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
Here we know that
now from kinematics we have
now from above all values we have
so final angular speed is -12.6 rad/s
Infrared radiation<span> lies between the </span>visible<span> and microwave portions of the electromagnetic spectrum. Infrared waves have wavelengths longer </span>than visible<span> and shorter </span>than<span> microwaves, and have </span>frequencies<span> which are lower </span>than visible<span> and </span>higher than<span> microwaves.</span>
Answer:
0.19m/s²
Explanation:
Initial velocity(u) = 50×1000/60×60
=13.88 m/s
Final velocity(v) = 36.5×1000/60×60
=10.13 m/s
Acceleration(a) = v-u/t
=10.13-13.88/19.5
a= -0.19m/s²
-a = 0.19m/s²
The magnitude of retar dation is 0.19m/s²
I think in parallel circuits.
