Answer:2.55 rad/s
Explanation:
Given
Diameter of ride=5 m
radius(r)=2.5 m
Static friction coefficient range=0.60-1
Here Frictional force will balance weight
And limiting frictional force is provided by Centripetal force

weight of object=mg
Equating two
f=mg




Answer:
The time taken will be 0.553 seconds.
Explanation:
We should start off by finding the force exerted by the rope on the 3kg weight in this case.
Weight of 5kg mass = 5 * 9.81 = 49.05 N
Weight of 3kg mass = 3 * 9.81 = 29.43 N
The force acting upward on the 3kg mass will equal the weight of the 5kg mass. Thus the resultant force acting on the 3kg mass is:
Total force = 49.05 - 29.43 = 19.62 N (upwards)
We can now find the acceleration:
F = m * a
19.62 = 3 * a
a = 6.54 m/s^2
We now use the following equation of motion to get the time taken to travel 1 meter:


t = 0.553 seconds
The average velocity of the man is determined as 1.25 m/s.
<h3>
Average velocity of the man</h3>
Total distance traveled by the man = 100 m
If 1 complete cycle = 4 s
20 complete cycle = ?
= 20 x 4 s
= 80 s
Average velocity = total distance/total time
Average velocity = 100 m/80 s
Average velocity = 1.25 m/s
Thus, the average velocity of the man is determined as 1.25 m/s.
Learn more about average velocity here: brainly.com/question/6504879
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Answer:
The tension in string P is 25 N, while that of Q is 85 N.
Explanation:
Considering the conditions for equilibrium,
i. Total upward force = Total downward force
+
= 110 N
ii. Taking moment about P,
clockwise moment = anticlockwise moment
110 × (2.5 - 0.8) =
× (3 - 0.8)
110 × 1.7 =
× 2.2
187 = 2.2
= 
= 85 N
From the first condition,
+
= 110 N
+ 85 N = 110 N
= 110 - 85
25 N
Therefore, the tension in string P is 25 N while that of Q is 85 N.