Answer:
I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).
The moment of inertia about the center of a sphere is 2 / 5 M R^2.
By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is I = 2/5 M R^2 + M R^2 = 7/5 M R^2
I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2
Answer:
The magnitude of the impulse is 1.33 kg m/s
Explanation:
please look at the solution in the attached Word file
Answer:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.
Explanation:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.
B, since it is the only one that actually conserves matter for certain. In each of the others, matter could still be imbalanced, since for A, for example, it could be 5 Carbons on the right and 5 Chlorines on the left, and that would not balance.
Answer:
is that rm in you profile pic
