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mina [271]
3 years ago
5

Kepler's second law: as a planet moves around its orbit, it sweeps out ______areas in ______ times.

Physics
1 answer:
natali 33 [55]3 years ago
6 0
<span>Kepler's second law: as a planet moves around its orbit, it sweeps out  EQUAL areas in EQUAL times.

</span>
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Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy
8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV

Energy of the second orbit in H atom .

E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV

Energy of the third orbit in H atom .

E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV

Energy of the fifth orbit in H atom .

E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV

Energy of the sixth orbit in H atom .

E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV

Energy of the seventh orbit in H atom .

E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV

During an absorption of energy electron jumps from lower state to higher state.So,  absorption will take place in :

1) n = 2 → n = 3

2) n=  5 → n = 6

Energy absorbed when: n = 2 → n = 3

E=E_3-E_2

E=(-1.51 eV) -(-3.40 eV)=1.89 eV

Energy absorbed when: n = 5 → n = 6

E'=E_6-E_5

E'=(-0.378 eV)-(-0.544 eV) =0.166 eV

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0
3 years ago
Describe what happens to the system inside of a refrigerator or freezer in terms of heat transfer, work, and conservation of ene
Over [174]

Answer: A. Work is done on the system and heat is transferred from the system for a net decrease in internal energy.

Explanation:

A refrigerator is a device which dispenses heat from the close system to a warmer area or in the surrounding. By dispensing the heat the internal temperature of the refrigerator drops. The system of refrigerator violates the second law of thermodynamics. As it performs the work to cool the region instead of heating the region. The work is done on the system and the internal energy decreases and the heat energy is liberated to the surrounding area. A refrigerator is an open system.

6 0
2 years ago
A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A
Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2)  Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

            Em₀ = U = mg y

Final. Low point just before the crash

           Emf = K = ½ m v²

          Em₀ = Emf

          m g y = ½ m v²

           v = √ 2 g y

Let's calculate

           v = √ (2 9.8 0.05)

           v = 0.99 m / s

1) the moment before the crash is

           p₀ = m v

           p₀ = 0.221 0.99

           p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

           p = 0

2) change of momentum

           Δp = p - p₀

            Δp = 0- 0.219

            Δp = -0.219 kg m / s

3) the reason is

     Δp / p = 1

In percentage form it is 100%

3 0
3 years ago
With a mass of 109 kg, Baby Bird is the smallest monoplane ever
OLga [1]

Total resultant velocity=5.11-3.27=1.84m/s

  • m_1=61.4kg
  • m_2=109kg
  • v_1=1.84m/s
  • v_2=?

\\ \sf\longmapsto ∆P=P

\\ \sf\longmapsto m_1v_1=m_2v_2

\\ \sf\longmapsto v_2=\dfrac{m_1v_1}{m_2}

\\ \sf\longmapsto v_2=\dfrac{61.4(1.84)}{109}

\\ \sf\longmapsto v_2=112.976/109

\\ \sf\longmapsto v_2\approx 1.3m/s

4 0
2 years ago
1 1.3.3 Test (CST): Marketing and
Angelina_Jolie [31]
I would have to say D
8 0
2 years ago
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