Answer:
Explanation:
It is given that,
Speed of one quoll around a curve, v = 3.2 m/s (maximum speed)
Radius of the curve, r = 1.4 m
On the curve, the centripetal force is balanced by the frictional force such that the coefficient of frictional is given by :
So, the coefficient of static friction between the quoll's feet and the ground in this trial is 0.74. Hence, this is the required solution.
The calculation of the centripetal acceleration of an object following a circular path is based on the equation,
a = v² / r
where a is the acceleration, v is the velocity, and r is the radius.
Substituting the known values from the given above,
4.4 m/s² = (15 m/s)² / r
The value of r from the equation is 51.14 m.
Answer: 51.14 m
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Can you add a picture or type the options down?
Answer:
When the tree is at rest, its leaves are also at rest. But when we bring the tree in motion by means of shaking it, due to the inertia of leaves they still tend to be in rest. Thus, force is acting on leaves vigorously with changing direction rapidly. This results in detaching of leaves from the tree.
Explanation:
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