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3 years ago

Neon has 10 protons and 10 electrons. How many additional electrons would it take to fill the outer (second) electron shell?

Physics

2 answers:

blondinia [14]3 years ago

6 0

The first shell has 2 in and the second shell has 8 in therefore ypuvwont need to add anymore :-) so the answer is 0

mote1985 [20]3 years ago

6 0

Answer:

A.) 0

Explanation:

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In an experiment of a simple pendulum, measurements show that the pendulum has length し 0.397 ± 0.006 m, mass M-0.3172 ± 0.0002

vichka [17]

Answer:

1.)1.265+or minus 0.0006m

2).0.71%

Explanation:

See attached file

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3 years ago

Which of the following is a small star that has reached the end of its stellar evolution?

Akimi4 [234]

D.) White Dwarf

It is the smallest star whose mass is approximately equal or greater than 1.4M
Here, M = mass of the Sun.

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3 0

3 years ago

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a sound with a constant frequency is produced by the siren on top of a firehouse. Compared to the frequency produced by the sire

polet [3.4K]

There's an effect called The Doppler Effect, that's the "up and down" sound.
That frequency is 1.8mm and 115 per minute.

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4 years ago

If a chemical change has taken place, which of these is most likely true?

Digiron [165]

Whats the possible answers?

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4 years ago

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Assume the average value of the vertical component of Earth's magnetic field is 42 μT (downward) in some region that has an area

Oliga [24]

Answer:

The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

Explanation:

Given that,

Magnetic field = 42 μT

Area $A=3.71\times10^{5}\ km^2$

$A=3.71\times10^{11}\ m^2$

We need to calculate the flux per unit area

$flux\ per\ unit\ area=\dfrac{42\times10^{-6}}{3.71\times10^{11}}$

$flux\ per\ unit\ area=1.132\times10^{-16}\ T/m^2$

We need to calculate the total earth's surface area

$A'=4\pi r^2$

$A'=4\times\pi\times(6.3781\times10^{6})^2$

$A'=5.1120\times10^{14}\ m^2$

We need to calculate the rest of earth's area

$A''=A-A'$

Put the value into the formula

$A''=5.1120\times10^{14}-3.71\times10^{11}$

$A''=5.10829\times10^{14}\ m^2$

We need to calculate the net magnetic flux through the rest of Earth's surface

$B'=5.10829\times10^{14}\times1.132\times10^{-16}$

$B'=5.782\times10^{-2}\ T$

Hence, The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

3 0

4 years ago