I - ok. Czechoslovakia is officially taken.
Answer:
the final angular velocity of the platform with its load is 1.0356 rad/s
Explanation:
Given that;
mass of circular platform m = 97.1 kg
Initial angular velocity of platform ω₀ = 1.63 rad/s
mass of banana = 8.97 kg
at distance r = 4/5 { radius of platform }
mass of monkey = 22.1 kg
at edge = R
R = 1.73 m
now since there is No external Torque
Angular momentum will be conserved, so;
mR²/2 × ω₀ = [ mR²/2 + ( R)² + R² ]w
m/2 × ω₀ = [ m/2 + ( )² + ]w
we substitute
w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1
w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )
w = 48.55 × [ 1.63 / ( 76.3908 ) ]
w = 48.55 × 0.02133
w = 1.0356 rad/s
Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s
Distance = 2AU / tan1.0
If you mean 1.0 is in degrees, then Distance = 114.58 AU
Y - yo = Vo*t - g * (t^2) / 2
Vo = - 9.0 m/s
t = 0.50 s
=> y - yo = -9.0 m/s * 0.5 s - 9.8 m/s^2 * (0.5s)^2 / 2 = - 4.5m - 1.225m = - 5.725 m.
Answer: option c) - 5.7