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
A group of cells together
Answer:Same magnitude
Explanation:
When ball is dropped from shoulder height h then velocity at the bottom is given by
if it makes elastic collision then it will acquire the same velocity and riser up to the same height
If m is the mass of ball then impulse imparted is given by
Thus impulse imparted by gravity and Floor will have same magnitude of impulse but direction will be opposite to each other.
You will get 20460000 as your answer which is broken down into, 2.046 x 10^7 as your number has to be between 1-10.
Answer:
Explanation:
From the question we are told that:
Frictional force 
Coefficient of kinetic friction 
Generally the equation for Normal for is mathematically given by
Therefore
