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The answer is Codominance, AB blood cells inherit both A and B blood types so they are codominance
Explanation:
It is given that,
Mass of person, m = 70 kg
Radius of merry go round, r = 2.9 m
The moment of inertia, 
Initial angular velocity of the platform, 
Part A,
Let
is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

Here, 


Solving the above equation, we get the value as :

Part B,
The initial rotational kinetic energy is given by :



The final rotational kinetic energy is given by :



Hence, this is the required solution.
Consider velocity to the right as positive.
First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right
Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left
Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s
Let v (m/s) be the velocity of the center of mass of the 2-block system.
Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s
Answer:
The center of mass is moving at 1.33 m/s to the left.
Answer:
B. decreases while his angular speed remains unchanged.
Explanation:
His angular speed will always be the same as the wheel's angular speed, which remains constant as it's in uniform motion. As for linear speed, which is defined as the product of angular speed and distance r to the center of rotation, and his distance to center is decreasing, his linear speed must be decreasing as well.