The average current density in the wire is given by:

where I is the current intensity and A is the cross-sectional area of the wire.
The cross-sectional area of the wire is given by:

where r is the radius of the wire. In this problem,
, so the cross-sectional area is

and the average current density is

Answer:
1.034m/s
Explanation:
We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

Substituting,


Part B)
For the Part B we need to apply conserving momentum equation, this formula is given by,

Where here
is the velocity after the collision.



Answer:
given,
mass of copper = 100 g
latent heat of liquid (He) = 2700 J/l
a) change in energy
Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (300 - 4)
Q = 11153.63 J
He required
Q = m L
11153.63 = m × 2700
m = 4.13 kg
b) Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (78 - 4)
Q = 2788.41 J
He required
Q = m L
2788.41 = m × 2700
m = 1.033 kg
c) Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (20 - 4)
Q = 602.90 J
He required
Q = m L
602.9 = m × 2700
m =0.23 kg