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Dmitry_Shevchenko [17]
3 years ago
6

Suppose you monitor a large number (many thousands) of stars over a period of 3 years, searching for planets through the transit

method. Which of the following are necessary for this program to detect an extrasolar planet around one of these stars?State all that apply:
1. You must be able to precisely measure variations in the planet's brightness with time.
2. You must be able to precisely measure variations in the star's brightness with time.
3. As seen from Earth, the planet's orbit must be nearly face-on (perpendicular to our line-of-sight).
4. The planet must have an orbital period of more than about 3 years.
5. As seen from Earth, the planet's orbit must be seen nearly edge–on (in the plane of our line-of-sight).
6. You must repeatedly obtain spectra of the star that the planet orbits.
7. The planet must have an orbital period of less than about 1 year.
8. You must repeatedly obtain spectra of the planet itself.
Physics
1 answer:
nasty-shy [4]3 years ago
4 0

Answer:

2. You must be able to precisely measure variations in the star's brightness with time.

5. As seen from Earth, the planet's orbit must be seen nearly edge–on (in the plane of our line-of-sight).

6. You must repeatedly obtain spectra of the star that the planet orbits.

Explanation:

The transit method is a very important and effective tool for discovering new exoplanets (the planets orbiting other stars out of the solar system). In this method the stars are observed for a long duration. When the exoplanet will cross in front of theses stars as seen from Earth, the brightness of the star will dip. To observe this dip following conditions must be met:

1. The orbit of the planet should be co-planar with the plane of our line of sight. Then only its transition can be observed.

2. The brightness of the star must be observed precisely as the period of transit can be less than a second as seen from Earth. Also the dip in brightness depends on the size of the planet. If the planet is not that big the intensity dip will be very less.

3. The spectrum of the star needs to be studied and observe during the transit and normally to find out the details about the planets.

4. Also, the orbital period should be less than the period of observation for the transit to occur at least once.

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A pendulum of mass 5.0 kg hangs in equilibrium. A frustrated student walks up to it and kicks the bob with a horizontal force of
Phoenix [80]

Answer:

The length is L = 6.206m and the angle is \theta = 37.752^o.

Explanation:

The period T of the pendulum is related to its length L by

T = 2\pi \sqrt{\dfrac{L}{g} },

where g =9.8m/s^2 is the acceleration due to gravity.

Solving for L we get

L = \dfrac{T^2g}{4\pi^2}

putting in T =5.0s and g =9.8m/s^2 we get:

L = \dfrac{(5.0s)^2*9.8m/s^2}{4\pi^2}

\boxed{L = 6.206m.}

There are two forces acting on the pendulum: The gravitational force mg and the F = 30N student's force. Therefore, the angular displacement \theta that these forces give is

sin(\theta ) = \dfrac{F}{mg}

\theta = sin^{-1}( \dfrac{F}{mg})

putting in F =30N, m =5.0kg, and g =9.8m/s^2 we get

\theta = sin^{-1}( \dfrac{30N}{5.0kg*9.8m/s^2})

\boxed{\theta = 37.752^o.}

4 0
3 years ago
Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of th
Strike441 [17]

Answer:

Biomedical engineering

Explanation:

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7 0
3 years ago
The speed of light is about 3.00 × 105 km/s. It takes approximately 1.28 seconds for light reflected from the
Ulleksa [173]
Just to correct you - the speed of light is 3.0 x 10^5 km/sec and not 105 km/sec as given by you (maybe it was just a typing mistake from your end).

The average distance between earth and moon would be - 384,000 kms. 

This is calculated by the formula -> Distance = Speed x Time.
 

8 0
3 years ago
A clarinet sounds as a closed pipe. if a clarinet sounds a note with a pitch of 375 hz, what are the frequencies of the lowest t
mote1985 [20]
So mathematical harmonics are based around a divergent set of fractions. Sigma(1/n)
with the 1st harmonic being... well 1, or 1 full wavelength.The second harmonic is exactly 1/2 the wavelength of the 1st with the third being 1/3 the wavelength. As Wavelengths go down, frequencies go up in a perfect ratio.

Second Harmonic has double the Frequency of the 1st or base note. Third Harmonic is triple and so on.

So the Harmonic set of 375 is.
1. 375
2. 375×2=750
3. 375×3= 1125
.
.
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etc (: I hope this helps.
8 0
3 years ago
The third floor of a house is 8m above street level. How much work is needed to move a 136kg refrigerator to the third floor?
jonny [76]

m = Mass of the refrigerator to be moved to third floor = 136 kg

g = Acceleration due to gravity by earth on the refrigerator being moved = 9.8 m/s²

h = Height to which the refrigerator is moved  = 8 m

W = Work done in lifting the object

Work done in lifting the object is same as the gravitational potential energy gained by the refrigerator. hence

Work done = Gravitation potential energy of refrigerator

W = m g h

inserting the values

W = (136) (9.8) (8)

W = 10662.4 J



8 0
3 years ago
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