Here stress is parallel to the surface of the body. So it's a Shear stress.
(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s²
<span>(B) </span>
<span>(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev </span>
<span>(C) </span>
<span>(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s</span>
Answer:
Magnitude of the force between the charges is F = 1.92×10^20N
Explanation:
Given the magnitude of force according to coulombs law
F =K[(q1*q2)/r2]
Where q1 and q2 are the charges
r is the distance between the charges
K is the coulombs constant
Substituting the given values, we have;
F = 8.98×10^9 × 1.5×10^6 × 3.2×10^4/1.5²
F = 43.1×10^19/2.25
F = 19.16×10^19N
F = 1.92×10^20N
