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2 years ago

This diagram shows a hydrogen atom. Explain why it looks like it does

Physics

1 answer:

aniked [119]2 years ago

4 0

<h3>Answer:</h3>

Make-up of Hydrogen

Hydrogen atoms are extremely simple. They are the first atom on the periodic table, this means that they only have 1 proton. The majority of hydrogen atoms don't even have a neutron. As for electrons, hydrogen atoms with no charge have 1 electron. All of this means that for almost all hydrogen atoms the nucleus is made of one singular proton.

Diagram

The diagram shows a hydrogen atom. It looks like one circle because it is just showing the singular proton found in the nucleus. Electrons are extremely small compared to protons, so the electron is not shown in the picture. It is very hard to accurately depict electrons on diagrams like this because of their size and the fact that we can never know where they are. So, the diagram may look too simple but it just depicts how most hydrogen nuclei look.

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A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

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4 years ago

Because white dwarfs are small, as their name implies, they are hard to see. What is a way astronomers have to find white dwarfs

Bumek [7]

Astronomers find white dwarfs that distinguish them from main sequence stars because white dwarfs get really hot, we can search for their ultraviolet radiation.

<h3>What is a white dwarf?</h3>

A white dwarf is a very hot star that radiated much energy in the form of ultraviolet radiation.

This UV radiation is initially very bright and then these stars become red with time.

In conclusion, Astronomers find white dwarfs they can search for their ultraviolet radiation.

Learn more about white dwarfs here:

brainly.com/question/19602278

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3 0

2 years ago

A car turns a certain curve of radius 24.98 m with constant linear speed of

Anastaziya [24]

Answer:

3525.19 kg

Explanation:

The computation of the mass of the car is shown below:

As we know that

Fc = m × V^2 ÷ R

m = Fc × R ÷ V^2

Provided that:

Fc = 34.652 kN = 34652 N

R = Radius = 24.98 m

V = speed = 15.67 m/s

So,

m = 34652 × 24.98 ÷ 15.67^2

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7 0

3 years ago

Elemental analysis of the unknown gas from part a revealed that it is 30.45% n and 69.55% o by mass. What is the molecular formu

Anna71 [15]

Answer:

the molecular formula for the gas is NO₂

Explanation:

since it contains

Nitrogen = n → 30.45%

Oxygen = o → 69.55%

and 30.45%+69.55% = 100% , then the gas only contains nitrogen and oxygen

Also we know that the proportion of oxygen over nitrogen is

proportion of oxygen over nitrogen = moles of oxygen / moles of nitrogen

since

moles = mass / molecular weight

then for a sample of 100 gr of the unknown gas

mass of oxygen = 69.55%*100 gr = 69.55 gr

mass of Nitrogen = 30.45%*100 gr = 30.45 gr

proportion of oxygen over nitrogen = (mass of oxygen/ molecular weight)/(mass of nitrogen / molecular weight of nitrogen ) = (69.55 gr/ 16 gr/mol) /( 30.45 gr /14 gr/mol) = 1.998 mol of O/ mol of N≈ 2 mol of O/ mol of N

therefore there are 2 atoms of oxygen per atom of nitrogen

thus the molecular formula for the gas is:

NO₂

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julia-pushkina [17]

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