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kobusy [5.1K]
2 years ago
9

A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a

nd is also at 20.0°C. Let Ecu be the electric field in the copper wire and Eag be the electric field in the silver wire. Use the resistivities at room temperature for copper. PCu = 1.72 x 10^-8.m and for silver PAE= 1.47 x 10^-8.m. What is the ratio Ecu/Eng?
Physics
1 answer:
IgorC [24]2 years ago
7 0

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

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Dafna1 [17]

1 watt = 1 joule/second

1 horsepower = 746 watts = 746 joule/second

   (150 horsepower) x (746 watt/HP) x (1 joule/sec  /  watt) x (10 sec)

=  (150 x 746 x 1 x 10)  joule  =  1,119,000 joules .   
if correct plz mark brainly
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Why does carpet tend to produce differences in static electricity more that hardwood or tile floors
Makovka662 [10]

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Levi and Clara are trying to move a very heavy box. Levi is pushing the box with a force of 30 N, and Clara is pulling the box w
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Three identical train cars, coupled together, are rolling east at speed v0. A fourth car traveling east at 2v0 catches up with t
aliina [53]

Answer:v_o

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suppose rest two cars also has same mass m

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Final momentum P_f

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8 0
2 years ago
A 15-cm-tall closed container holds a sample of polluted air containing many spherical particles with a diameter of 2.5 μm and a
TEA [102]

Answer:

t = 224 s

Explanation:

given,

length of container = 15 cm = 0.15 m

diameter of spherical particle = 2.5 μm

mass of particle = 1.9 x 10⁻¹⁴ kg

viscosity of air = μ = 1.18 x 10⁻⁵ kg/m.s

time taken by the particle to stop = ?

radius of particle = 2.5/2 = 1.25 μm

volume of particle = \dfrac{4}{3}\pi r^3

                              =\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3

Density =\dfrac{mass}{volume}

Density =\dfrac{1.9 \times 10^{-14}}{\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3}

ρ = 2322 kg/m³

terminal velocity

v_t = \dfrac{2}{9}\ \dfrac{gR^2(\rho - \rho_{air})}{\mu}

v_t = \dfrac{2}{9}\ \dfrac{9.8 \times (1.25 \times 10^{-6})^2(2322 - 1)}{1.18 \times 10^{-5}}

v_t = 6693 x 10⁻⁷ m/s

t = \dfrac{d}{v_t}

t = \dfrac{0.15}{6693 \times 10^{-7}}

t = 224 s

7 0
2 years ago
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