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.
Answer:

Explanation:
m = Mass of object = 
mg = Weight of object = 20 N
g = Acceleration due to gravity = 
v = Final velocity = 15 m/s
u = Initial velocity = 0
d = Distance moved by the object = 150 m
= Angle of slope = 
f = Force of friction
fd = Work done against friction
The force balance of the system is

The work done against friction is
.
Answer:
Ф = 2.179 eV
Explanation:
This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.
E = K + Ф
the energy of the photons is given by the Planck relation
E = h f
we substitute
h f = K + Ф
Ф= hf - K
the speed of light is related to wavelength and frequency
c = λ f
f = c /λ
Φ =
let's reduce the energy to the SI system
K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J
calculate
Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹
Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹
Ф = 3.4571 10⁻¹⁹ J
we reduce to eV
Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
Ф = 2.179 eV
From the measured wavelength from diagram, the frequency of the sound is 6660 Hz.
<h3>What is the frequency of a wave?</h3>
The frequency of a wave is the number of complete oscillation per second completed by a wave.
Frequency is related to wavelength and speed by the following formula:
- Frequency = velocity/wavelength
Velocity of sound in air = 330 m/s
The measured wavelength = 5.0 cm = 0.05 m
Frequency = 330/0.05 = 6660 Hz
Therefore, based on the measured wavelength from diagram, the frequency of the sound is 6660 Hz.
Learn more about frequency of sound at: https://brainly.in/question/15373132
#SPJ1