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3 years ago

The absolute magnitude of a star________.

1 answer:

8 0

The absolute magnitude of a star is how bright it would appear to us
if it were located ten parsecs (about 32.6 light years) from us. So it's
a way of treating all stars equally ... on a "level playing field" ... and it
describes each star's actual brightness.

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Which element does not have the same number of electrons in its outermost shell as the other elements in its group?(1 point)

Setler79 [48]

Helium (He) does not have the same number of valence electrons as other elements in its group.

The periodic table is divided into groups with the last number of the group coinciding with the number of electrons that an element in the group has in its outermost or valence shell.

Helium is in group 18 which means that it should have the same number of valence electrons as :

Neon
Argon
Krypton
Xenon and,
Radon

Yet Helium only has 2 valence electrons. We can therefore conclusively say that Helium does not have the same number of valence electrons as other elements in its group.

More information is available at brainly.com/question/20944279.

4 0

3 years ago

Read 2 more answers

Why is it important to understand the concept of density ?

N76 [4]

Density applies to many if not all aspects of life. With density you can explain why ice floats. You can explain why oxygen is on the earth, and not floating around in space( or being replaced by another gas). You can also explain why heat rises while cold air sinks.

8 0

2 years ago

Which letter on the diagram below represents the trough of a wave?

Nikitich [7]

A would be the wavelength, C would be a crest, D would be the amplitude, leaving B which is the trough.

8 0

3 years ago

Read 2 more answers

An electron is released a short distance above earth's surface. a second electron directly below it exerts an electrostatic forc

slamgirl [31]

The mass of an electron is 9.109 x 10⁻³¹ kg

The weight of the electron is (mass) x (g) = 8.926 x 10⁻³⁰ Newton

The charge on an electron is -1.602 x 10⁻¹⁹ Coulomb

The repelling force between the two electrons is (K · q₁ · q₂ / r²) =

(8.98755 x 10⁹ N-m²/C²) x (1.602 x 10⁻¹⁹ C)² / D²

In order for the bottom one to just exactly hold the top one up at a distance 'D', the repelling force has to be exactly equal to the weight of the upper electron.

8.926 x 10⁻³⁰ N = (8.98755 x 10⁹ N-m²/C²)·(1.602 x 10⁻¹⁹ C)² / D²

We have to solve THAT ugly mess for ' D '.

Clean up the units first:

Cancel the C² on the right side, then divide each side by Newton:

8.926 x 10⁻³⁰ = (8.98755 x 10⁹ m²) x (1.602 x 10⁻¹⁹)² / D²

Now, let's multiply both sides by (D² x 10²⁹) :

D² x 8.926 x 10⁻¹ = (8.98755 m²) x (1.602)²

Divide each side by (0.8926):

D² = (8.98755 x 1.602²) / (0.8926) meter²

D² = 25.84 m²

Take the square root of each side:

D = 5.08 meters

I am shocked, impressed, and amazed !

Are you shocked, impressed, or amazed ?

3 0

3 years ago

A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro

grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

Φ $._{E}$ = ∫ E. dA = $q_{int}$ / ε₀

For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

∫ E dA = $q_{int}$ / ε₀

The area of a sphere is

A = 4π r²

E 4π r² = $q_{int}$ / ε₀

E = (1 /4πε₀ ) q / r²

Having the solution of the problem let's analyze the points:

A ) r = 3R / 4 = 0.75 R.

In this case there is no charge inside the Gaussian surface therefore the electric field is zero

E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

E = (1 /4πε₀ ) Q / (1.25R)²

E = (1 /4πε₀ ) Q / R2 1 / 1.56²

E₀ = (1 /4π ε₀ ) Q / R²

$E_{B}$ = Eo /1.56 ²

$E_{B}$ = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

$E_{B}$ =(1 /4π ε₀ ) Q 1/(2R)²

$E_{B}$ = (1 /4π ε₀ ) q/R² 1/4

$E_{B}$ = Eo 1/4

$E_{B}$ = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0

3 years ago