Question: A thin, uniform rod is bent into a square<span> of... A thin, uniform rod is bent into a </span>square<span> of side length a. </span>If the total mass is M<span>, </span>find the moment of inertia about an axis through the center and perpendicular to the plane of the square<span>. </span>Use the parallel-axis theorem<span>.</span>
Answer:
a = 8 m/s^2, Ffriction = 10 N, μk = 0.205
Explanation:
a. Force = Mass*Acceleration,
(since you didn't add the units..."5 block"....for the mass, I will assume it to be in kg, per SI units)
40 N = 5 kg*acceleration,
a = 40/5 = 8 m/s^2
b. As you know newtons second law (F=m*a) is actually in the form Fnet = m*a. Which means that if the friction force comes into play, it would be Fapplied - Ffriction = m*a.
Fapplied - Ffriction = m*a,
40 - Ffriction = 5*6,
40 - Ffriction = 30,
Ffriction = 40 - 30 = 10 N
c. The coefficient of kinetic friction is calculated by the formula "Ffriction = μk*Fnormal".
10 = μk*Fnormal (Fnormal = m*g = 5*9.8)
10 = μk*49,
μk=10/49 ≈ 0.205
Answer:
The period of a pendulum does not depend on the mass of the ball, but only on the length of the string. Two pendula with different masses but the same length will have the same period. Two pendula with different lengths will different periods; the pendulum with the longer string will have the longer period.
Explanation:
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