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SVETLANKA909090 [29]
3 years ago
5

What causes an electric current in a wire?

Physics
2 answers:
Rina8888 [55]3 years ago
7 0
I think the correct answer from the choices listed above is option B. It is the electrons that  flow because of electrical attraction and repulsion causes an electric current in a wire. Electrons, which continuously move in wire, are called Electric Current. Hope this answers the question.
m_a_m_a [10]3 years ago
7 0

Answer: The correct option is B.

Explanation:

The rate at which the electrons flow constitutes the electric current. The force is requires to make current flow through the wire.

There is a flow of the electrons in a wire due to the potential difference in the circuit.

The electrons always flow from high potential to the low potential. The electric force between the charged particles are important in electricity.

Unlike charges attract each other.The electrons are attracted towards the positive terminal of the battery. So, there are electrical attraction and electrical repulsion. Electrons flow because of electrical attraction and repulsion.

Therefore, the correct option us B.

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A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa
NemiM [27]

Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

mV_1+3mV_2=(m+3m)V_f=4mV_f

V_1+3V_2=4V_f and making V_f the subject then

V_f=\frac {V_1+3V_2}{4} and since V_1 is initial velocity of car, value given as 4 m/s, V_2 is the initial velocity of the three cars stuck together, value given as 2 m/s and v_f is the final velocity which is unknown. By substitution

V_f=\frac {4+(3\times2)}{4}=2.5 m/s

(b)

Initial kinetic energy is given by

\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ

Final kinetic energy is given by

\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

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