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alukav5142 [94]
3 years ago
11

Which of the following provides alternating current? Select all that apply.

Physics
1 answer:
Zielflug [23.3K]3 years ago
7 0
Answer is power outlet in your house
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the element has 7 valance electrons is CI:Chlorine

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Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

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3 years ago
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The resistivity of gold is at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 m
IgorC [24]

Answer:

0.0360531138247 V/m

Explanation:

\rho = Resistivity of gold = 2.44\times 10^{-8}\ \Omega .m (General value)

I = Current = 940 mA

d = Diameter = 0.9 mm

A = Area = \dfrac{\pi}{4}d^2

E = Electric field

Resistivity is given by

\rho=\dfrac{EA}{I}\\\Rightarrow E=\dfrac{\rho I}{A}\\\Rightarrow E=\dfrac{2.44\times 10^{-8}\times 940\times 10^{-3}}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}\\\Rightarrow E=0.0360531138247\ V/m

The  electric field in the wire is 0.0360531138247 V/m

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Besides ambitious, it would be 184.6 mi
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Which of the following best describes the electromagnetic spectrum?
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D.) Travels in a pattern of particle motion....
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