7m
Substitute into the formula:
F = 9*10^9 * -0.00067 * 0.00096/0.7²
F = 9*10⁹*-6.7*10⁻⁴*9.6*10⁻⁴/0.49
F = -578.88*10⁹⁻⁸/0.49
F = -578.88*10/0.49
F = -5788.8/0.49
F = -11,813.87N
<em>Hence the force between the two charges is -11,813.87N </em>
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Answer:
C because you can test it your self
We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.