Explanation:
The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

At t = 1 s,
Current,

So, the current at t = 1 s is 3 A.
For lowest current,

Hence, this is the required solution.
' C ' is the only correct statement on the list. We don't know anything about diagram-x or diagram-y because we can't see them.
"Constant velocity" is practically a definition for zero acceleration.
Explanation:
It s given that,
Mass of a planet, 
Radius of a planet, 
(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :



(2) The escape velocity is given by :


v = 7324.61 m/s
Hence, this is the required solution.
Answer:
Explanation:
Explanation: total displacement =3√2m. and total distance covered=14m. I hope this is right and helps u.