What method has detected the most extrasolar planets so far?

Physics

1 answer:

Sedaia [141]3 years ago

5 0

<h2>Answer: The Transit method</h2>

Detecting extrasolar planets by direct observation (with a telescope) is a complicated task. This is because any planet constitutes an extremely dim light source compared to the star around which it orbits.

So, to detect this extremely dim source is quite difficult due to the glare of the star's light that dulls it.

In this sense, scientists and astronomers have made several methods to find these extrasolar planets, among which the most successful has been the transit method.

This method is based on <u>astronomical transit</u>, a phenomenon in which a body (a planet in this case) passes in front of a larger one (the star), blocking (eclipsing) its vision to some extent.

It should be noted that this is the method currently used in the search for extrasolar planets. Space agencies such as ESA (Europe) and NASA (USA) have put into orbit satellites with extremely sensitive photometric sensors to observe even the smallest variations of intensity of a star due to the passage of a planet.

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Which element could provide one atom to make an ionic bond with nitrogen?

aev [14]

Answer:

boron

aluminum

gallium

indium

thallium

Explanation:

Any of these could work. Nitrogen has 5 valence electrons so you just needed to pick an element that has 3 valence electrons that nitrogen could borrow. This periodic table shows valence electron counts:

8 0

2 years ago

An apparatus like the one Cavendish used to ﬁnd G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg.

Ber [7]

Answer:

The magnitude of gravitational force between two masses is $4.91\times 10^{-9}\ N$ .

Explanation:

Given that,

Mass of first lead ball, $m_1=5.2\ kg$

Mass of the other lead ball, $m_2=0.046\ kg$

The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m

We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

$F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N$