1) Current in the wire: 0.0875 A
The current in the wire is given by:

where
Q is the charge passing a given point in the conductor
t is the time elapsed
In this problem, we have
Q = 420 C is the total charge passing through a given point in a time of
t = 80 min = 4800 s
So, the current is

2) Drift velocity of the electrons: 
The drift velocity of the electrons in the wire is given by:

where
I = 0.0875 A is the current
is the number of free electrons per cubic meter
A is the cross-sectional area
is the charge of one electron
The radius of the wire is

So the cross-sectional area is

So, the drift velocity is

Explanation:it is beause they are sharper and also have less surface area and therefore more pressure
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
In hot gases , the atoms keeps colliding with each other and sometimes the energy liberated during collision takes the electron to a higher level,thus, .The object is a cloud of hot gas and finally the electron returns back emitting photon
<h3><u>Answer;</u></h3>
= 1.256 m
<h3><u>Explanation;</u></h3>
We can start by finding the spring constant
F = k*y
Therefore; k = F/y = m*g/y
= 0.40kg*9.8m/s^2/(0.76 - 0.41)
= 11.2 N/m
Energy is conserved
Let A be the maximum displacement
Therefore; 1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2
Thus; A = sqrt((1.20 - 0.55)^2 + m/k*v^2)
= sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)
= 0.846 m
Thus; the length will be 0.41 + 0.846 = 1.256 m