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2 years ago

Electric current originates from which part of an atom? *

Engineering

1 answer:

yanalaym [24]2 years ago

3 0

Answer: Electric current originates from positively charged protons negatively charged electrons of an atom.

Explanation:

The movement of ions (positive or negative) from one point to another is called electric current.

An atom has three sub-atomic particles. These are protons, neutrons and electrons.

Protons are positively charged, neutrons have no charge and electrons are negatively charged. Protons and neutrons reside inside the nucleus of an atom whereas electrons revolve around the nucleus.

So, protons and electrons are responsible for originating electric current form an atom as these are the charged particles.

Thus, we can conclude that electric current originates from positively charged protons negatively charged electrons of an atom.

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3 years ago

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While there are many ways to solve this problem, one strategy is to calculate the volume of any metal's unit cell given its theo

IgorC [24]

Answer:

$V_{c}=1.396x10^{-28} m^{3} /unit.cell$

Explanation:

z = number of atoms

M = Molar mass of zirconium

N = Avogadro’s number

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d = density

$z=12x\frac{1}{6}+2x\frac{1}{2}+3=6$

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3 years ago

At a festival, spherical balloons with a radius of 140.cm are to be inflated with hot air and released. The air at the festival

Tpy6a [65]

Answer:

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Explanation:

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3 years ago

Takt time is the rate at which a factory must produce to satisfy the customer's demand. a)- True b)- False

laila [671]

Answer: a)True

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3 years ago

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jeka57 [31]

Answer:

a) at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

b) daylight (d) = 0.50 μm

Incandescent ( i ) = 1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

Values are gotten from the table named: blackbody radiation functions

a) Calculate the band emission fractions for the visible region

at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

attached below is a detailed solution to the problem

b)calculate wavelength corresponding to the maximum spectral intensity

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For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

3 0

2 years ago