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Hitman42 [59]
3 years ago
10

How does battery generate electric current in a circuit?

Physics
1 answer:
sveticcg [70]3 years ago
3 0

Answer:

C. It creates negative electric charges and pushes them into the circuit

Explanation:

A battery has three main parts. The Cathode (Positive), the Electrolytes (energy source), and the Anode (negative). The anode takes energy from the electrolytes and creates more electrons. And because subatomic particles with the same charge don't like being in the same place, they move through the circuit to reach the cathode because opposites attract.

I hope this helps!

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Answer:

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Explanation:

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3 years ago
Two wires are parallel and one is directly above the other. Each has a length of 50.0 m and a mass per unit length of 0.020 kg/m
ivann1987 [24]

Answer: The time required for the impluse passing through each other is approximately 0.18seconds

Explanation:

Given:

Length,L = 50m

M/L = 0.020kg/m

FA = 5.7×10^2N

FB = 2.5×10^2N

The sum of distance travelled by each pulse must be 50m since each pulse started from opposite ends.

Ca(t) + CB(t) = 50

Where CA and CB are the velocities of the wire A and B

t = 50/ (CA + CB)

But C = Sqrt(FL/M)

Substituting gives:

t = 50/ (Sqrt( FAL/M) + Sqrt(FBL/M))

t = 50/(Sqrt 5.7×10^2/0.02) + (Sqrt(2.5×10^2/0.02))

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t = 50/280.15

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4 0
3 years ago
Read 2 more answers
A fixed coil of wire with 10 turns and an area of 0.055 m2 is placed in a perpendicular magnetic field. This field oscillates in
Blababa [14]

Answer:

Part a)

Average EMF for half cycle is

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Part b)

For one complete cycle we will have

E_{avg} = 0

Part c)

Maximum induced EMF will be at

t = 0.025 s and 0.075 s

minimum induced EMF is at

t = 0.05s and 0.1 s

Explanation:

As we know that magnetic field is oscillating in direction as well as magnitude

so induced EMF is given as

E = NBA\omega sin(\omega t)

Part a)

For average value of EMF from positive maximum to negative maximum which is equal to half cycle

so we have

E_{avg} = NBA\omega \frac{2}{T}\int_0^{T/2} sin(\omega t) dt

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E_{avg} = 2.64 V

Part b)

For one complete cycle we will have

E_{avg} = NBA\omega \frac{1}{T}\int_0^T sin(\omega t) dt

E_{avg} = 0

Part c)

Maximum induced EMF will be at

t = \frac{T}{4} and \frac{3T}{4}

here we know

T = \frac{1}{f} = 0.1 s

t = 0.025 s and 0.075 s

minimum induced EMF is at

t = \frac{T}{2} and T

so it is

t = 0.05s and 0.1 s

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sertanlavr [38]

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