Tangential acceleration of a point on the rim of the flywheel during this spin-up process is 0.2548 m/s².
Tangential acceleration is defined as the rate of change of tangential velocity of the matter in the circular path.
Given,
Radius of flywheel (r) = 1.96 cm = 0.0196m
Angular acceleration (α)= 13.0 rad/s²
The tangential acceleration formula is at=rα
where, α is the angular acceleration, and r is the radius of the circle.
using the formula; at=rα = (13.0 rad/s²) (0.0196m) = 0.2548 m/s².
The tangential acceleration is 0.2548 m/s².
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The answers to the problem are as follows:
MA= 5
IMA= input distance/ output distance, 5
<span>AMA= output force/ input force, 5/2
</span>
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<span>To relate or measure the by the quantity of something, not against the quantity</span>
The best possible Answer is c
