A cat dozes on a stationary merry-go-round, at a radius of 5.7 m from the center of the ride. Then the operator turns on the rid
1 answer:
Answer:
0.76
Explanation:
we are given:
radius (r) =5.7 m
speed (s) = 1 revolution in 5.5 seconds
acceleration due to gravity (g) = 9.8 m/s^{2}
coefficient of friction (Uk) = ?
we can get the minimum coefficient of friction from the equation below
centrifugal force = frictional force
m x r x ω^{2} = Uk x m x g
r x ω^{2} = Uk x g
Uk =
where ω (angular velocity) =
= = 1.14
Uk = = 0.76
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Answer:
757,93 feets
Explanation:
We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.
The angle of depression, its O in the graph.
Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.
ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).
Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.
Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.
We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:
This its equal to the distance from the boat to the shoreline.
Answer:
Radial acceleration of moon is
Explanation:
Given :
Time period sec
Distance from center of moon to planet m
From the equation of radial acceleration,
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So
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