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Diano4ka-milaya [45]
2 years ago
14

A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr

oss-section of the wire? (e=1.60×10−19 C) A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cross-section of the wire? ( C) 1.5×1023 1.6×1018 6.3×1015 1.6×1017 3.7×1015
Physics
1 answer:
Zarrin [17]2 years ago
3 0

Answer:

(1.6 × 10¹⁸) /s

Explanation:

Current flowing in a wire is given by

I = (Q/t)

where Q = total charges on the electrons flowing in the wire

t = time

But Q = nq

where n = number of electrons flowing in the wire

q = charge on one electron = 1.602 × 10⁻¹⁹ C

So, I = nq/t

(n/t) = (I/q)

(n/t) = number of electrons per second, for any cross sectional Area.

(n/t) = (I/q)

I = 260 mA = 0.26 A

q = 1.6 × 10⁻¹⁹ C

(n/t) = (0.26/(1.602×10⁻¹⁹)) = (1.62 × 10¹⁸) /s = (1.60 × 10¹⁸) /s

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A 1-m-long monopole car radio antenna operates in the AM frequency of 1.5 MHz. How muchcurrent is required to transmit 4 W of po
Zanzabum

Answer:

The current needed to transmit Power of 4 W is 28.47 A

Solution:

As per the question:

Length of the antenna, L_{a} = 1 m

Frequency, \vartheta = 1.5 MHz = 1.5\times 10^{6} Hz

Power transmitted, P_{t} = 4 W

Now,

For a monopole antenna:

\lambda_{a} = \frac{c}{\vartheta}

where

\lambda_{a} = wavelength transmitted by the antenna

c = speed of light in vacuum

\lambda_{a} = \frac{3\times 10^{8}}{1.5\times 10^{6}} = 200 m

Now,

Since, the value of \lambda_{a} >> L_{a} thus the monopole is a Hertian monopole.

The resistance is calculated as:

R = \frac{1}{2}(\frac{dL_{a}}{\lambda_{a}})^{2}\times 80\pi^{2}

R = \frac{1}{2}(\frac{1}{200)^{2}\times 80\pi^{2} = 9.869\times 10^{- 3} = 9.869 m\Omega

P_{radiated} = P_{t}

P_{radiated} = \frac{R}{I^{2}}

Now, the current I is given by:

I = \sqrt{\frac{2P_{t}}{R}} = \sqrt{\frac{2\times 4}{9.869\times 10^{- 3}}} = 28.47 A

5 0
2 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
1 year ago
A truck covered 2/7 of a journey at an average speed of 40
slavikrds [6]

Answer:8h

Explanation:

8 0
2 years ago
Read 2 more answers
U-238 has protons and146 neutrons. A particular isotope of plutonium has 94 protons, neutrons, and a mass number of 241. Thorium
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#1

^{238}U

so mass number = 238

mass number = protons + neutrons

given that

neutrons = 146

238 = protons + 146

protons = 92

#2

^{241}Pu

so mass number = 241

mass number = protons + neutrons

given that

Protons = 94

241 = 94 + neutrons

neutrons = 147

#3

^ATh

A = mass number

Protons = 90

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A = protons + Neutrons

A = 90 + 137 = 227

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Answer:

5.22 x 10^5 V

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