Answer:

Explanation:
Let the length of the string is L.
Let T be the tension in the string.
Resolve the components of T.
As the charge q is in equilibrium.
T Sinθ = Fe ..... (1)
T Cosθ = mg .......(2)
Divide equation (1) by equation (2), we get
tan θ = Fe / mg




As θ is very small, so tanθ and Sinθ is equal to θ.


Answer:
Mass of ion will be
Explanation:
We have given ion is triply charged that is 
Radius r = 36 cm = 0.36 m
Velocity of the electron 
Magnetic field B = 0.55 T
We know that radius of the path is given by 

A. when it reaches the top of its flight
potential energy increases as the height of the object does, and the highest point of a ball's flight is when it reaches the top.
This is the luster of the mineral