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ollegr [7]
3 years ago
7

when electrons are excited to differnt energy levels, the average radii from the nucleus also changes. Rank the following electr

on energy states according to the average distance of the electron from the nucleus. Rank from largest to smallest distances..
Physics
1 answer:
Gnoma [55]3 years ago
5 0

Answer:

The problem can be solved using the equation below

Explanation:

ΔE = E1 − E2

ΔE = 2.179*〖10〗^(-18)*(⁡〖1/〖n_1〗^2 〗-⁡〖1/〖n_2〗^2 〗 )

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You are making a 568b utp crossover cable that will be used to cascade two switches on an ethernet network. you have attached an
Elina [12.6K]

For a 568B crossover cable that already has wht-org and org on pins 1 and 2 of the connector at one end, the connector at the OTHER end should have wht-grn and grn on pins 1 and 2 respectively.

The wht-org and org at that end should drop to pins 3 and 6 respectively.

You're welcome, and good luck.

6 0
3 years ago
Why do we need to square the period for pendulum experiment? How it is related to obtaining gram ?
andreev551 [17]

Answer:

The period of a pendulum does not depend on the mass of the ball, but only on the length of the string. Two pendula with different masses but the same length will have the same period. Two pendula with different lengths will different periods; the pendulum with the longer string will have the longer period.

Explanation:

6 0
3 years ago
You are looking at yourself in a plane mirror, a distance of 3 meters from the mirror. Your brain interprets what you are seeing
Maslowich

Answer:6

Explanation:

5 0
3 years ago
Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
kumpel [21]

Answer:\frac{D_A}{v_B-v_A}

Explanation:

Given

car A had a head start of D_A

and it starts at x=0 and t=0

Car B has to travel a distance of D_A and d_a

where d_a is the distance travel by car A in time t

distance travel by car A is

d_a=v_A\times t

For car B with  speed v_B

d_B=D_A+d_a

v_B\times t=D_A+v_A\times t

t=\frac{D_A}{v_B-v_A}

7 0
4 years ago
What mass of water must evaporate from the skin of a 70.0 kg man to cool his body 1.00 ∘C? The heat of vaporization of water at
romanna [79]

Answer:

100 cc

Explanation:

Heat released in cooling human body by t degree

= mass of the body x specific heat of the body x t

Substituting the data given

Heat released by the body

= 70 x 3480 x 1

= 243600 J

Mass of water to be evaporated

= 243600 / latent heat of vaporization of water

= 243600 / 2420000

= .1 kg

= 100 g

volume of water

= mass / density

= 100 / 1

100 cc

1 / 10 litres.

6 0
3 years ago
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