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Mariulka [41]
3 years ago
5

Consider the four quantum numbers of an electron in an atom, n, l, ml, and ms. The energy of an electron in an isolated atom dep

ends on:______a. l, ml, and msonly.b. n only.c. n and l only.d. n, l, and mlonly.e. all four quantum numbers.
Physics
1 answer:
Effectus [21]3 years ago
6 0

Answer:

The energy of an electron in an isolated atom depends on b. n only.

Explanation:

The quantum number n, known as the principal quantum number represents the relative overall energy of each orbital.

The sets of orbitals with the same n value are often referred to as an electron shell, in an isolated atom all electrons in a subshell have exactly the same level of energy.

The principal quantum number comes from the solution of the Schrödinger wave equation, which describes energy in eigenstates E_n, and for the case of an hydrogen atom we have:

E_n=-\cfrac{13.6}{n^2}\, eV

Thus for each value of n we can describe the orbital and the energy corresponding to each electron on such orbital.

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Explanation:

after 5 seconds, the velocity is (5s)(3m/s²) = 15m/s

The displacement after 5s is

x=vo + (1/2)at²

x = 0 + (1/2)(3m/s²)(5s)(5s)

x= 37.5 m

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Where is the frequency of ultrasound in relation to the range of human ability to hear
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ultra sounds have frequency higher than the upper audible limit of human hearing, for healthy, young adults.

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someone help pls. Two students, Mia and Peter, leave school to meet at the local coffee shop. Peter decides to jog to the coffee
cluponka [151]

Answer:

1) The distance further it takes Peter to arrive at the Coffee shop than Mia is 1.24 km

2) Mia's average speed is 6.00 km/hour

Peter's average speed is 8.48 km/hour

4) Mia's average velocity = Peter's average velocity = 6.00 km/hour

Explanation:

The given information from the diagram are;

The distance Peter jogs from school to the flower shop = 2.00 km

The distance Peter jogs from the Flower shop to the Coffee shop = 2.24 km.

The distance Mia walks from school directly to the Coffee shop = 3.00 km

The time it takes both Peter and Mia to arrive at the coffee shop = 30 minutes = 0.5 hour

1) The total distance Peter travels from school to the Coffee shop = 2.00 km + 2.24 km = 4.24 km

The distance Mia travels from school to the Coffee shop = 3.00 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 4.24 km - 3.00 km = 1.24 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 1.24 km

2) Average \ speed = \dfrac{Total \ distance \ traveled}{Total \ time \ taken \  in \ the \ journey}

Therefore, \ Mia's \ average \ speed = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Mia's average speed = 6.00 km/hour

Peter's \ average \ speed = \dfrac{4.24 \ km}{0.5 \ hour}= 8.48 \ km/hour

Peter's average speed = 8.48 km/hour

4) Average \ velocicty = \dfrac{Displacement }{Time  \ taken}

The displacement from the School to the Coffee shop is 3.00 km for both Mia and Peter

The time it takes both Peter and Mia to arrive at the Coffee shop from the school is 30 minutes = 0.5 hour

Therefore, \ Mia's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Mia's average velocity = 6.00 km/hour

Peter's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Therefore, Peter's average velocity is also = 6.00 km/hour

6 0
2 years ago
What is the potential energy of a 30 Newton ball that is on the ground
Juli2301 [7.4K]
The potential energy of a 30N ball on the ground will be zero. With respect to height, h. Potential energy will be calculated like this. P=mgh. So if its on the ground relatively speaking the h=0. Thus inputting into the above formula. P=0.
3 0
2 years ago
These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the neg
Kay [80]

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C

Calculating the common potential:

\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\

Calculating the charge after redistribution:

When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2        

\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\  \\  \to               q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C

6 0
3 years ago
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