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Vlad1618 [11]
4 years ago
12

Which planet do most known extrasolar planets most resemble?

Physics
1 answer:
tatuchka [14]4 years ago
6 0
The awnser is A>Most known exoplanets resemble gas giants known as "hot Jupiters" as a result of being large objects orbiting close to they're host star.
Most known exoplanets that we have detected are likely or confirmed to be gas giants. This is because one of the most used detection techniques is the Transit technique.

This technique involves looking at a distance star's light curve (the changes in the brightness of star over a period of time). Most stars dim and brighten over time by a small bit. But if the light curve shows a periodic and large dip in brightness, this is a sign that something large is passing (or transiting) in front of it.

The reason most detected exoplanets are gas giants is simply because they're the easiest to detect. They cause a larger dip in the curve than a smaller planet would. A small planet's dip are masked by the star's normal dimming and brightening.

Imagine having a flashlight shining on a wall. If you pass a large object through the light beam, it has a large shadow. If you pass a smaller object through, it has a smaller shadow. The change in the light you see on the wall is similar to what you will see in a star's light curve. Now if you pass an object closer to flashlight, as opposed to closer to the wall, this changes the shadow as well.

This is why most exoplanets are "hot Jupiters". They're large gas giants that are hot because they pass very close to the star, resulting in a larger dip that is easier to see in the curve. They also have shorter orbits so they're more likely to be seen in a few weeks or month's time.


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Please answer these questions for me! I am giving MAX points! Brainliest!
guajiro [1.7K]

I think you have them all marked correctly

7. They carry hereditary material from parent

8. 50 percent

1. Nutrition

2. X-linked dominant


If Im wrong Im sorry. its been like 4 years since i took biology but i remember some things

3 0
3 years ago
Read 2 more answers
If two firecrackers produce a sound level 85dB. What level does the explosion of one firecracker produce?
mars1129 [50]

Level of sound is related to intensity by this equation

L = 10 Log \frac{I}{I_o}

given that

L = 85 dB

also we know that

I_o = 10^{-12} W/m^2

now we have

85 = 10 Log \frac{I}{10^{-12}}

By solving above equation

I = 10^{8.5} * 10^{-12}

I = 3.16 * 10^{-4} W/m^2

now above is the intensity due to two firecrackers and if we wish to find the sound level of one firecracker then we will find its half level intensity

I_1 = \frac{I}{2}

I_1 = 1.58 * 10^{-4}

now again by above formula

L = 10 Log \frac{1.58 * 10^{-4}}{10^{-12}}

L = 82 dB

8 0
3 years ago
Fission and fusion reactions both involve making changes to an atom's _____.
spayn [35]

Option (A) is correct. Fission and fusion reactions both involve making changes to an atom's nuclei.

Nuclear fission is a process in which a bigger nucleus breaks into two or more smaller nuclei with the liberation of large amount of energy.

Nuclear fusion is a process in which two smaller nuclei fuse together to form a bigger nucleus. A tremendous amount of energy is released during fusion.

Thus both fission and fusion involve the changes in the nuclei of atoms. Rest of the atom like the electronic configuration etc remains same.

6 0
3 years ago
The mass of a satellite in geostationary orbit is 1,847 kg. The mass of Earth is 5.97*1024 kg. Because of the force Earth exerts
ozzi

Answer:

411.88 N

Explanation:

Given that:

mass of satellite m = 1847 kg

mass of the earth M = 5.97 × 10²⁴ kg

centripetal acceleration a = 0.223 m/s²

The magnitude of the force exerted by the earth on the satellite = the centripetal force that is being exerted by the satellite.

∴

m \dfrac{v^2}{r} = G \dfrac{mM}{r^2}

ma = G \dfrac{mM}{r^2}

G \dfrac{mM}{r^2}= 1847 \times 0.223

G \dfrac{mM}{r^2}= 411.88 \ N

5 0
3 years ago
A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 2.5 m away. He then wants to take a p
Temka [501]

Answer:

0.004 m away from the film

Explanation:

u = Object distance

v = Image distance

f = Focal length = 50 mm

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{2.5}\\\Rightarrow \frac{1}{v}=\frac{98}{5} \\\Rightarrow v=\frac{5}{98}=0.051\ m

The image distance is 0.051 m

When u = 50 cm

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.5}\\\Rightarrow \frac{1}{v}=18\\\Rightarrow v=\frac{1}{18}=0.055\ m

The image distance is 0.055 m

The lens has moved 0.055-0.051 = 0.004 m away from the film

3 0
3 years ago
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