The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.
- The force per unit length between two parallel thin current-carrying
and 
wires at distance ' r ' is given by 
....(1) .
- If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.
A schematic of the information provided in the question can be seen in the image attached below.
From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3
Using equation (1) , we get
I₃ = 2.4 A and the current is pointing in the downward direction
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Answer:
mechanical energy per unit mass is 887.4 J/kg
power generated is 443.7 MW
Explanation:
given data
average velocity = 3 m/s
rate = 500 m³/s
height h = 90 m
to find out
total mechanical energy and power generation potential
solution
we know that mechanical energy is sum of potential energy and kinetic energy
so
E = 
×m×v² + m×g×h .............1
and energy per mass unit is
E/m = ×v² + g×h
put here value
E/m = ×3² + 9.81×90
E/m = 887.4 J/kg
so mechanical energy per unit mass is 887.4 J/kg
and
power generated is express as
power generated = energy per unit mass ×rate×density
power generated = 887.4× 500× 1000
power generated = 443700000
so power generated is 443.7 MW
We know, momentum = mass * speed
25kgm/s = 2 kg * s
s = 25/2 = 12.5 m/s
Answer:
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