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Shtirlitz
3 months ago
14

A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele

ctrons per cubic meter. (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?
Physics
2 answers:
Maru [3.3K]3 months ago
5 0

Answer:

a)n= 3.125 x 10^{19 electrones.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) ver explicación

Explanation:

La corriente 'I' = 5A =>5C/s

diámetro 'd'= 2.05 x 10^{-3 m

radio 'r' = d/2 => 1.025 x 10^{-3 m

número de electrones 'n'= 8.5 x 10^{28}

a) La cantidad de electrones que pasan por la bombilla cada segundo se determina mediante:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

Como sabemos que: Q= ne

donde e es la carga del electrón, es decir, 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrones.

b) La densidad de corriente 'J' en el cable se calcula como

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) La velocidad típica 'V_{d' de un electrón se expresa como:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) De acuerdo con estas ecuaciones,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

Si utilizaras un cable de doble diámetro, ¿cuáles de las respuestas anteriores cambiarían? ¿Aumentarían o disminuirían?

kicyunya [3.2K]3 months ago
4 0

Answer:

a) 3.1205*10^19 electrons per second

b) 1.51*10^6 A/m^2

c) 1.11*10^-4 m/s

Explanation:

a) The number of electrons can be determined from the current flowing through the wire using this equation:

I=5.00A=5.00\frac{C}{s}\\\\1C=6.241*10^{18}e\\\\I=5.00(6.241*10^{18}e)/s=3.1205*10^{19}\frac{e}{s}

3.1205*10^19 electrons every second

b) The formula used to calculate the current density is as follows:

J=\frac{I}{A}

I: electric current within the wire

A: wire's cross-sectional area

J=\frac{I}{\pi r^2}=\frac{5.00A}{\pi(\frac{2.05}{2}*10^{-3})^2}=1.51*10^6\frac{A}{m^2}

c) The drift speed of the electrons can be calculated using the relevant formula:

I=nqv_dA\\\\v_d=\frac{I}{nqA}

n: density of free electrons

q: elementary charge of the electron = 1.6*10^{-19}C

v_d=\frac{5.00C/s}{(8.5*10^{28}m^{-3})(1.6*10^{-19}C)(3.30*10^{-6}m^2)}=1.11*10^{-4}\frac{m}{s}

d) Increasing the dia of the wire does not affect the amount of electrons that flow through the light bulb each second.

The current density will be lower since J=I/A. When A increases, J decreases.

The velocity at which electrons drift decreases as well, due to the presence of Area in the denominator of the drift speed formula.

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inna [3103]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

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to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

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implying a change in velocity is du = v

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shear stress acting on the surface is expressed as

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substituting the values

= 1.75 ×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

the area between the air and puck is given by

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area = \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v

now applying Newton's second law

force = mass × acceleration

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solving for t = \frac{0.1 v * 0.03}{1.37*10^{-3} v}

the time needed after impact for the puck is 2.18 seconds

3 0
3 months ago
How many electrons must be removed from a neutral, isolated conducting sphere to give it a positive charge of 8.0 x 10 8 C? [Q=n
inna [3103]

The new charge of the ball will amount to 8x10^8C after removing 5x10^27 electrons.

Explanation:

Initially, if the sphere is electrically neutral, its charge stands at 0C.

When an electron with a charge of (-1.6*10^-19 C) is taken away, we effectively add a positive charge, leading to:

1.6*10^-19 C as the sphere's new charge.

For a total of N electrons removed, the sphere's overall charge now becomes:

N*1.6*10^-19 C.

To calculate N when:

N*1.6*10^-19 C = 8.0x 10^8 C.

We find that N is: (8.0/1.6)x10^(8 + 19) = 5x10^27 electrons.

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Complete the paragraph to describe the relationship between kinetic energy and braking distance. Use . A car moves at a speed of
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ke prop to v^2

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2 months ago
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A nerve signal is transmitted through a neuron when an excess of Na+ ions suddenly enters the axon, a long cylindrical part of t
Yuliya22 [3333]

Answer:

1.32.225 N/C, moving away from the point charge

2. 8.972*10^-12 C

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Explanation:

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E = k*|q|/r²

Where:

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E = (8.98755*10⁹)*(8.972*10^-12)/0.05² = 32.255 N/C

A positive point charge produces an electric field that radiates outward, while a negative point charge creates an electric field directed inward.

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