I believe the correct answer is B
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:
The answer is 312.5j
Explanation:
The kinetic energy (KE):
KE=1/2*m*v^2
M= mass of the object
v= velocity of the object
We have;
m=25g
v=5m/s
KE=1/2*25g*5^2m/s
KE =312.5j
Answer:
63
Explanation:
it would take around 63 if ur asking for 1200/19.1
The frequencies are missing in the question. The three successive resonance frequencies within the volume are
.
Solution :
Let the volume be : v
The frequency for one end open and one end closed is given by :
So, 
Therefore,
,
, 
So,
and 
Therefore, the ratio of
which is not a whole number.
Now the frequency for open volume and closed at both end

So,
,
, 
From above formulae we can see that ratio of is not a whole number that is a identification for the frequency of the volume at one end open and one end closed.
Also ratio of the consecutive frequency of the volume at open from both side and closed from both side is always a whole number.