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Anton [14]
3 years ago
10

A 12-V battery is connected across a 100-Ω resistor. How many electrons flow through the wire in 1.0 min?

Physics
1 answer:
Vanyuwa [196]3 years ago
4 0

Answer:

The quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.

Explanation:

Given;

emf of the battery, V = 12 V

resistance of the resistor, R = 100-Ω

time of current flow, t = 1 min

charge of 1 electron = 1.602 x 10¹⁹ C

The current through this circuit is given by;

I = V / R

I = (12) / (100)

I = 0.12 A

The quantity of charge or electron flowing the wire in the given time is calculated as;

Q =It

where;

I is the current flowing through the wire

t is the time of current flow = 1 x 60s = 60 s

Q = 0.12 x 60

Q = 7.2 C

1.602 x 10⁻¹⁹ C --------------- 1 electron

7.2 C -----------------------------? electron

= \frac{7.2 }{1.602*10^{-19}} \\\\= 4.5*10^{19} \ electrons

Therefore, the quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.

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JulsSmile [24]

Answer:

b. 600,000 J

Explanation:

Applying the law of conservation of energy,

The thermal energy created = Kinetic energy of the suv.

Q' = 1/2(mv²)............... Equation 1

Where Q' = Thermal energy, m = mass of the suv, v = velocity of the suv.

From the question,

Given: m = 3000 kg, v = 20 m/s

Substitute these values into equation 1

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Q' = 600000 J

Hence the right option is b. 600,000 J

5 0
3 years ago
Suppose the sphere is electrically neutral. Is it attracted to
Elden [556K]

Answer:

No, it is not attracted.

Explanation:

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8 0
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A square coil of wire with 15 turns and an area of 0.40 m2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped
Drupady [299]

Answer:

The magnitude of the average induced emf is 90V

Explanation:

Given;

area of the square coil, A = 0.4 m²

number of turns, N = 15 turns

magnitude of the magnetic field, B = 0.75 T

time of change of magnetic field, t = 0.05 s

The magnitude of the average induced emf is given by;

E = -NAB/t

E = -(15 x 0.4 x 0.75) / 0.05

E = -90 V

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Let the distance between the towns be d and the speed of the air be s.

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d

s−15

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3d=7∗(s−15)5d=7∗(s+15)

3d=7s−1055d=7s+105

subtract first equation from second equation we get

2d=210d=105km

Substitute the value of d in the above equations for s.

5∗105=7s+1057s=420s=60km/hr

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