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3 years ago

Which types of orbitals are found in the principal energy level n = 3?

Physics

2 answers:

Aleonysh [2.5K]3 years ago

7 0

Answer:

Explanation:

Within each shell of an atom there are some combinations of orbitals. In the n=1 shell you only find s orbitals, in the n=2 shell, you have s and p orbitals, in the n=3 shell, you have s, p and d orbitals and in the n=4 up shells you find all four types of orbitals.

denis23 [38]3 years ago

7 0

B. s, p, d
Within each shell of an atom there are some combinations of orbitals. In the n=1 shell you only find s orbitals, in the n=2 shell, you have s and p orbitals, in the n=3 shell, you have s, p and d orbitals and in the n=4 up shells you find all four types of orbitals.

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Calculate the index of refraction for a medium in which the speed of light is 2.1x 108 m/s. The speed of light in vacuum is 3x10

strojnjashka [21]

Answer:

n = 1.42

Explanation:

The refractive index for a medium is given by the ratio of the speed of light in vacuum to the speed of light in a medium.

$n=\dfrac{c}{v}\\\\n=\dfrac{3\times 10^8}{2.1\times 10^8}\\\\n = 1.42$

So, the refractive index of the medium is 1.42.

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3 years ago

A spaceship moves radially away from Earth with acceleration 29.4 m/s 2 (about 3g). How much time does it take for sodium street

Gemiola [76]

Answer:

doppler shift's formula for source and receiver moving away from each other:

λ'=λ°√(1+β/1-β)

Explanation:

acceleration of spaceship=α=29.4m/s²

wavelength of sodium lamp=λ°=589nm

as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm

using doppler shift's formula:

λ'=λ°√(1+β/1-β)

putting the values:

700nm=589nm√(1+β/1-β)

after simplifying:

β=0.17

by this we can say that speed at that time is: v=0.17c

to calculate velocity at an acceleration of a=29.4m/s²

we suppose that spaceship started from rest so,

v=v₀+at

where v₀=0

so v=at

as we want to calculate t so:-

t=v/a v=0.17c ,c=3x10⁸ ,a=29.4m/s²

putting values:

=0.17(3x10⁸m/s)/29.4m/s²

t=1.73x10⁶

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3 years ago

Which game represents a wave

alisha [4.7K]

Dominoes does because when you hit one, it knocks over the next one, and so on, so forth. The same type of pattern happens in a wave.

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3 years ago

One property that makes electromagnetic waves differ from other types of waves is that they can

IgorC [24]

Electromagnectic Waves Travel In A Vacuum

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3 years ago

Read 2 more answers

Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

Setler [38]

Answer:

The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

Explanation:

Given that,

Diameter = 10 light year

Orbital speed = 180 km/s

Suppose determine the mass of the massive object at the center of the Milky Way galaxy.

Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.

We need to calculate the radius of the orbit

Using formula of radius

$r=\dfrac{d}{2}$

$r=\dfrac{15\times9.461\times10^{15}}{2}$

$r=7.09\times10^{16}\ m$

We need to calculate the mass of the massive object at the center of the Milky Way galaxy

Using formula of mass

$M=\dfrac{v^2r}{G}$

Put the value into the formula

$M=\dfrac{(180\times10^3)^2\times7.09\times10^{16}}{6.67\times10^{-11}}$

$M=3.44\times10^{37}\ Kg$

Hence, The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

5 0

3 years ago