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r-ruslan [8.4K]
3 years ago
15

Deserve to be reborn I need your opinions ?

Physics
1 answer:
kvv77 [185]3 years ago
6 0
I say Albert Einstein gets reborn
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Gold is called noble gas why
slamgirl [31]
This is what i found.
Gold<span> is a </span>noble<span> element because "in chemistry, the </span>noble<span>metals are metals that are resistant to corrosion and oxidation in moist air (unlike most base metals)."</span>
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4 years ago
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How many kilohertz are<br> 750 megahertz?
e-lub [12.9K]

Answer:

750,000

Explanation:

I just looked it up on Google lol. You apparently multiple by 1000

Hope this helps :p

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How are wavelength and frequency of a wave related?
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Wavelength is the length from one peak of a wave to the next, while frequency is the amount of waves present in a unit of time. The greater the wavelength the smaller the frequency. The smaller the wavelength, the bigger the frequency.
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A 1kg pool ball travels to the right at 2m/s and hits another stopped pool ball bounces back to the left at 1m/s what is the vel
Burka [1]
Ptotal=Ptotal —> m1v1+m2v2=m1v1’+m2v2’ —> (1kg)(2m/s)+(1kg)(0m/s)=(1kg)(-1m/s)+(1kg)(v2’) —> v2’=3m/s

answer: v=3m/s
5 0
3 years ago
The surface temperature of a planet depends on both the distance of the planet from the Sun and on how the panet's atmosphere di
BaLLatris [955]

Answer:

Aphelion: 6404 W/m2

Perihelion: 14978 W/m2

Explanation:

The solar energy flux depends on the solar power output divided by the surface of a sphere with a radius equal to the distance to the Sun.

\Phi sol = \frac{Psol}{4 * \pi * d^2}

The distances we need are the aphelion and perihelion of Mercury.

Planetary orbits are ellipses. In an ellipse the eccentricity is related to linear eccentricity and the length of the semi major axis:

e = \frac{c}{a}

Where

e: eccentricity

c: linear eccentricity

a: semi major axis

The linear eccentricity is equal to the distance of the focus of the center of the ellipse.

c = a * c =

a = 0.39 AU = 5.83e10 m

c = 5.83e10e * 0.21 = 1.22e10 m

In planetary orbits the Sun is in one of the fucuses. With this we can calculate the prihelion and aphelion as:

Ap = a + c = 5.83e10 + 1.22e10 = 7.05e10 m

Pe = a - c = 5.83e10 - 1.22e10 = 4.61e10 m

And the solar energy fluxes will be:

\Phi Ap = \frac{4e26}{4 * \pi * 7.05e10^2} = 6404 W/m2

\Phi Pe = \frac{4e26}{4 * \pi * 4.61e10^2} = 14978 W/m2

4 0
3 years ago
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