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BabaBlast [244]
3 years ago
14

Explain how Copernicus concluded that stars were farther away than planets. Draw a diagram showing how this principle applies to

another example.
Physics
1 answer:
Rainbow [258]3 years ago
8 0
<span>Copernicus decided this with more of an educated guess than anything. For example is when your standing right next to a plane it's huge Right? Well when it's flying it looks really small. He used the same reasoning for stars. Since it looks small it must be farther away.</span>
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All the terrestrial planets have
Hoochie [10]
They have solid, rocky surfaces they also have molten heavy-metal core, few moons , and topological features
6 0
3 years ago
An artillery shell is fired at a target 200 m above the ground. When the shell is 100 m in the air, it has a speed of 100 m/s. W
Marat540 [252]

Explanation:

Below is an attachment containing the solution.

3 0
3 years ago
Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,
REY [17]

Answer:

a) 238U, 40K and 87Rb, b)   235U and to a lesser extent 40K , c)  he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

8 0
4 years ago
True or False: Newton's Second Law of Motion states acceleration depends on the objects mass and on the force acting on the obje
natulia [17]

That is true, i think.

8 0
3 years ago
Atomic hydrogen produces a well-known series of spectral lines in several regions of the electromagnetic spectrum. Each series f
navik [9.2K]

Answer:

n₁ = 3

Explanation:

The energy of the states in the hydrogen atom is explained by the Bohr model, the transitions heal when an electron passes from a state of higher energy to another of lower energy,

       ΔE = E_{nf} - E₀ = - k²e² / 2m (1 / n_{f}²2 - 1 / n₀²)

The energy of this transition is given by the Planck equation

       E = h f = h c / λ

       h c / λ = -k²e² / 2m (1 / no ²- 1 / no²)

       1 / λ = Ry (1/ n_{f}² - 1 / n₀²)

Let's apply these equations to our case

     λ = 821 nm = 821 10⁻⁹ m

     E = h c / λ

     E = 6.63 10⁻³⁴ 3 10⁸/821 10⁻⁹

     E = 2.423 10⁻¹⁹ J

Now we can use the Bohr equation

Let's reduce to eV

       E = 2,423 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹) = 1,514 eV

       E_{nf} - E₀ = -13.606 (1 /  n_{f}² - 1 / n₀²)   [eV]

Let's look for the energy of some levels

n         E_{n} (eV)          E_{nf} - EE_{ni} (eV)

1         -13,606           E₂-E₁ = 10.20

2        -3.4015           E₃-E₂ = 1.89

3        -1.512              E₄- E₃ = 0.662

4        -0.850375

We see the lines of greatest energy for each possible series, the closest to our transition is n₁ = 3 in which a transition from infinity (n = inf) to this level has an energy of 1,512 eV that is very close to the given value

8 0
3 years ago
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