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Angular acceleration is 80 rad/s²
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Number of revolutions undergone is 1.02</h2>
Explanation:
We have equation of motion v = u + at
Initial angular velocity, u = 0 rad/s
Final angular velocity, v = 32 rad/s
Time, t = 0.40 s
Substituting
v = u + at
32 = 0 + a x 0.40
a = 80 rad/s²
Angular acceleration is 80 rad/s²
We have equation of motion s = ut + 0.5 at²
Initial angular velocity, u = 0 rad/s
Angular acceleration, a = 80 rad/s²
Time, t = 0.4 s
Substituting
s = ut + 0.5 at²
s = 0 x 0.4 + 0.5 x 80 x 0.4²
s = 6.4 rad
Angular displacement = 6.4 rad

Number of revolutions undergone is 1.02
Answer:
F = -307.4 N
Explanation:
It is given that,
Mass of the baseball, m = 0.145 kg
Initial speed of the baseball, u = 60 m/s
Final speed of the baseball, 
Time of contact, 
(a) It is assumed to find the horizontal component of average force. It is given by :
F = -307.4 N
So, the horizontal component of average force is 307.4 N. Hence, this is the required solution.
Answer:
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Answer:
V = 2.87 m/s
Explanation:
The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.
Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:
Centripetal acceleration = V^2 / r
where r is the distance of water from the pivot or shoulder.
For our case, r will be 0.65 + 0.19 = 0.84 m
and solving the above equation we get:
9.81 = V^2 / 0.84
V^2 = 8.2404
V = 2.87 m/s