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bezimeni [28]
3 years ago
9

A copper wire has a circular cross section with a radius of 1.25 mm. If the wire carries a current of 3.70 A, find the drift spe

ed in mm/s of the electrons in this wire. (Note: the density of charge carriers (electrons) in a copper wire is n = 8.46 × 1028 electrons/m3)
Physics
1 answer:
yulyashka [42]3 years ago
8 0

Answer:

0.0557 mm / s

Explanation:

r = 1.25 mm = 1.25 x 10^-3 m, i = 3.7 A, n = 8.46 x 10^28 per cubic metre,

e = 1.6 x 10^-19 C

Let vd be the drift velocity

use the formula for the current in terms of drift velocity

i = n e A vd

3.7 = 8.46 x 10^28 x 1.6 x 10^-19 x 3.14 x 1.25 x 1.25 x 10^-6 x vd

vd = 5.57 x 10^-5 m/s

vd = 0.0557 mm / s

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mr_godi [17]

Answer:

1.5 \Omega

Explanation:

The resistance of a wire is given by the equation:

R=\rho \frac{L}{A}

where

\rho is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

In this problem, we have a wire of platinoid, whose resistivity is

\rho = 3.3\cdot 10^{-7} \Omega m

The length of the wire is

L = 7.0 m

And its radius is

r=\frac{0.14 cm}{2}=0.07 cm = 7\cdot 10^{-4} m, so the cross-sectional area is

A=\pi r^2=\pi(7\cdot 10^{-4})^2=1.54\cdot 10^{-6}m^2

Solving for R, we find the resistance of the wire:

R=(3.3\cdot 10^{-7})\frac{7.0}{1.54\cdot 10^{-6}}=1.5 \Omega

3 0
3 years ago
Which type of solar radiation is the most powerful?
cluponka [151]

Answer: Gamma rays

Explanation:

4 0
4 years ago
Read 2 more answers
A high-speed K0 meson is traveling at β = 0.90 when it decays into a π + and a π − meson. What are the greatest and least speeds
san4es73 [151]

Answer:

greatest speed=0.99c

least speed=0.283c

Explanation:

To solve this problem, we have to go to frame of center of mass.

Total available energy fo π + and π - mesons will be difference in their rest energy:

E_{0,K_{0} }-2E_{0,\pi }  =497Mev-2*139.5Mev\\

                       =218 Mev

now we have to assume that both meson have same kinetic energy so each will have K=109 Mev from following equation for kinetic energy we have,

K=(γ-1)E_{0,\pi }

K=E_{0,\pi}(\frac{1}{\sqrt{1-\beta ^{2} } } -1)\\\frac{1}\sqrt{1-\beta ^{2} }}=\frac{K}{E_{0,\pi}}+1\\   {1-\beta ^{2}=\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta ^{2}=1-\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta = +-\sqrt{\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\\\\beta =+-\sqrt{1-\frac{1}{(\frac{109Mev}{139.5Mev+1)^2}}

u'=+-0.283c

note +-=±

To find speed least and greatest speed of meson we would use relativistic velocity addition equations:

u=\frac{u'+v}{1+\frac{v}{c^{2} } } u'\\u_{max} =\frac{u'_{+} +v_{} }{1+\frac{v}{c^{2} } } u'_{+} \\u_{max} =\frac{0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } 0.828\\ u_{max} =0.99c\\u_{min} =\frac{u'_{-} +v_{} }{1+\frac{v}{c^{2} } } u'_{-}\\u_{min} =\frac{-0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } -0.828c\\u_{min} =0.283c

6 0
3 years ago
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Effectus [21]

Current is the amount of charged passed divided by elapsed time.

I = Q/Δt

I = current, Q = charge, Δt = elapsed time

We also know an electron has a charge of 1.6×10⁻¹⁹C, so let us find the total charge by multiplying this unit of charge by the total number of electrons:

Q = 1.6×10⁻¹⁹(3×10²⁰) = 48C

We also have Δt = 6s, so let's plug these values in to solve for I:

I = 48/6

I = 8A

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3 years ago
A box of bananas weighing 40 N rests on a horizontal surface. The coefficient of static friction between the box and the surface
ale4655 [162]

Answer:

0.4x40 \div 0.2 = 80

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