A particle falling at its terminal speed vterm is in dynamic equilibrium with no net force. Use Newton's second law for this par
1 answer:
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Answer:
Explanation:
For this interesting problem, we use the definition of centripetal acceleration
a = v² / r
angular and linear velocity are related
v = w r
we substitute
a = w² r
the rectangular body rotates at an angular velocity w
We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A
the distance OB = L₂
the distance AB = L₁
the sides of the rectangle
It is indicated that the acceleration in in A and B are related
we substitute the value of the acceleration
w² r_A = n r_B
the distance from the each corner is
r_B = L₂
r_A =
we substitute
\sqrt{L_1^2 + L_2^2} = n L₂
L₁² + L₂² = n² L₂²
L₁² = (n²-1) L₂²
Answer:
The frictional force is
Explanation:
From the question we are told that
The coefficient of kinetic force is μk = 0.35
The normal force felt by the puck is
Generally the frictional force that acts on the puck is mathematically represented as
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