How is angular momentum conserved in this proton+water scenario?
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hey mate!!!!!
good question
good answer :-
Oxygen, the proton is approaching one of the Hydrogen arms horizontally and the water molecule is not spinning. Total angular momentum of the system is zero.
The proton will repel the Hydrogen and vice-versa. The proton will decelerate and apply torque to the water molecule. It seems that total angular momentum is now nonzero.
hope it helps✌️✌️✌️
good question
good answer :-
Oxygen, the proton is approaching one of the Hydrogen arms horizontally and the water molecule is not spinning. Total angular momentum of the system is zero.
The proton will repel the Hydrogen and vice-versa. The proton will decelerate and apply torque to the water molecule. It seems that total angular momentum is now nonzero.
hope it helps✌️✌️✌️
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Consider a universe consisting of only one water molecule and one proton.
In the reference frame of the Oxygen, the proton is approaching one of the Hydrogen arms horizontally and the water molecule is not spinning. Total angular momentum of the system is zero.
The proton will repel the Hydrogen and vice-versa. The proton will decelerate and apply torque to the water molecule. It seems that total angular momentum is now non zero.
In the reference frame of the Oxygen, the proton is approaching one of the Hydrogen arms horizontally and the water molecule is not spinning. Total angular momentum of the system is zero.
The proton will repel the Hydrogen and vice-versa. The proton will decelerate and apply torque to the water molecule. It seems that total angular momentum is now non zero.
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