(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: T= 715 N
Explanation:
The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:
T = mv² / r
At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.
So the kinetic energy will be the following:
K = 1/2 m v² = 15. 0 J
Solving for v², and replacing in the expression for T:
T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N
Answer:
E = 24000 N/C = 24 KN/C
Explanation:
The electric field experienced by a test charge is given by the following formula:
where,
E = Electric Field = ?
F = Force of attraction = 3 x 10⁻⁶ N
q = Charge on piece of lint = 1.25 x 10⁻¹⁰ C
Therefore, using these values in the equation, we get:
<u>E = 24000 N/C = 24 KN/C</u>
