Answer:
The ball reaches Barney head in 
Explanation:
From the question we are told that
The rise velocity is 
The height considered is 
The horizontal velocity of the large object is 
Generally from kinematic equation

Here s is the distance of the object from Barney head ,
u is the velocity of the object along the vertical axis which is equal but opposite to the velocity of the helicopter
So

So

= 
Solving the above equation using quadratic formula
The value of t obtained is 
Answer:

Explanation:
We are given that
Diameter of wire=d=4.12 mm
Radius of wire=r

Current=I=8 A
Drift velocity=
We have to find the density of free electrons in the metal
We know that
Density of electron=
Using the formula
Density of free electrons=
By using Area of wire=

Density of free electrons=
The same number of electron shells, the atomic number is given to each individual element, same as the atomic mass, and properties have nothing to do with groups