The displacement ∆S of the particle during the interval from t = 2sec to 4sec is; 210 sec
<h3>How to find the displacement?</h3>
We are given the velocity equation as;
s' = 40 - 3t²
Thus, the speed equation will be gotten by integration of the velocity equation to get;
s = ∫40 - 3t²
s = 40t - ¹/₂t³
Thus, the displacement between times of t = 2 sec and t = 4 sec is;
∆S = [40(4) - ¹/₂(4)³] - [40(2) - ¹/₂(2)³]
∆S = 210 m
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Answer:
Yes
Explanation:
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Answer:
The current through the coil is 2.05 A
Explanation:
Given;
number of turns of the coil, N = 1
radius of the coil, r = 9.8 cm = 0.098 m
magnetic moment of the coil, P = 6.2 x 10⁻² A m²
The magnetic moment is given by;
P = IA
Where;
I is the current through the coil
A is area of the coil = πr² = π(0.098)² = 0.03018 m²
The current through the coil is given by;
I = P / A
I = (6.2 x 10⁻² ) / (0.03018)
I = 2.05 A
Therefore, the current through the coil is 2.05 A