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Sindrei [870]
3 years ago
6

Classify the planets as inner planets of outer planets explain your answer​

Physics
1 answer:
rusak2 [61]3 years ago
7 0

Answer: Planets A and B have a rocky mantle and an iron core. They are at a distance of 1 AU or less from the Sun. This means they are relatively close to the Sun. There are the properties of inner (terrestrial) planets. So planet A and B are inner planets.

Explanation:

Planet C is composed if the gases hydrogen and helium. It has a high mass and is much farther from 1 AU from the sun. These properties are consistent with the outer planets. So, planet C is an outer (gas giant) planet.

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MODERN PHYSICS
frosja888 [35]

Answer:

D. n=6 to n=2

Explanation:

Given;

energy of emitted photon, E = 3.02 electron volts

The energy levels of a Hydrogen atom is given as;  E = -E₀ /n²

where;

E₀ is the energy level of an electron in ground state =  -13.6 eV

n is the energy level

From the equation above make n, the subject of the formula;

n² = -E₀ / E

n² = 13.6 eV / 3.02 eV

n² = 4.5

n = √4.5

n = 2

When electron moves from higher energy level to a lower energy level it emits photons;

E = E_0(\frac{1}{n_1^2}-\frac{1}{n_2^2} )\\\\\frac{1}{n_1^2}-\frac{1}{n_2^2} = \frac{E}{E_o} \\\\\frac{1}{4} -\frac{1}{n_2^2} = \frac{3.02}{13.6} \\\\\frac{1}{4} -\frac{1}{n_2^2} =0.222\\\\\frac{1}{n_2^2} = 0.25 - 0.22\\\\\frac{1}{n_2^2} = 0.03\\\\n_2^2 = 33.33\\\\n_2 = \sqrt{33.33} = 6

For a photon to be emitted, electron must move from higher energy level to a lower energy level. The higher energy level is 6 while the lower energy level is 2

Therefore,  The electron energy-level transition is from n = 6 to n = 2

3 0
3 years ago
The metal wire in an incandescent lightbulb glows when the lights is switch on and stops glowing when switched off. This simple
lutik1710 [3]

Answer:

When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat.  

Explanation:

4 0
3 years ago
Read 2 more answers
An object is inside a room that has a constant temperature of 292 K. Via radiation, the object emits three times as much power a
maw [93]

Answer: Maybe he has the heater on

Explanation: If there is that much temperature then it has to be from something heated.

8 0
3 years ago
Ok sooo I need help I don’t understand this lol
Burka [1]

Answer:

I dont either but I'm sorry and hopefully you figured it out

3 0
3 years ago
A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th
Marysya12 [62]

Answer:

Mininmum acceleration required = 2.74 m/s^{2}

Explanation:

Given,

Lift off speed = 125km/hr.

We know km/hr conversion factor to m/s is \frac{5}{18},

Therefore,

Lift off speed required = 125 \times\frac{5}{18} m/s

Lift off speed required = 34.72222 m/s

Length of Takeoff Run given = 220 m.

So,

the flight needs to achieve its takeoff speed at a constant acceleration before it reaches the end of the 220 m runway.

Consider the formula from kinematics,

v^{2}-u^{2}=2as,

where,

v - final velocity of particle,

u - initial velocity of particle

a - the constant acceleration at which the particle is moving with

s - distance travelled by particle

So at minimum acceleration,it reaches the takeoff speed at the end of the runway.

Therefore in the formula,

we use, v = Lift off speed = 34.72222 m/s;

u = 0 m/s (plane starts from rest);

a = minimum acceleration of the plane

s = 220 m;

Substituting these values in the formula, we get,

(34.72222)^{2}-0 = 2\times a \times 220

a = \frac{(34.72222)^{2} }{2\times220}

a = 2.74 m/s^{2}.

Therefore,the minimum acceleration of the plane required = 2.74 m/s^{2}.

7 0
3 years ago
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