Answer:
The maximum current, in amperes, that a conductor can carry continuously under the conditions of abuse without exceeding its temperature rating.
kinetic energy=1/2mv^2.
which is 4320000=1/2×m×23^2.
which is 4320000=1/2×m×529.
4320000=264.5m.
m=4320000/264.5.
m=16332.70~16333g
Answer: 2940 J
Explanation: solution attached:
PE= mgh
Substitute the values:
PE= 10kg x 9.8 m/s² x 30 m
= 2940 J
The force per unit length between the two wires is
Explanation:
The magnitude of the force per unit length exerted between two current-carrying wires is given by
where
is the vacuum permeability
are the currents in the two wires
r is the separation between the two wires
For the wires in this problem, we have
r = 2.00 cm = 0.02 m
Substituting into the equation, we find
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B is the best answer for the question