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swat32
3 years ago
7

Write the sequence of filling electrons in the subshell in order of increasing energy.​

Physics
1 answer:
tankabanditka [31]3 years ago
5 0

Answer:

In a way of energy “1s” subsell contains most energy. because 1 s subsell contains the electrons of K orbital which is nearest from the nucleus. so the order of energy will be 1s>2s>2p> 3s> 3p> 4s> 3d> 4p> 5s> 4d> 5p> 6s> 4f> 5d> 6p> 7s and the increasing order of thair energies will be- 4s <3d<4p<5s<4d<5p<6s<4f.

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Lubov Fominskaja [6]
-- Put the rod into the freezer for a while.  As it cools,
it contracts (gets smaller) slightly.

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it expands (gets bigger) slightly.

-- Bring the rod and the cylinder togther quickly, before the
rod has a chance to warm up or the cylinder has a chance
to cool off.

-- I bet it'll fit now.

-- But be careful . . . get the rod exactly where you want it as fast
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8 0
3 years ago
An ac generator with Em = 223 V and operating at 399 Hz causes oscillations in a series RLC circuit having R = 222 Ω, L = 147 mH
Doss [256]

Answer:

Xc= 17.267 Ω,   Z= 415.5 Ω,   I= 0.537 A

Explanation:

Em = 223 V

f= 300 Hz, R = 222 Ω, L = 147 mH,  C = 23.1 μF

a)

Capacitive reactance = Xc=?

Xc= \frac{1}{2\pi fC}

Xc=1/2pi *399*23.1*10^-6

Xc= 17.267 Ω

b).

Z=\sqrt{ R^2 + (Xl - Xc)^2}

Xl= 2π * f * L  

Xl= 2π * 399 * 147 * 10^{-3}

Xl= 368.5 Ω

Z=\sqrt{ R^2 + (Xl - Xc)^2} = \sqrt{222^2 + (368.5 - 17.267)^2}

Z= 415.5 Ω  

c).

Current:

I= V / Z= Em / Z

I= 223/415.5

I= 0.537 A

3 0
4 years ago
Explain how machines can be useful if the output is always less than the input work
Anni [7]

Because: Some of the work done by the machine is used to overcome the friction created by the use of the machine. ... Work output can never be greater than work input. Machines allow force to be applied over a greater distance, which means that less force will be needed for the same amount of work.

6 0
3 years ago
A lead nucleus is spherical with a radius of about 7 ✕ 10⁻¹⁵ m. The nucleus contains 82 protons (and typically 126 neutrons). Be
Komok [63]

Answer:

∈=3.1584x10^{26} \frac{V}{m}

Explanation:

Using the Gauss Law to determine the electric field of the net flux at the surface of the nucleus

∈=\frac{P}{E_{o}}

The P is the charge density and 'Eo' is the constant of permittivity in free space

to find P

P=\frac{q}{V}

V=\frac{4}{3}*\pi*r^3

V=\frac{4}{3}\pi*(7x10^{-15})^3

V=2.932x10^{-14} m^3

P=\frac{82C}{2.932x10^{-14}m^3}=2.7965x10^{15} \frac{C}{m^3}

So replacing

∈=\frac{2.7965x10^{15}\frac{C}{m^3}}{8.8542x10^{-12}\frac{C^2}{N*m^2}}

∈=3.1584x10^{26} \frac{V}{m}

3 0
4 years ago
Austin buys a new moped he travel 3 km south and then 4 km east and then 3km north how far does he need to go to get back to whe
Ira Lisetskai [31]
The first thing to do is to define the origin of the coordinate system as the point at which the moped journey begins.
 Then, you must write the position vector:
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 Rewriting
 r = 4i
 To go back to where you started, you must go
 d = -4i
 That is to say, must travel a distance of 4Km to the west.
 Answer
 West, 4km.
4 0
3 years ago
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