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3 years ago

TEST

Physics

2 answers:

satela [25.4K]3 years ago

8 0

R =8-0 jevzfh z9834vdjn t the Danzig

Viefleur [7K]3 years ago

7 0

Yes yes sir yes I’ll dm is the day I get that I wanna go home and get my

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Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

Circular velocity for the particle in the Encke Division:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

Circular velocity for the particle in D Ring:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago

How does the speed of the different molecules that make up the air depend on their masses?

koban [17]

Larger molecules will move slower and smaller molecules will move faster. Did this answer your question?

8 0

3 years ago

A car is travelling at 18m/s accelerates ti 30m/s in 3seconds. what's the acceleration of the car

Luden [163]

Answer: a = 4 m/s²

Explanation:

a = Δv/t = (30 - 18) / 3 = 4 m/s²

3 0

3 years ago

An electron is confined to a one dimensional infinite potential well 150 pm. How much energy must it absorb if it is to jump to

igor_vitrenko [27]

Answer:

\Delta E=1.22\times 10^{-22}J

Explanation:

The energy of electron in any state is given by $E=\frac{n^2h^2}{8mL^2}$ here h is planck's constant n is state of electron L is the infinte potential well m is the mass of electron

We know that $h=6.67\times 10^{-34}$

Potential well dimension = $150pm=150\times 10^{-12}m$

Mass of electron $=9.1\times 10^{-31}kg$

So energy required to electron to jump from ground state to 3rd state

$\Delta E=\frac{h^2}{8mL^2}\left ( 3^2-1^2 \right )$

$\Delta E=\frac{\left ( 6.67\times 10^{-34} \right )^2}{8\times 9.1\times 10^{-31}(150\times 10^{-12})^2}\left ( 9-1 \right )$

$\Delta E=1.22\times 10^{-22}J$

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3 years ago

Put each event in order of its occurrence, according to the big bang theory. please note that more accurate calculations of the

GalinKa [24]

I think the events are already chronologically arranged correctly. It always have to start with the simplest elements, the hydrogen and helium. In fact, these are the elements that compose the stars. The nuclear fusion of hydrogen atoms to helium is what gives the stars and the sun their energy. So, this is followed by letter b, which is the formation of stars. Then, when these stars explode a stated in c, they produce even heavier elements. As a result, this big bang theory tells us that the universe is expanding because of this 'explosion'. In the end, it led to the formation of the sun, the planets and the universe. Thus, the arrangement is: a, b, c and d.

6 0

3 years ago