Apoint on a wheel of radius 40 cm that is rotating at a constant 5.0 revolutions per second is located 0.20 m from the axis of rotation. what is the acceleration of that point due to the spin of the wheel?
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Ответ:
Explanation:
The centripetal acceleration of a point moving by circular motion is given by:
where
r is the distance from the axis of rotation
The point on the wheel makes 5.0 revolutions per second, so the frequency is
and the angular velocity is
While the distance of the point from the axis of rotation is
Substituting, we find the acceleration:
Ответ:
Answer
Correct answer is C ( 43.9 degrees)
Explanation:
In this problem we have given
weight of photo frame=4.16N
Tension=3N
It is clear that the photo frame is stationary, It means the wires will balance the force due to gravity (the weight).
It means the vertical components of the tension in the wires is equal the total weight.
Because the wires are symmetric, therefore weight of photo frame will be shared equally between wires.
It mean weight supported by each wire=4.16/2
=2.08 N
From figure it is clear that
sin(angle) = 2.08 / 3
angle = arcsin(2.08/3)
angle = 43.9 degrees
Hope this answer will help you