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2 years ago

Which of the following groups contain three elements with stable electron configurations?

Physics

2 answers:

Wewaii [24]2 years ago

6 0

Answer:

B. helium, xenon, neon

Explanation:

barxatty [35]2 years ago

3 0

B is the correct answer because Helium, Xenon, and Neon are all Noble Gases at Group 18, which means they contain the full 8 electrons they need meaning they all have stable electron configurations.

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when a electric current is passed through an insulated wire that is coiled around an iron core like a nail what is created?

Anika [276]

electromagnet

When a electric current is passed through an insulated wire that is coiled around an iron core, like a nail, an electromagnet is created.

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2 years ago

Sophie says that geologic maps do not matter because she gets no benefits from them. Why is Sophie wrong? a. She can see the ran

astra-53 [7]

The Answer is Option C
I think...
Sorry If i am wrong...

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2 years ago

Read 2 more answers

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ???? 0 ν0 . Find the mi

Allisa [31]

Answer:

<h2>E = 2.8028*10⁻¹⁹ Joules</h2>

Explanation:

The minimum energy needed to eject electrons from a metal with a threshold frequency fo is expressed as E = hfo

h = planck's constant

fo = threshold frequency

Given the threshold frequency fo = 4.23×10¹⁴ s⁻¹

h = 6.626× 10⁻³⁴ m² kg / s

Substituting this value into the formula to get the energy E

E = 4.23×10¹⁴ * 6.626 × 10⁻³⁴

E = 28.028*10¹⁴⁻³⁴

E = 28.028*10⁻²⁰

E = 2.8028*10⁻¹⁹ Joules

7 0

3 years ago

Alan leaves Los Angeles at 8:00 A.M. to drive to San Francisco 400 mi away. He manages to travel at a steady 50 mph in spite of

Fudgin [204]

Answer:

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

Explanation:

let 'd' be distance b/w Los Angeles and San Francisco i.e 400 mi

considering ,

Alan's speed $v_A$ =50mph

Beth's speed $v_B$ =60mph

->For Alan:

The time required $t_A$ = d/ $v_A$ = 400/50 => 8h

-> For beth:

The time required $t_B=\frac{d}{v_B} =\frac{400}{60} =>6\frac{2}{3} h$ => 6h 40m

Alan will reach at 8:00 a.m +8h = 4:00p.m.

Beth will reach at 9:00 a.m +6h 40m= 3:40p.m.

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

3 0

2 years ago

A paperweight consists of a 8.55-cm-thick plastic cube. Within the plastic a thin sheet of paper is embedded, parallel to opposi

ella [17]

Answer:

$\mu=1.5322$

Explanation:

Given:

Thickness of the paperweight cube, $x=8.55\ cm$

apparent depth from one side of the inbuilt paper in the plastic cube, $i=4.02\ cm$

apparent depth from the other side of the inbuilt paper in the plastic cube, $i'=1.56\ cm$

Now as we know that refractive index is given as:

$\rm \mu=\frac{real\ depth}{apparent\ depth}$

Let the real depth form first side of the slab be, $d$

Then the depth from the second side of the slab will be, $d'=x-d=8.55-d$

Since refractive index for an amorphous solid is an isotropic quantity so it remains same in all the direction for this plastic.

$\mu=\mu'$

$\frac{d}{i} =\frac{d'}{i'}$

$\frac{d}{4.02}=\frac{8.55-d}{1.56}$

$d=6.1597\ cm$

Now the refractive index:

$\mu=\frac{d}{i}$

$\mu=\frac{6.1596}{4.02}$

$\mu=1.5322$

7 0

2 years ago